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2 years ago

A bullet with a mass of 0.1 kg is travelling at

Physics

1 answer:

mel-nik [20]2 years ago

6 0

The block with the bullet lodged in the block is now travelling at 2.133 m/s.

<h3>What is momentum conservation principle?</h3>

When there is no external force acting on the system, the momentum remains conserved.

For inelastic collision, after collision both objects travel with common speed.

m1u1 + m2u2 =(m1 +m2)v

Substitute initial velocity of bullet u1 =320 m/s , initial velocity of block u2 =0, mass of bullet m1 = 0.1 kg and mass of block m2 = 14.9 kg.

Solve for the final velocity of bullet,

0.1 x 320 + 14.9 x 0 = (0.1 +14.9) x v

v = 2.133 m/s

Thus, the block with the bullet lodged in block now travelling at 2.133 m/s.

Learn more about momentum conservation principle.

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How fast does a 500 Hz wave travel if its wavelength is 0.5 m?

olga_2 [115]

Answer:

250 m/s

Explanation:

4 0

3 years ago

Read 2 more answers

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

3 years ago

The law of reflection says that the angle of incidence is

nekit [7.7K]

The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.

hope this helps :)

4 0

3 years ago

Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interv

GuDViN [60]

I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time.

This is sneaky. At first, I thought it must be both 'c' and 'd'. But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)

The answer is momentum.

Momentum is (mass x speed). Change in momentum is (force x time).

No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it. If the time is the same for any car, then more momentum
will always require more force.

4 0

3 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

3 years ago