This is an inelastic collision, for that reason the conservation of momentum law allows us to determine the final velocity of both, the bullet and block together after the collision:

$m_{bullet} v_{bullet0} +m_{block} v_{block} =(m_{bullet}+m_{block})v_{f} \\v_{f} =\frac{m_{bullet} v_{bullet0} }{m_{bullet}+m_{block}} \\v_{f} =\frac{0.0047kg*678m/s}{4.8kg+0.0047kg}\\v_{f} =0.66 m/s$

3 0

3 years ago

When people do a bungee jump theyare asked how much they weigh. Why?

Katena32 [7]

Answer:

The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.

hopefully this'll help

have a nice day!!! :D

3 0

2 years ago

Suppose a 65.5 kg gymnast climbs a rope. What is the tension in the rope if she climbs at a constant speed

natali 33 [55]

The tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Given:

Mass of gymnast, m = 65.5 kg

The speed 'v' of gymnast is constant

Solution:

Consider the free-body diagram of the system as shown below.

Balancing forces along the vertical axis we get:

ΣFy = 0

Thus, we get:

F = ma - (1)

where, m is mass of gymnast

a is acceleration of gymnast (a = 0m/s², as the speed is constant)

Also,

F = T - mg -(2)

where, T is tension in the rope

g is acceleration due to gravity

Equating (1) & (2), we get:

ma = T - mg

Re-arranging the equation, we get:

T = m(a+g)

Applying values in above equation we get:

T = (65.5 kg)(0 m/s²+9.8 m/s²)

T = 641.9 N

Therefore, the tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Learn more about tension here:

<u>brainly.com/question/14294135</u>

#SPJ4

7 0

2 years ago