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zhuklara [117]
3 years ago
10

An object’s momentum includes its __________.

Physics
1 answer:
sveta [45]3 years ago
7 0

Answer: Momentum can be defined as "mass in motion

Explanation:

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wolverine [178]
1. Always be the bigger person
2. Violence is never the answer
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An object appears red when viewed with white light. What color would it appear when viewed with blue light?
zvonat [6]

Answer:

Colour of any object is by the colour of light it reflects i.e if white light is incident on the object,it will reflect blue color.so it will appear blue.But if red light is incident on it,it will not reflect that and absorb it.so as it will not reflect any light it will appear black.

Explanation:

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3 years ago
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A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

brainly.com/question/11411375

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#LearnwithBrainly

6 0
3 years ago
Find the final velocity of a car that accelerates at +2 m/s2 for +4m from an
Stolb23 [73]

Answer:

Explanation:

according to third equation of motion

2as=vf²-vi²

vf²=2as+vi²

vf=√2as+vi²

vf=√2as+vi

vf=√2*2*4+3

vf=√16+3

vf=4+3=7

so final velocity is 7 m/s

5 0
3 years ago
A 66.0−kg short-track ice skater is racing at a speed of 10.0 m/s when he falls down and slides across the ice into a padded wal
dexar [7]

Answer:

3300J

Explanation:

Work done is the energy that is lost by the skater

Formula for workdone = 1/2*mV^2

m = 66kg

V = 10m/s

Work done = 1/2 * 66 * 10^2

= 3300J

7 0
3 years ago
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