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spayn [35]
4 years ago
7

A 15.0 kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its

velocity become 3.20 m/s.
a) What is the change in kinetic energy?
b) How much work was done on the cart?
c) How far did the cart move while the force acted? ...?
Physics
1 answer:
mezya [45]4 years ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below are the answers:

a. <span>DE=421.875-76.8=-345.075 joules (negitive sign means that the energy is transfering out of your system ie slowing down) 

b. </span><span>work=DE (work done is change in energy)=-345.075 joules 

c. </span><span>W=f*d </span>

<span>d=w/f </span>

<span>d=345.075/10 </span>

<span>d=34.5075 meters</span>
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he answer is 540 because you need to do the area of the wall times six. 15 times 6 times 6 equals 540

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Why does the earth bulge at the equator?
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3 years ago
A jeweler is determining the optical properties of an unknown blue gemstone. She uses an angle of incidence of 62°, and measures
Juliette [100K]

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3 0
4 years ago
A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward ever
kiruha [24]

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier (V_s) = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave =  (V_s - V_w)

where:

V_s = velocity of the skier

V_w = velocity of the wave

The wavelength (\lambda) can be written as:

\lambda = (V_s-V_w)T

\lambda = (V_s-V_w) 0.450m ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = (V_s + V_w)

The wavelength (\lambda) can be written as:

\lambda = (V_s+V_w)T

\lambda = (V_s+V_w) 0.342m   ------------------> Equation 2

Equating equation (1) and equation (2) and substituting  V_s  = 14.3 m/s ; we have:

(V_s-V_w) 0.450m  =  (V_s-V_w) 0.342m

0.450m(V_s)-0.450m(V_w)   =  0.342m(V_s)+0.342m(V_w)

Collecting the like terms; we have:

0.450m(V_s) - 0.342m(V_s) =  0.342m(V_w)+0.450m(V_w)

(V_s)(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.108m =  (V_w)0.792m

1.5444m^2/s =  (V_w)0.792m

(V_w) = \frac{1.5444m^2/s}{ 0.792m}

(V_w) = 1.95 m/s

b)

The Wavelength of the wave can be calculated using :  ( \lambda }) = (V_s-V_w) 0.450m

({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)

(\lambda) = (12.35)0.450m

(\lambda)= 5.5575 m

λ ≅ 5.56 m

5 0
3 years ago
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