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aivan3 [116]
3 years ago
11

ListenA person on a ledge throws a ball vertically downward, striking the ground below the ledge with 200 joules of kinetic ener

gy. The person then throws an identical ball vertically upward at the same initial speed from the same point. What is the kinetic energy of the second ball when it hits the ground? [Neglect friction.]
A. 200 J
B. 400 J
C. less than 200 J
D.more than 400 J
Physics
1 answer:
polet [3.4K]3 years ago
8 0

Answer:

A. 200 J

Explanation:

The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.

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A. is a neutral substance a pH of 7 describes something of neutral pH where anything less than 7 is an acid and higher is a base

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A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?
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GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
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Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
2 years ago
A car moving with a speed of 35 m/s sees a child standing in the
zlopas [31]

Answer:

2100 N

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2 years ago
How to solve 30(a)<br><br> Give solution asap.
expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s  te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

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t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0
3 years ago
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