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andriy [413]
2 years ago
8

A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.

Physics
1 answer:
s344n2d4d5 [400]2 years ago
8 0

Answer:the answer is wave.

Explanation:

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What is the value of the activation energy of the uncatalyzed reaction in reverse?
alexgriva [62]
It would be: Activation Energy = 300 KJ

Hope this helps!
5 0
3 years ago
Write two ways to be protected from the energy crisis?​
Arada [10]
Possible Solutions to the Problem of Global Energy Crisis:

1. Move Towards Renewable Resources.
2. Buy Energy-Efficient Products.
5 0
2 years ago
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
Why are movie scenes in outer space often unrealistic?
kifflom [539]

Answer:

see explanation

Explanation:

by my calculations, i believe it may be because they have to pretend to be in space.

8 0
2 years ago
Read 2 more answers
An object of mass 3.4 kg is moving in a straight line with kinetic energy 59.177 J. A force is applied in the direction of its m
rusak2 [61]

Answer:

Its momentum is multiplied by a factor of 1.25

Explanation:

First, we <u>calculate the initial velocity of the object</u>:

  • K = 0.5 * m * v₁²
  • 59.177 J = 0.5 * 3.4 kg * v₁²
  • v₁ = 5.9 m/s

With that velocity we can <u>calculate the initial momentum of the object</u>:

  • p₁ = v₁ * m
  • p₁ = 20.06 kg·m/s

Then we <u>calculate the velocity of the object once its kinetic energy has increased</u>:

  • (59.177 J) * 1.57 = 0.5 * 3.4 kg * v₂²
  • v₂ = 7.4 m/s

And <u>calculate the second momentum of the object</u>:

  • p₂ = v₂ * m
  • p₂ = 25.16 kg·m/s

Finally we <u>calculate the factor</u>:

  • p₂/p₁ = 1.25
3 0
2 years ago
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