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tankabanditka [31]
3 years ago
14

A 0.85 kg soccer ball is booted straight up in the air. If it left the soccer player's foot at a height of 1.0 m and reaches a h

eight of 47.0 m, what was its kinetic energy immediately after it was kicked?​
Physics
1 answer:
Daniel [21]3 years ago
7 0

Answer:

383 J

Explanation:

m = 0.85 kg, h = 1.0 m, H = 47.0 m

Find KE at h

At h, total energy = KE + mgh

At H, KE = 0, total energy = mgH

Energy conservation:

KE + mgh = mgH

so KE = mgH - mgh = mg(H - h) = 0.85*9.8*(47.0-1.0) = 383 J

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A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
A 250 kg cart is at the top of a hill that is 32 m high, what is its potential energy?
atroni [7]

Answer:

<h2>80,000 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 250 × 10 × 32

We have the final answer as

<h3>80,000 J</h3>

Hope this helps you

3 0
3 years ago
A gun is fired on a day when the speed of sound is 335 m/s and an echo is heard 0.75 seconds later. How far away is the object t
cricket20 [7]

Answer:

v= 335 m/s

2∆t= 0.75 s

∆x= v.∆t → ∆x= 335×½×0.75 = 125.625 m

8 0
3 years ago
Can someone help me?
vlabodo [156]
1. Science.
2. evidence
6 0
3 years ago
Read 2 more answers
HEY CAN ANYONE ANSWER DIS RQ PLS!!!!!
zalisa [80]

Answer: all of the above and yes

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5 0
4 years ago
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