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Alexeev081 [22]
3 years ago
12

A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo

n impact?
Physics
1 answer:
Drupady [299]3 years ago
6 0

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

You might be interested in
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
3 years ago
The question is on the picture
Firlakuza [10]

Answer:

<em> think 2 also if not im so sorry  but i think it is :)</em>

7 0
2 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
ivolga24 [154]

Answer:

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

Explanation:

The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed

m_{tackler}*v_{tackler}=(m_{tackler}+m_{receiver})V     (1)

We solve in order to find the receiver mass:

m_{receiver}={m_{tackler}*v_{tackler}/V}-m_{tackler}    

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

5 0
3 years ago
A football player collides with another player. The force of the collision is
iragen [17]

Answer:

Impulse = 1000 Ns

Explanation:

Given the following data;

Force of collision = 1000 kg•m/s.

Time = 1 seconds

To find the impulse;

Mathematically, the impulse experienced by an object or body is given by the formula;

Impulse = force * time

Substituting into the formula, we have;

Impulse = 1000 * 1

Impulse = 1000 Ns

4 0
2 years ago
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