Which of the following would be an example of mechanical energy?

A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average

alexgriva [62]

I think it is 5 m/s

4 0

3 years ago

Read 2 more answers

The image shows mountains in Alaska.

zalisa [80]

Answer:

A syncline is visible.

Explanation:

A syncline is visible.

8 0

2 years ago

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A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

Learn more about work:

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4 0

3 years ago

A small economy car (low mass) and a limousine (high mass) are pushed from rest across a parking lot, equal distances with equal

Studentka2010 [4]

Answer:

The car that receives more kinetic energy is the small economy car.

Explanation:

K.E = 0.5*mv²

Where;

K.E is the kinetic Energy

M is the mass of an object

V is the velocity of the moving object

But F = m(v/t), from Newton's second law of motion

If equal forces were applied to the two cars, then the velocity of each car will be calculated as follows.

v = (Ft/m)

v² = (Ft/m)²

Substitute in the value of v² into Kinetic energy equation

K.E = 0.5*mv²

K.E = 0.5*m(Ft/m)² = (0.5*F²t²)/m

Also assuming equal distance, equal force and assuming equal time for both cars.

The above equation will reduce to, K.E = k/m

Where k = 0.5*F²t², which is equal in both cars.

Thus, Kinetic energy will depend only on the mass of each car.

From the above expression, Kinetic Energy received by each car is inversely proportional to the mass of the car.

A small economy car (low mass) will receive more kinetic energy while a limousine (high mass) car will receive less kinetic energy.

Therefore, the car that receives more kinetic energy is the small economy car.

6 0

3 years ago

A 123-turn circular coil of radius 2.41 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi

klasskru [66]

Answer:

67.44 V

Explanation:

Number of turns N =123

Radius = 2.41 cm =0.0241 m

The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T

Time interval dt = 0.151 sec

We know that the induced emf $e=-NA\frac{dB}{dt}$

Area $A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2$

Putting all these values in emf equation $e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V$ here negative sign indicates that it opposes the cause due to which it is produced

8 0

3 years ago