The impulse needed to stop the 32g mass traveling at a velocity of 38 m/s is -1.216 kg·m/s.
To find the impulse required to stop a mass, we can use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it. The equation for impulse is given by:
Impulse = Change in momentum
The momentum of an object is defined as the product of its mass and velocity:
Momentum = mass × velocity
Given:
Mass (m) = 32g = 0.032 kg (since 1 g = 0.001 kg)
Velocity (v) = 38 m/s
To find the momentum of the mass before stopping, we can multiply the mass and velocity:
Initial momentum = mass × velocity
Initial momentum = 0.032 kg × 38 m/s
Now, since the mass comes to a stop, the final velocity (vf) is 0 m/s. Therefore, the final momentum is zero.
According to the impulse-momentum principle, the change in momentum is equal to the impulse applied. Thus, the impulse required to stop the mass can be calculated as:
Impulse = Final momentum - Initial momentum
Impulse = 0 - (0.032 kg × 38 m/s)
Impulse = -1.216 kg·m/s
The negative sign indicates that the impulse is in the opposite direction to the initial momentum.
Therefore, the impulse needed to stop the 32g mass traveling at a velocity of 38 m/s is -1.216 kg·m/s.
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Q2.
The volume of 380 g of ice is 410 cm³.
Calculate the density of the ice in g/cm³.
Show your working
Explanation:
You are given g and cm^3 and you want g / cm^3 :
380 g / 410 cm^3 = .927 gm/cm^3
6. Describe why field investigations can differ from classroom experiments.
Field investigations and classroom experiments can differ in several ways due to the unique characteristics and limitations of each setting.
Here are some reasons why field investigations can differ from classroom experiments:Real-world context: Field investigations take place in the natural environment, allowing students to observe and study phenomena in their natural setting. This context provides a more authentic experience and helps students understand the complexities and interactions of the real world. In contrast, classroom experiments often involve controlled conditions that may not accurately reflect real-world scenarios.Complexity and unpredictability: Field investigations often deal with complex and unpredictable variables, such as weather, terrain, and natural processes. This complexity can make it challenging to control and manipulate variables compared to classroom experiments, where conditions can be tightly controlled.Scale and scope: Field investigations can involve larger scales and broader scopes than classroom experiments. For example, studying the ecosystem of a forest or the geological features of a landscape requires observing and collecting data over a large area, which may not be feasible within a classroom setting.Resources and equipment: Classroom experiments often have access to a controlled and well-equipped laboratory, whereas field investigations may require specialized equipment, transportation, and logistical planning to conduct research in the field. This can add logistical challenges and resource constraints to field investigations. Ethical considerations: Field investigations may involve interactions with living organisms and ecosystems, raising ethical considerations related to environmental impact and the well-being of organisms. Classroom experiments, on the other hand, can be designed with ethical considerations in mind, ensuring the well-being and safety of participants.Overall, field investigations provide students with valuable opportunities to engage with the natural world, understand its complexity, and develop skills in observation, data collection, and critical thinking. Classroom experiments, on the other hand, offer controlled environments for testing specific hypotheses and concepts. Both approaches have unique benefits and play important roles in science education, providing complementary learning experiences for students.
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Two satellites I and II move in a circular orbit around the Earth in the plane of the equator. Both engines are turned off. 1) Mark in the picture the forces with which the Earth acts on each of the satellites, observe the ratio of these forces and assuming that both satellites have the same mass! 2) Mark the speed of each satellite in the picture, observing the average of these speeds! 3) Using formulas , explain whether the speed of satellites depends on their mass! 4) Explain which satellite will complete one orbit around the Earth in a longer time - I or II
The forces acting on each satellite are the force of gravity from the Earth and the normal force from the Earth's surface. The ratio of these forces is equal to the ratio of the masses of the satellites. Since the satellites have the same mass, the ratio of the forces is 1:1.
The speed of each satellite is the same. This is because the satellites are in circular orbits, and the speed of a satellite in a circular orbit is constant.
The speed of a satellite does not depend on its mass. This is because the force of gravity from the Earth is proportional to the mass of the satellite, but the centripetal force required to keep the satellite in a circular orbit is also proportional to the mass of the satellite. Therefore, the ratio of the forces is equal to 1, and the speed of the satellite is constant.
The satellite with the larger radius will complete one orbit around the Earth in a long time. This is because the centripetal force required to keep a satellite in a circular orbit is proportional to the square of the radius of the orbit. Therefore, the satellite with the larger radius will experience a larger centripetal force, and it will take longer to complete one orbit.
Here are the formulas used to calculate the speed and period of a satellite in circular orbit:
Speed:
v = sqrt(GM/r)
where:
v is the speed of the satellite
G is the gravitational constant
M is the mass of the Earth
r is the radius of the orbit
Period:
T = 2pi sqrt(r/GM)
U
where:
T is the period of the satellite
pi is approximately 3.14159
r is the radius of the orbit
G is the gravitational constant
M is the mass of the Earth
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if the air was solid and we were liquid how would this work
Answer:
Chaos.
Explanation:
53 J/kg. °C
Question 1 A person fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet? The specific heat of silver is 234 J/kg .°C.
The temperature change of the bullet moving with a speed of 200 m/s is 85.47 °C.
What is temperature change?This is process whereby the degree of hotness of a body (or medium) changes.
To calculate the temperature change of the bullet, we use the formula below
Formula:
Δt = v²/2c....................... Equation 1Where:
Δt = Temperature change of the bulletv = Velocity of the bulletc = Specific heat capacity of the bulletFrom the question,
Given:
v = 200 m/sc = 234 J/kg.°CSubstitute these values into equation 1
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A horse is tied with a long rope at a pole. What distance will the horse run around the pole with rope tightly stretched, if it takes one and a half round?
Answer:
The horse will run a distance of 1.5 times the length of the rope around the pole with the rope tightly stretched.
Explanation:
The distance the horse will run around the pole with the rope tightly stretched can be calculated using the formula for the circumference of a circle:
Circumference = 2 × π × radius
Since the horse takes one and a half rounds, we need to multiply the circumference by 1.5 to get the total distance the horse runs around the pole.
Let's assume that the length of the rope is the radius of the circle, and the horse is tied at the center.
Therefore, the distance the horse will run around the pole with the rope tightly stretched is:
Distance = 1.5 × 2 × π × radius
Since the horse is tied with a long rope, we need to use the length of the rope as the radius of the circle.
Let's assume that the length of the rope is 'L'. Then the radius of the circle is equal to L/2π.
Substituting this value in the formula, we get:
Distance = 1.5 × 2 × π × (L/2π)
Simplifying the expression, we get:
Distance = 1.5 × L
Therefore, the horse will run a distance of 1.5 times the length of the rope around the pole with the rope tightly stretched.
What is the mass of a person that has 4336500 joules of potential energy standing at the top of Mt. Everest at 8850 meters?
The mass of the person standing at the top of Mt. Everest with 4336500 joules of potential energy is approximately 49.1 kilograms.
To find the mass of the person standing at the top of Mt. Everest, we can use the formula for potential energy:
Potential Energy (PE) = mass (m) x acceleration due to gravity (g) x height (h)
We know that the potential energy (PE) is 4336500 joules, the height (h) is 8850 meters, and the acceleration due to gravity (g) is 9.8 m/s^2. So, we can rearrange the formula to solve for the mass (m):
m = PE / (g x h)
Substituting the given values, we get:
m = 4336500 J / (9.8 m/s^2 x 8850 m)
m ≈ 49.1 kg
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What is a pendulum?
A.)A pendulum is a grandfather clock.
B.)A pendulum is a mass suspended on the bottom of a string.
C.)A pendulum is several weights tied together and arranged in a circle.
D.) A pendulum is the time it takes for a bob to swing back and forth one time.
B) A pendulum is a mass suspended on the bottom of a string.
A pendulum is a simple mechanical device that consists of a mass (known as the bob) suspended from a fixed point by a string, wire, or rod. When the bob is pulled to one side and released, it swings back and forth under the influence of gravity, forming a regular pattern of motion. The time it takes for the pendulum to complete one full swing (i.e., from one extreme position to the other and back again) is known as its period. The period of a pendulum is affected by the length of the string and the strength of gravity. The longer the string, the longer the period, and the stronger the gravity, the shorter the period.Pendulums have a wide range of practical applications, such as timekeeping, as seen in grandfather clocks. They are also used in scientific experiments to measure time intervals and gravitational acceleration.
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A disc of mass 3kg and radius 50cm rotate on a horizontal plane about a fixed point through it's center into an angular velocity of 40rad if it is brought to rest in 5sec calculate it's angular acceleration
Explanation:
A disc of mass 3kg and radius 50cm rotate on a horizontal plane zcity of 40rad if it is brought to rest in 5sec calculate it's angular acceleration
A spring-loaded toy gun is used to shoot a ball of mass M straight up in the air. The ball is not attached to the spring. The ball is pushed down onto the spring so that the spring is compressed a distance S below its unscratched point. After release, the ball reaches a maximum height 35, measured from the unscratched position of the spring (see diagram). a) Find the spring constant of the spring. Height Spring b) Find the equilibrium point of the ball when it is sitting on the spring with no forces other than gravity and the spring acting on it. Clearly indicate the point you are using as the origin of your coordinate system and what direction is positive. | 1fa = 11 c) Now, the ball is glued onto the spring so that it oscillates up and down rather than flying off the spring. The spring is again compressed the same distance 5 below its unscratched point. Write an equation for the position of the ball as a function of time after it is released. Clearly indicate the point you are using as the origin of your coordinate system and what direction is positive.
A) To find the spring constant of the spring, we can use the conservation of energy. The spring constant of the spring is 1378.8 N/m.
B) The equilibrium point is at a height of 0 meters.
C) The frequency of the oscillation is: f = 1/T ≈ 6.25 Hz
y(t) = 0.05 cos (2 π × 6.25 where y is in meters and t is in seconds.
The positive direction is from the equilibrium point.
a) To find the spring constant of the spring, we can use the conservation of energy. At the maximum height, the ball has no kinetic energy, so all the energy stored in the spring has been transferred to potential energy in the ball. The potential energy stored in a spring is given by:
PE = (1/2) k S^2
where k is the spring constant and S is the distance the spring is compressed. The potential energy stored in the spring must be equal to the potential energy of the ball at its maximum height. Using the given values, we can set up the equation:
(1/2) k S^2 = M g H
where M is the mass of the ball, g is the acceleration due to gravity, and H is the maximum height reached by the ball. Solving for k, we get:
k = 2 M g H / S^2
Substituting the given values, we get:
k = 2 × M × 9.81 m/s^2 × 0.35 m / (0.05 m)^2 = 1378.8 N/m
Therefore, the spring constant of the spring is 1378.8 N/m.
b) The equilibrium point of the ball when it is sitting on the spring with no forces other than gravity and the spring acting on it is the unscratched point of the spring. We can choose this point as the origin of our coordinate system, and take the upward direction as positive. Therefore, the equilibrium point is at a height of 0 meters.
c) When the ball is glued onto the spring and oscillates up and down, its motion can be described by a simple harmonic motion equation:
y(t) = A cos (ω t)
where y is the position of the ball, A is the amplitude of the oscillation, ω is the angular frequency, and t is the time. The amplitude of the oscillation is equal to the initial compression of the spring, which is 0.05 meters. The angular frequency is given by:
ω = 2 π f
where f is the frequency of the oscillation. The frequency of the oscillation is related to the period of the oscillation T by:
T = 1/f
The period of the oscillation can be found using the formula for the period of a simple harmonic motion:
T = 2 π √(m/k)
where m is the mass of the ball and k is the spring constant. Substituting the given values, we get:
T = 2 π √(0.1 kg / 1378.8 N/m) ≈ 0.16 s
Therefore, the frequency of the oscillation is:
f = 1/T ≈ 6.25 Hz
Substituting these values in the equation for the position of the ball, we get:
y(t) = 0.05 cos (2 π × 6.25 t)
where y is in meters and t is in seconds. The positive direction is upward from the equilibrium point.
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A tennis player tosses a tennis ball straight up and then catches it after 2.21 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude
_____ m/s2
direction
---Select---
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude
_____ m/s
direction
---Select---
(c) Find the initial velocity of the ball.
____ m/s upward
(d) Find the maximum height it reaches.
____ m
(a) To determine the acceleration of the ball while it is in flight, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the ball is thrown straight up, so its final velocity at the highest point is 0 m/s. The initial velocity is unknown, the acceleration is due to gravity and is approximately -9.8 m/s^2 (negative since it acts in the opposite direction of motion), and the time of flight is 2.21 s.
Using the equation, we can solve for the acceleration:
0 = u - 9.8 * 2.21
u = 9.8 * 2.21
u ≈ 21.658 m/s
Therefore, the acceleration of the ball, while it is in flight, is approximately 21.658 m/s^2 in the upward direction.
(b) When the ball reaches its maximum height, its velocity is 0 m/s. This occurs when the ball is momentarily at rest before falling back down. Therefore, the magnitude of the velocity when the ball reaches its maximum height is 0 m/s.
(c) To find the initial velocity of the ball, we can use the equation:
v = u + at
At the highest point, the final velocity is 0 m/s, the acceleration is -9.8 m/s^2 (due to gravity), and the time is 2.21 s.
0 = u - 9.8 * 2.21
u = 9.8 * 2.21
u ≈ 21.658 m/s upward
Therefore, the initial velocity of the ball is approximately 21.658 m/s upward.
(d) The maximum height reached by the ball can be determined using the equation for vertical displacement:
s = ut + (1/2)at^2
At the highest point, the final displacement is 0 m, the initial velocity is 21.658 m/s upward, and the time of flight is 2.21 s.
0 = 21.658 * 2.21 + (1/2) * (-9.8) * (2.21)^2
0 = 47.864 + (-5.5294)
5.5294 = 47.864
Therefore, there seems to be an error in the calculations as the equation does not hold true. Please check the given values and equations to ensure accuracy.
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Q1: A person uses a screwdriver to turn a screw and insert it into a piece of wood. The person applies a force of 20 newtons to the screwdriver and turns the handle of the screwdriver a total distance of 0.5 meter. How would these numbers be different if the person inserted a nail with a hammer instead of the screw with the screwdriver?
A: The force applied would be the same, but the distance would be shorter.
B: The force applied would be greater, but the distance would be shorter.
C: The force applied would be less, but the distance would be greater.
D: The force applied would be the same, but the distance would be greater.
------------------------------------
Q2: What are people trying to increase when using simple machines?
A: applied mechanical force
B: the distance over which a force is applied
C: the energy needed to complete a task
D: mechanical advantage
------------------------------------
Q3: What do a Class 1 lever and a Class 2 lever have in common?
Both levers place the fulcrum in between the applied force and the object being lifted.
A:Both levers place the fulcrum in between the applied force and the object being lifted.
B:Both levers have an output force in the same direction as the input force.
C:Both levers would lift an object on the same side of the fulcrum as the applied force.
D: Both levers result in a larger output force from a smaller input force.
------------------------------------
What makes this lever effective?
A: The rock has more mass than the man.
B: The man has more mass than the rock.
C: The man is farther from the fulcrum than the rock is.
D: The man is closer to the fulcrum than the rock is.
------------------------------------
What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?
96.1%
0.961%
104%
4%
------------------------------------
A pulley system is used to lift an object. Which factor could affect the efficiency of the system?
the friction of the individual pulleys
the mass of the person pulling
the mass of the object
the distance the weight is lifted
------------------------------------
Based on the mass and speed data, which object will have the greatest amount of kinetic energy?
10 kg moving at 2 m/s
5 kg moving at 2 m/s
10 kg moving at 5 m/s
2 kg moving at 2 m/s
------------------------------------
Which statement accurately explains why the kinetic energy of the first rider is greater?
The first rider is taller.
The first rider is in front.
The potential energy of the first rider is greater.
The first rider has more mass.
------------------------------------
Which two factors affect the kinetic energy of an object?
mass and speed
speed and weight
mass and height above the ground
weight and height above the ground
------------------------------------
A single billiard ball, traveling at 20 m/s, strikes a cluster of 15 balls on a billiard table. What can be said about the balls on the table after the collision?
None of the balls will have a velocity greater than 20 m/s.
The kinetic energy will be divided equally among the 15 balls.
Each ball will have a velocity equal to 20 m/s.
The kinetic energy of the 15 balls will be greater than the kinetic energy of one ball.
------------------------------------
Where do the forces that make atoms interact come from?
from the electric fields of charged subatomic particles
from the electric fields of neutral molecules
from the electric fields of neutral subatomic particles
from the electric fields of charged molecules
------------------------------------
Q1: A: The force applied would be the same, but the distance would be shorter.
Q2: D: mechanical advantage
Q3: A: Both levers place the fulcrum in between the applied force and the object being lifted.
Q4: C: The man is farther from the fulcrum than the rock is.
Q5: 96.1%
Q6: the friction of the individual pulleys
Q7: 10 kg moving at 5 m/s
Q8: The first rider has more mass.
Q9: mass and speed
Q10: The kinetic energy of the 15 balls will be greater than the kinetic energy of one ball.
Q11: from the electric fields of charged subatomic particles
A fluid, density ρ=960 kg/m3 is flowing steadily through the below tube. The section diameters are d1=10 cm and d2=8 cm. The gauge pressure at 1 is P1=200kN/m2 The velocity at 1 is u1=5 m/s. The tube is horizontal (z1=z2). What is the gauge pressure at section 2?
Answer:
211,520 N/m^2
Explanation:
To calculate the gauge pressure at section 2, we can apply Bernoulli's equation, which states that the total energy of a fluid in a horizontal flow remains constant. Bernoulli's equation is expressed as:
P1 + 0.5 * ρ * u1^2 + ρ * g * z1 = P2 + 0.5 * ρ * u2^2 + ρ * g * z2
Given the information provided:
P1 = 200 kN/m^2
u1 = 5 m/s
d1 = 10 cm = 0.1 m (converted to meters)
d2 = 8 cm = 0.08 m (converted to meters)
z1 = z2 (since the tube is horizontal)
ρ = 960 kg/m^3 (density of the fluid)
We can calculate the velocity at section 2 (u2) using the continuity equation, which states that the mass flow rate is constant in an incompressible fluid:
A1 * u1 = A2 * u2
A1 = (π/4) * d1^2 (area at section 1)
A2 = (π/4) * d2^2 (area at section 2)
Substituting the values and solving for u2:
(π/4) * d1^2 * u1 = (π/4) * d2^2 * u2
(0.785) * (0.1)^2 * 5 = (0.785) * (0.08)^2 * u2
0.03925 = 0.03925 * u2
u2 = 1 m/s
Now we can substitute all the known values into Bernoulli's equation:
200 kN/m^2 + 0.5 * 960 kg/m^3 * (5 m/s)^2 = P2 + 0.5 * 960 kg/m^3 * (1 m/s)^2
Simplifying the equation:
200000 N/m^2 + 0.5 * 960 kg/m^3 * 25 m^2/s^2 = P2 + 0.5 * 960 kg/m^3 * 1 m^2/s^2
200000 N/m^2 + 12000 N/m^2 = P2 + 480 N/m^2
212000 N/m^2 = P2 + 480 N/m^2
P2 = 211520 N/m^2
Therefore, the gauge pressure at section 2 is 211,520 N/m^2.
A paragraph about which planet other than earth that humans should explore first
The planet other than the Earth that humans should explore first is Mars. Mars is a planet that has water on its surface which is important for human life.
Earth is the third planet in our solar system and it is called a terrestrial planet. The 3/4th of the surface of the earth is made up of water and its atmosphere has gases that are essential for living. Due to overpopulation and overexploitation of resources, much research was made for an alternate planet that meets the basic need of human life like water and air.
The alternate planet for living other than Earth is Mars. Mars is the fourth planet in our solar system and the distance between the Sun and Mars is quite large hence, the solar winds do not affect the planet. Mars is a planet full of water and has a thicker atmosphere which is suitable for human life.
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When Jose plays his guitar, the friction between his fingers and the strings allows him to pluck the strings. The friction creates some heat and the vibration of the strings creates the sound. The original amount of energy he applies to the strings is 1,000 joules. The energy of the vibrating strings is measured and is found to be 800 joules. Was Energy Lost?
Yes, energy was lost in this scenario. The original amount of energy applied to the strings by Jose was 1,000 joules. However, the measured energy of the vibrating strings is only 800 joules.
Yes, energy was lost in this scenario. The original amount of energy applied to the strings by Jose was 1,000 joules. However, the measured energy of the vibrating strings is only 800 joules. This discrepancy indicates that 200 joules of energy were lost. The energy loss can be attributed to various factors. Firstly, friction between Jose's fingers and the strings converts some of the applied energy into heat energy. This heat energy dissipates into the surrounding environment, resulting in a loss of energy from the system. Additionally, there may be other forms of energy loss involved, such as air resistance or sound energy radiated away from the vibrating strings. These energy losses contribute to the discrepancy between the original applied energy and the measured energy of the vibrating strings. Therefore, in this case, the difference between the initial and measured energy values indicates that some energy was lost in the form of heat, sound, or other forms of energy dissipation.
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Determine the force if 45N makes an angle of 20 degrees at northwest
Answer:
To determine the force acting at an angle of 20 degrees to the northwest, we need to break it down into its horizontal and vertical components. Since "northwest" is at an angle of 45 degrees to both north and west, we can find the horizontal and vertical components of the force by using trigonometry.
The horizontal component of the force can be found by multiplying the force by the cosine of the angle:
F_horizontal = 45 N * cos(20°) ≈ 42.9 N
The vertical component of the force can be found by multiplying the force by the sine of the angle:
F_vertical = 45 N * sin(20°) ≈ 15.4 N
Therefore, the force acting at an angle of 20 degrees to the northwest can be resolved into a horizontal component of about 42.9 N to the west, and a vertical component of about 15.4 N to the north.
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Number 3 please I’m not sure what to say
The shape and position of a track can impact speed by influencing factors such as acceleration distance, surface friction, centripetal force, banking angle, altitude changes, and straight sections for maximum speed. Factors such as track length, surface condition, curvature, and elevation changes can either facilitate or hinder the speed of moving objects.
The shape and position of a track can have a significant impact on the speed of a moving object, such as a vehicle or an athlete. Here are a few key factors to consider:
1. Track Length: The length of the track can influence speed. In general, longer tracks provide more distance for acceleration and allow for higher maximum speeds. A longer track also means that an object has more time to maintain its speed before decelerating.
2. Track Surface: The surface of the track plays a crucial role in determining speed. A smooth and well-maintained track offers less friction, allowing objects to move faster. On the other hand, a rough or uneven surface can increase resistance, slowing down the object.
3. Track Curvature: The curvature of a track affects speed through centripetal force. When an object moves along a curved track, it experiences a force directed toward the center of the curve, known as centripetal force. To maintain a constant speed while curving, the object must exert a force perpendicular to its velocity. As the curvature of the track increases, so does the required centripetal force, which can limit the maximum speed that can be maintained.
4. Track Banking: Banking refers to the angle at which a track is inclined or tilted along its curves. Properly banked tracks can assist in maintaining speed while going around curves. The banking angle is designed to counteract the effect of centripetal force and allows the object to navigate the curve more efficiently. Without proper banking, the object may experience lateral forces that can slow it down.
5. Track Altitude and Elevation Changes: Changes in track altitude, such as hills or inclines, can influence speed. When an object moves uphill, it must work against gravity, which can decrease its speed. Conversely, when moving downhill, gravity can aid in increasing the object's speed.
6. Track Straightaways: Straight sections of a track provide an opportunity for objects to reach and maintain their maximum speed. These sections allow for uninterrupted acceleration and reduce the need for constant course adjustments.
Therefore, It's important to note that the specific effect of the track's shape or position on speed will depend on the nature of the moving object, the forces acting upon it (e.g., gravity, air resistance), and other factors such as the object's mass and the power or force applied to it. Different types of tracks, such as those used in athletics, motorsports, or cycling, will have their own unique characteristics that can impact speed in varying ways.
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Please explain thoroughly and show work if needed! Thank youuu :) -If the period of a pendulum decreases from 8 seconds to 2 seconds, how will that affect the frequency of the pendulum?
Answer:
Physics
Explanation:
The period of a pendulum is the time it takes for the pendulum to complete one full oscillation, or swing back and forth. The frequency of a pendulum, on the other hand, is the number of oscillations it completes in one second. The frequency is calculated by taking the reciprocal of the period, or 1/period.
If the period of a pendulum decreases from 8 seconds to 2 seconds, it means that the pendulum is swinging back and forth more quickly, as it is taking less time to complete one full oscillation. To find the new frequency of the pendulum, we can use the formula:
frequency = 1 / period
Initially, when the period was 8 seconds, the frequency was:
frequency = 1 / 8 = 0.125 Hz
After the period decreased to 2 seconds, the new frequency can be calculated as:
frequency = 1 / 2 = 0.5 Hz
So, the frequency of the pendulum increases from 0.125 Hz to 0.5 Hz when the period decreases from 8 seconds to 2 seconds. This means that the pendulum is oscillating at a faster rate, completing more oscillations in one second.
Which one is the definition of the transverse wave?
Select one:
a. Particle oscillation is parallel to the direction of wave velocity.
b. Particle motion is perpendicular to the direction of wave propagation.
c. Wave carries the particles in parallel direction from one region to another
d. Wave transports energy but not matter from one region to another region.
A sound wave is traveling with a frequency of 880Hz. It has a wavelength of 0.75. What is the speed of the sound wave
The speed of the sound wave is 660 meters per second.
To calculate the speed of the sound wave, we need to use the formula:
Speed = Frequency x Wavelength
Here, the frequency of the sound wave is given as 880Hz, and the wavelength is given as 0.75. To get the answer, we just need to plug these values into the formula and solve for the speed:
Speed = 880 x 0.75
Speed = 660 meters per second
It's important to note that the speed of sound depends on the medium through which it is traveling. In air, the speed of sound is approximately 343 meters per second at standard temperature and pressure. However, this value can change depending on factors such as temperature, humidity, and altitude.
Understanding the speed of sound is important in various fields, such as music, engineering, and physics. For example, in music, the speed of sound determines the pitch of a note, while in engineering, it can be used to design and optimize acoustic systems. In physics, it's used to study the properties of waves and to explain phenomena such as Doppler effect and sonic booms.
Therefore, the speed of the sound wave is 660 meters per second.
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A motorcycle, travelling cast, starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s.Calculate a) Its acceleration b) Its final velocity c) At what time the motorcycle had covered half the total distance d) What distance the motorcycle had covered in half the total time.
The motorcycle had covered a distance of 16 meters in half the total time.
a) To calculate the acceleration, we can use the formula:
a = (v - u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.
Given:
u = 0 m/s (initial velocity)
v = ? (final velocity)
t = 4 s (time)
s = 64 m (distance)
Using the equation of motion:
s = ut + 1/2at^2
We can rearrange the equation to solve for acceleration:
a = 2s / t^2
a = 2(64) / (4)^2
a = 128 / 16
a = 8 m/s^2
Therefore, the acceleration of the motorcycle is 8 m/s^2.
b) To find the final velocity, we can use the formula:
v = u + at
v = 0 + (8)(4)
v = 32 m/s
Therefore, the final velocity of the motorcycle is 32 m/s.
c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:
s = ut + 1/2at^2
32 = 0 + 1/2(8)t^2
16 = 4t^2
t^2 = 4
t = 2 s
Therefore, the motorcycle had covered half the total distance at 2 seconds.
d) To calculate the distance covered in half the total time, we use the formula:
s = ut + 1/2at^2
s = 0 + 1/2(8)(2)^2
s = 0 + 1/2(8)(4)
s = 0 + 16
s = 16 m
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A cell In a deaf aid supplies a current of 25.0 mA through a resistance of 400w when the wearer turns up the volume,the resistance is changed to 100w and the current rises to 60mA what is the emf and internal resistance of the cell
Explanation:
See drawing
Using V= I * resistance
v = (.025 A) * (r+400 Ω) <====given
and
v = (.060A) * ( r+100 Ω)
equate the two equations
.025(r + 400) = .060(r+100) <==== solve for r = 114 .3 Ω
then use this value in either of the equations to calculate v = 12.9 v
The moon has a mass of 7.35 E 22 kg and is located 3.84 E 8 meters from Earth. If a car on Earth has a mass of 3,450 kg, what is the gravitational force between this car and the moon?
Please help and show work :)
A housecat accelerates (from rest) at 0.21 m/s^2 for 7 seconds. How fast is the cat going after 7 seconds? Hint: look at the kinematic equations.
Answer:
1.47 m/s
Explanation:
v = v0 + at
where
v0 = initial velocity (zero in this case)
a = acceleration = 0.21 m/s^2
t = time = 7 seconds
Plugging in these values, we get:
v = 0 + (0.21 m/s^2)(7 s)
v = 1.47 m/s
Draw vectors 9 and 10. (answer key)
Answer:
(9) - 10 N, Up
(10) - 5 N at 37 degrees
Explanation:
Refer to the attached image.
Solve the following problem and show your work with the three steps.
What is the wavelength of an electron moving at 2.5 × 107 m/s if the mass of the electron is 9.11 x
10-31 kg?
The wavelength of the electron moving at 2.5 × 10^7 m/s is approximately 2.901 × 10^-11 m.
To find the wavelength of an electron, we can use the de Broglie wavelength equation:
λ = h / (m * v)
where:
λ is the wavelength
h is Planck's constant (approximately 6.626 x 10^-34 J·s)
m is the mass of the electron
v is the velocity of the electron
Now, let's substitute the given values into the equation and solve for λ.
Step 1: Determine the values of the given variables:
v = 2.5 × 10^7 m/s
m = 9.11 × 10^-31 kg
h = 6.626 × 10^-34 J·s
Step 2: Substitute the values into the equation:
λ = (6.626 × 10^-34 J·s) / (9.11 × 10^-31 kg * 2.5 × 10^7 m/s)
Step 3: Simplify and calculate:
λ = 6.626 × 10^-34 J·s / (2.2785 × 10^-23 kg·m/s)
λ = (6.626 / 2.2785) × (10^-34 / 10^-23) J·s / (kg·m/s)
λ ≈ 2.901 × 10^-11 m
Therefore, the wavelength of the electron moving at 2.5 × 10^7 m/s is approximately 2.901 × 10^-11 m.
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what are the properties of a magnet
Answer: A magnet is an object that produces a magnetic field and has the ability to attract certain materials like iron, cobalt, and nickel. The properties of a magnet are:
Magnetic Field: A magnet produces a magnetic field that surrounds it and can attract or repel other magnets or magnetic materials.
North and South Pole: Every magnet has a north pole and a south pole, which are opposite in polarity and attract each other while repelling poles of the same polarity.
Retentivity: A magnet has the property of retentivity, which means it can retain its magnetism even after the magnetizing force is removed.
Coercivity: The coercivity of a magnet is its ability to resist demagnetization.
Magnetic Domains: The atoms in a magnet align themselves in groups called magnetic domains, which help create the overall magnetic field of the magnet.
Curie Temperature: The Curie temperature is the temperature at which a magnet loses its magnetism.
Magnetic Flux: Magnetic flux is the amount of magnetic field passing through a specific area.
Overall, the properties of a magnet allow it to attract or repel other magnets or magnetic materials, and this is useful in a wide range of applications such as electric motors, generators, and magnetic storage devices.
a car travelling at 18 km/hr accelerates uniformly at 2m per seconds square. calculate its velocity in km/hr in 5 seconds
1 km = 1000 m
1 hr = 3600 s
So, 18 km/hr = (18 * 1000) / (3600) m/s = 5 m/s
And 2 m/s^2 = 2 m/s^2
Now, we can use the formula for final velocity (v) when an object starts with an initial velocity (u) and accelerates at a constant rate (a) for a given time (t):
v = u + at
Plugging in the values, we get:
v = 5 + (2 * 5) m/s v = 15 m/s
To convert this back to km/hr, we use the inverse conversions: v = (15 * 3600) / (1000) km/hr v = 54 km/hr
Therefore, the car’s velocity in km/hr after 5 seconds is 54 km/hr.
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency of 400 Hz. The sound he hears reflected back from the wall is at a frequency of 408 Hz. At what is the speed is the boy riding his bicycle toward the wall? Assume the speed of sound in air is 340 m/s.
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a car is traveling at a speed of 30m/s .it decelerate at a rate of 12m/s .calculate the time taken for the car to come to rest.
approximately 2.5 seconds
To calculate the time taken for the car to come to rest, we can use the equation of motion:
v = u + at
Description:
v = final velocity (0 m/s since the car comes to rest)
u = initial velocity (30 m/s)
a = acceleration (deceleration in this case, which is -12 m/s²)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (0 - 30) / (-12)
Simplifying the equation further:
t = 30 / 12
t ≈ 2.5 seconds
Therefore, the time taken for the car to come to rest is approximately 2.5 seconds.