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Crank
2 years ago
6

5. A ball weighing 10 kg rolls 200 m down a frictionless incline with a 50 degree angle to the horizontal. If the ball’s initial

velocity was 0 m/s, how much does the mechanical energy of the system change by the time the ball reaches its destination? A) It increased by 12%. B) It increases by 58%. C) It decreases by 12%. D) It does not change.
Physics
1 answer:
sdas [7]2 years ago
3 0

Answer:

D) It does not change

Explanation:

Since there is no friction in the inclined plane. Therefore, there is no loss in the total mechanical energy of the system. So according to the law of conservation of energy we can write:

Total Mechanical Energy at Start = Total Mechanical Energy at End + Frictional Loss

Total Mechanical Energy at Start = Total Mechanical Energy at End + 0

Total Mechanical Energy at Start = Total Mechanical Energy at End

It means there is no change in the total mechanical energy of the system.

Therefore, the correct option is:

<u>D) It does not change</u>

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a man exerts 700 newtons of force to move a piece of furniture 4 meters. if it takes him 2 seconds to move the furniture, how mu
dybincka [34]

                               Work = (force) x (distance)

The work he did:    Work = (700 N) x (4m)  =  2,800 joules

The rate at which
he did it (power):    Work/time =  2,800 joules/2 sec

                                                =  1,400  joules/sec

                                                =  1,400 watts

                                                =  1.877... horsepower (rounded)
 
6 0
3 years ago
Assume that the home construction industry is perfectly competitive and in long-run competitive equilibrium. It follows that: A.
olga nikolaevna [1]

Answer:

B. Marginal cost equals long-run average total cost.

Explanation:

The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.

4 0
3 years ago
What type of device forms images by changing the speed at which light
Nady [450]

Answer:

Any Lens

Explanation:

I Hope it's right if not so Sorry :)

3 0
3 years ago
Read 2 more answers
MATHPHYS HELP
rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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