All categories

3 years ago

When a particle vibrates and passes to the next vibration what happens?

1 answer:

4 0

Im not sure how detailed your answer is supposed to be but here it is: The particle that vibrates contains energy. That exact amount of energy is passed on to the next particle. This occurs because as you must know, energy is neither created nor destroyed. The same amount of energy contained in the vibration is simply passed on.

You might be interested in

Which of the following is a true statement?

forsale [732]

C. Electromagnetic waves do not always need a medium to travel.

Since they are transverse waves they do not need a material medium.

4 0

3 years ago

Read 2 more answers

In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let’s find out by doing a

mixer [17]

Answer:

0.07142 m/s in the opposite direction of the bullet

Explanation:

$m_1$ = Mass of rifle = 70 kg

$m_2$ = Mass of bullet = 0.01 kg

$v_1$ = Velocity of rifle

$v_2$ = Velocity of bullet = 500 m/s

As the momentum of the system is conserved

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.01\times 500}{70}\\\Rightarrow v_1=-0.07142\ m/s$

The recoil velocity of the rifle is -0.07142 m/s

The negative sign shows the opposite direction of the rifle.

3 0

3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

4 years ago

An object has a net charge of +5 Coulombs.

exis [7]

The net charge of the electron will be there because there is no exponents there

3 0

3 years ago

A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st

sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2 ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck : m1 = 2000 Kg v1 = 5 m/s u1 = 0

car : m2 = 1000 kg v2 = 0 u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = $\frac{2000}{1000}$ x 5

= 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0

3 years ago