In heating coil of resistance 20ohms, Connected to a 220 volts Source is used to boil a in Certain quantity of water container of heat Capacity 100j/k for 2 minute, If the initial temp. of water is 40°c. Calculate the mass of water in the container (4.2x10³j/k =SHC of water) Assume 100°c of water.​

Answer:

(9) - 10 N, Up

(10) - 5 N at 37 degrees

Explanation:

Refer to the attached image.

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

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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️

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.

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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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.

Film 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

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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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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.

As 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.

The 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.

Assuming 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.

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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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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Explanation:

You are given g   and  cm^3 and you want   g / cm^3   :

380 g / 410 cm^3 = .927 gm/cm^3

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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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.

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.

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.

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

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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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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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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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.

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.

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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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) 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: 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

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

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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