Answer:
I don't actually know, but it has to do with the research they id
Explanation:
to seem precise
7. Which statement accurately describes the charge of the nucleus of an atom? (2 points)
O The nucleus can be either positively charged or neutral.
The nucleus never has an electrical charge.
The nucleus always has a positive charge.
The charge of a nucleus can change from positive to negative.
The nucleus never has a negative charge, and its charge cannot change from positive to negative.
The nucleus of an atom has a charge that can either be positively charged or neutral, as stated in the sentence. Positively charged protons and neutral neutrons make up the nucleus.
As a result, it can either be neutral or have a positive charge when there are an equal amount of protons and neutrons. The other claims made in the question are untrue. The charge of the nucleus is always positive and cannot ever become negative.
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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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Type of law that applies to an action that took place before the law was passed
Answer:
Black's Law Dictionary defines a retroactive law as a law “that looks backward or contemplates the past, affecting acts or facts that existed before the act came into effect.” While Congress often considers legislation that would apply retroactively, the Constitution imposes some limited constraints on such laws.
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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HELPPPPP!!! Please
A cone with an apex radius of 2 theta = 60 degrees rolls without slipping on a horizontal boundary with the apex of the cone held at rest as shown in the figure. What is the apex of the cone at the same height as the center of the base of the cone which is moving with speed. Determine the angular speed of the cone and the angular acceleration of the cone
A canister filled with 3.5 mol of single-atom helium gas has a temperature of
300 K. What is the approximate total internal energy of the gas? (Recall that the equation for kinetic energy due to translation in a gas is: 3/2 nRT; the
equation for kinetic energy due to rotation of a molecule in a gas is: nRT, and R= 8.31 J/(mol.K).)
R= 8.31 J/(mol-K).)
A. 13,100 J
B. 5800 J
C. 15,400 J
D. 8200 J
The answer is C
Please dont get mad at me if this is not right im pretty sure it is
hope this helps!
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
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
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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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.
an object starting from rest travels 20m in first 2sec and 160m in next 4 sec. what will be the velocity after 7sec from the start??
please answer fast .
Answer:
The velocity of the object 7 seconds from the start of its journey is 70 ms⁻¹.
Explanation:
To find the velocity of the object after 7 seconds after the start of its journey, we first needs to consider the acceleration of the object in the each part of the motion: the initial 2 seconds and the subsequent 4 seconds.
First leg of journeyAs we have been given the displacement (s), the initial velocity (u), the final velocity (v) and the time (t), we can use the following SUVAT equations to find the acceleration (a) and the final velocity (v) of the object during the first leg of its journey (initial 2 seconds):
[tex]\textsf{Given:} \quad s=20\;\textsf{m}, \quad u=0\;\textsf{ms}^{-1}, \quad t=2\;\textsf{s}[/tex]
[tex]\begin{aligned}\textsf{Using:} \quad s&=ut+\dfrac{1}{2}at^2\\\\\implies 20&=0(2)+\dfrac{1}{2}a(2^2)\\20&=2a\\a&=10\; \sf ms^{-2}\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Using:} \quad v&=u+at\\\\\implies v&=0+10(t)\\v&=20\; \sf ms^{-1}\end{aligned}[/tex]
Therefore, the acceleration of the object for the initial 2 seconds of its journey is 10 ms⁻² and its final velocity is 20 ms⁻¹.
Second leg of journeyThe initial velocity of the second leg of the journey is equal to the final velocity of the previous leg of the journey. Therefore, as we have been given the displacement (s) and the time (t), and have calculated the initial velocity (u), we can use the following SUVAT equation to find the acceleration (a) of the object during the second leg of its journey (next 4 seconds):
[tex]\textsf{Given:} \quad s=160\;\textsf{m}, \quad u=20\;\textsf{ms}^{-1}, \quad t=4\;\textsf{s}[/tex]
[tex]\begin{aligned}\textsf{Using:} \quad s&=ut+\dfrac{1}{2}at^2\\\\\implies 160&=20(4)+\dfrac{1}{2}a(4^2)\\160&=80+8a\\8a&=80\\a&=10\; \sf ms^{-2}\end{aligned}[/tex]
Therefore, the acceleration of the object for the next 4 seconds of its journey is 10 ms⁻².
We can observe that the acceleration for both legs of the journey is the same, and so the object is moving with constant acceleration of 10 ms⁻².
Velocity after 7 secondsAssuming the acceleration remains constant through the entire journey of the object, to calculate the velocity of the object after 7 seconds from the start, use the following SUVAT equation:
[tex]u=0\; \textsf{ms}^{-1}, \quad a=10\;\textsf{ms}^{-2}, \quad t=7\;\textsf{s}[/tex]
[tex]\begin{aligned}\textsf{Using:} \quad v&=u+at\\\\\implies v&=0+10(7)\\v&=70\; \sf ms^{-1}\end{aligned}[/tex]
Therefore, the velocity of the object 7 seconds from the start of its journey is 70 ms⁻¹.
How do you measure the direction, speed, and position of a moving object?
pls help!
Answer:
Explanation:velocity is aterm used under the branchof the kinematics of motionwhich defines the correlationbetween the position of the body and direction of motion with respect to time interval. thus the velocity is ameasure of both speed and direction of motion
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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PLEASE HELP and show work thank you !
Suppose that two identical mass planets are sitting a million miles apart. At that distance, the planets have a gravitational force of 1,000,000 N. If the planets are moved to two million miles apart, what is the new gravitational force between them?
Answer:
F1 = K M1 M2 / R1^2
F2 = K M1 M2 / R2^2
F2 / F1 = (R1 / R2)^2 = 1/4
The new force is 1,000,000 N / 4 = 250,000 N
The new gravitational force between the planets, when they are two million miles apart, is approximately 2.66972 × [tex]10^(-10)[/tex] N.
To solve this problem, we can use the inverse square law of gravity, which states that the gravitational force between two objects is inversely proportional to the square of the distance between them.
Let's denote the initial distance between the planets as "r1" (1 million miles) and the initial gravitational force as "F1" (1,000,000 N). The final distance between the planets will be "r2" (2 million miles), and we need to find the new gravitational force, which we'll denote as "F2."
The inverse square law formula for gravitational force is:
F = G * (m1 * m2) / [tex]r^{2}[/tex]
where:
F is the gravitational force,
G is the gravitational constant ≈ 6.67430 × [tex]10^(-11)[/tex] N m²/kg²,
m1 and m2 are the masses of the two objects (in this case, the masses of the planets), and
r is the distance between the centers of the two objects.
Since the two planets are identical, their masses are the same. Let's denote the mass of each planet as "m."
Now, we can set up two equations using the given information:
For the initial setup:
F1 = G * (m * m) / [tex]r1^2[/tex]
For the final setup:
F2 = G * (m * m) / [tex]r2^2[/tex]
Since both planets have the same mass (m), we can set these two equations equal to each other:
G * (m * m) / [tex]r1^2[/tex] = G * (m * m) / [tex]r2^2[/tex]
Now, we can solve for F2:
F2 = (G * (m * m) / [tex]r1^2[/tex]) * [tex]r2^2[/tex]
Substitute the given values:
F2 = (6.67430 × [tex]10^(-11)[/tex] N m²/kg² * (m * m) / (1 million mile)^2 * (2 million mile[tex]s)^2[/tex]
Note: We need to convert the distances from miles to meters before plugging into the equation. 1 mile is approximately 1609.34 meters.
r1 = 1 million miles * 1609.34 m/mile ≈ 1.60934 × [tex]10^9[/tex] meters
r2 = 2 million miles * 1609.34 m/mile ≈ 3.21868 × [tex]10^9[/tex] meters
F2 = (6.67430 × [tex]10^(-11)[/tex] N m²/kg² * (m * m) / (1.60934 × [tex]10^9[/tex] meter[tex]s)^2[/tex]) * (3.21868 × [tex]10^9[/tex] meters[tex])^2[/tex]
Now, we see that the mass (m) cancels out, leaving us with
F2 = 6.67430 × [tex]10^(-11)[/tex] N m²/kg² * 3.21868 × [tex]10^9[/tex] meters)^2 / (1.60934 × [tex]10^9[/tex] meters[tex])^2[/tex]
F2 = 6.67430 × [tex]10^(-11)[/tex] N m²/kg² * (3.21868 / 1.6093[tex]4)^2[/tex]
F2 = 6.67430 × [tex]10^(-11)[/tex] N m²/kg² * 4
F2 ≈ 2.66972 × [tex]10^(-10)[/tex] N
Hence, the new gravitational force between the planets, when they are two million miles apart, is approximately 2.66972 × [tex]10^(-10)[/tex] N.
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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
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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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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Draw vectors 9 and 10. (answer key)
Answer:
(9) - 10 N, Up
(10) - 5 N at 37 degrees
Explanation:
Refer to the attached image.
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.
Can momentum be hidden to human eyes like how kinetic energy can be hidden as heat?
Answer:
Yes and no
Explanation:
Momentum in a typical mechanical system made up of macroscopic elements cannot be "hidden" from human sight. But momentum can be concealed in other systems. For instance, in an electromagnetic system, where the electromagnetic field is often inaudible to human vision, momentum can be transferred to it.
Why might someone choose to use film, rather than digital methods, for photography? (1 point)
Some film methods allow you to print a picture from anywhere without electricity.
Film allows you to study a picture immediately.
Digital photographs are more difficult to store in large quantities.
Digital images are more difficult to share.
Someone might choose to use film rather than digital methods, for photography because some film methods allow you to print a picture from anywhere without electricity. Option (A) is right.
Understanding Film PhotographyFilm Photography is the practice of capturing images using photographic film, which is a light-sensitive material coated with an emulsion containing silver halide crystals.
Advantages
With film, it is possible to develop and print pictures even in locations without access to electricity or digital devices. Film can be processed in darkroom environments, allowing photographers to produce physical prints in various situationLearn more about film photography here:
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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
CALCULATE THE WORK AND POWER (image below)
Answer:
(1) - [tex]W=20,000 \ J[/tex]
(2) - [tex]W=8.00 \times 10^6 \ J[/tex]
(3) - [tex]W=7500\ J[/tex] for part a and [tex]P=375 \ W[/tex] for part b
Conceptual:
What is work?
- Work is simply energy transferred from one place to another. Work is a scalar quantity and is measured in Joules, J.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formulas used to calculate work:}}\\\\W=F\Delta r\cos\theta \ \text{(Constant} \ \vec F)\\\\W=\int\limits^{r_2}_{r_1} {F\cos\theta} \, dr \ \text{(Varible} \ \vec F)\end{array}\right}[/tex]
** "F" is the applied force on the object, "Δr" is the displacement of the object, "θ" is the measured angle between the applied force and direction of displacement. For our purposes we are assuming θ is 0, which would make the cosine equal 1.**
What is power?
- Power is the measure of work being done per second, J/s. A joule per second is a Watt, W.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formula used to calculate power:}}\\\\P=\frac{W}{t} \end{array}\right}[/tex]
Step-by-step:
Question (1):
Given:
[tex]F= 1000 \ N\\\\\Delta r=20 \ m\\\\Eqn\rightarrow W=F \Delta r[/tex]
Plug the known values into the equation.
[tex]\Longrightarrow W=(1000)(20)\\\\\therefore \boxed{\boxed{W=20,000 \ J}}[/tex]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Question (2):
Given:
[tex]F= 8000 \ N\\\\\Delta r=1000 \ m\\\\Eqn\rightarrow W=F \Delta r[/tex]
Plug the known values into the equation.
[tex]\Longrightarrow W=(8000)(1000)\\\\\therefore \boxed{\boxed{W=8.00 \times 10^6 \ J}}[/tex]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Question (3):
Given (part a):
[tex]F= 500 \ N\\\\\Delta r=15 \ m\\\\Eqn\rightarrow W=F \Delta r[/tex]
Plug the known values into the equation.
[tex]\Longrightarrow W=(500)(15)\\\\\therefore \boxed{\boxed{W=7500\ J}}[/tex]
Given (part b):
[tex]W= 7500 \ J\\\\t=20 \ s\\\\Eqn\rightarrow P=\frac{W}{t}[/tex]
Plug the known values into the equation.
[tex]\Longrightarrow P=\frac{7500}{20} \\\\\therefore \boxed{\boxed{P=375 \ W}}[/tex]
Thus, all given questions are solved.
B. Calculate the total resistance of the circuit below. (4 points)
c. In the circuit diagram above, meters 1 and 2 are connected as shown. Write 2 - 3 sentences identifying each type of meter and how it is connected with the 30.0 Ω resistor in the circuit. (4 points)
d. In the circuit diagram above, predict which resistors (if any) will stop working when the switch is opened. Write 2 - 3 sentences explaining your reasoning. (4 points)
B. The equivalent resistance of the two resistors is 20.0 ohms.
C. The voltmeter will measure the voltage across the 30.0 ohm resistor.
D. The 30.0 ohm resistor will stop working.
How to determine resistance?B. The total resistance of the circuit is 60.0 ohms. This is because the 30.0 ohm resistor and the 60.0 ohm resistor are in parallel, and the equivalent resistance of two resistors in parallel is equal to the product of the resistors divided by the sum of the resistors.
R_T = 1/(1/R_1 + 1/R_2 + ...)
In this case, the product of the resistors is:
30.0 ohms × 60.0 ohms = 1800 ohms,
and the sum of the resistors is:
30.0 ohms + 60.0 ohms = 90.0 ohms.
Therefore, the equivalent resistance of the two resistors is 1800 ohms / 90.0 ohms = 20.0 ohms.
C. Meter 1 is an ammeter, and it is connected in series with the 30.0 ohm resistor. This means that the ammeter will measure the current flowing through the 30.0 ohm resistor.
Meter 2 is a voltmeter, and it is connected in parallel with the 30.0 ohm resistor. This means that the voltmeter will measure the voltage across the 30.0 ohm resistor.
D. When the switch is opened, the 30.0 ohm resistor will stop working. This is because the switch is in series with the 30.0 ohm resistor, and when the switch is opened, the circuit is broken.
The 60.0 ohm resistor will continue to work, because it is in parallel with the switch, and the current will continue to flow through the 60.0 ohm resistor even when the switch is opened.
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You are a musician and want to know the frequency of a guitar string. You measure the length of the string to be 0.6 meters and the tension in the string to be 60 Newtons. If the mass of the string is 0.01 kg, what is the frequency of the string? (Units: frequency -hertz (Hz), length -meters (m), tension -newtons (N), mass -kilograms (kg))
Answer:
Explanation:
I frequency = 278 Hz
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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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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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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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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1. Explain home a dial test indicator (DTI) reading is adjusted when the gange is not positioned at right angle to the contact surface.
When using a dial test indicator (DTI), it is essential to ensure that the gauge is positioned at a right angle to the contact surface for accurate readings.
However, in certain situations, it may be challenging to achieve a perfect right angle alignment. In such cases, adjustments can be made to compensate for the misalignment and obtain accurate measurements.To adjust the DTI reading when the gauge is not positioned at a right angle to the contact surface, the following steps can be taken:Determine the misalignment angle: Measure the angle at which the DTI is misaligned from the right angle position. This can be done using a protractor or by estimating the deviation visually.Calculate the correction factor: Based on the misalignment angle, calculate the correction factor using trigonometric functions such as sine or cosine. The correction factor accounts for the difference between the actual displacement and the displacement measured by the DTI.Apply the correction factor: Multiply the correction factor by the DTI reading to adjust the measurement. This compensates for the misalignment and provides a more accurate reading.It's important to note that adjusting the DTI reading can introduce some degree of error, especially if the misalignment is significant. Therefore, it is always preferable to position the gauge at a right angle to the contact surface whenever possible to obtain the most precise measurements.
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