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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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An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above
ryzh [129]

Answer:

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v  that is

v² = u² - 2gh      ............1        for upward speed

v² = u² + 2gh     ............2        for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9  = 26.76

v² = 8² + 2(9.8) 1.9   = 101.24

v = 5.173  m/s    ..............3

v = 10.061 m/s   ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173  = 2.827 m/s     .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s     ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

6 0
2 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
2 years ago
When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b
inysia [295]
We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x  0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons
5 0
3 years ago
A small ball of charge Q and mass m has a velocity v at infinity. It collides head-on with a ball of the same charge and mass wh
Hatshy [7]

Answer:

Explanation:

Kinetic energy of ball in motion = 1/2 m v² . Potential energy = 0

Let the minimum distance between the balls be d  on collision.

Electric potential energy at that time= k Q²/d , Here kinetic energy is converted into potential energy . So

1/2 m v² = kQ²/d

d =2 k Q² / mv²,= 18 x 10⁹ x Q²/ m v².

8 0
3 years ago
A ball is thrown straight up. If the launch velocity is 15 m/s, at what SPEED will the ball return to the thrower’s hand?
IgorC [24]

Answer:

Vf = 15 m/s

Explanation:

First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)

h = height =?

Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)

Vi = Initial Velocity = 15 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²

h = (-225 m²/s²)/(-19.6 m/s²)

h = 11.47 m

Now, we consider downward motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 11.47 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²

Vf = √(224.812 m²/s²)

<u>Vf = 15 m/s</u>

8 0
3 years ago
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