A bullet with a mass of 12 g moving horizontally strikes a fixed block of wood and penetrates a distance of 5.2 cm. The speed of the bullet just before the collision is 640 m/s. What is the average force that the wood exerted on the bullet

F = 47261.5 N = 47.26 KN

Explanation:

First, we will calculate the deceleration of the bullet by using the third equation of motion:

where,

a = deceleration = ?

s = distace covered = 5.2 cm = 0.052 m

vf = final speed = 0 m/s

vi = initial speed = 640 m/s

Therefore,

negative sign shows deceleration.

Now for the force, we will use Newton's Second Law of Motion:

F = ma

where,

F = Force = ?

m = mass of bullet = 12 g = 0.012 kg

Therefore,

F = (0.012 kg)(3938461.5 m/s²)

F = 47261.5 N = 47.26 KN

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//