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3 years ago

What travels by vibrating particles? Mechincal Waves or ElecrtoMagnetic Waves.

Physics

1 answer:

Naddik [55]3 years ago

5 0

Answer:mechanical waves.

Explanation:

Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.

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Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

OLga [1]

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

The position of the two cars at time t is given by the following functions:

$x_A(t) = \alpha t + \beta t^2$

with

$\alpha = 2.60 m/s\\\beta = 1.20 m/s^2$

Substituting,

$x_A(t)=2.60t+1.20 t^2$

And

$x_B(t)=\gamma t^2 - \delta t^3$

with

$\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3$

Substituting,

$x_B(t)=2.80t^2-0.20t^3$

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

$x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m$

$x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m$

So, car A is initially ahead.

The two cars are at the same point when their position is the same. Therefore, when

$x_A(t)=x_B(t)$

which means when

$2.60t+1.20t^2 = 2.80t^2-0.20t^3$

Re-arranging the equation, we find

$0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0$

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

$0.20t^2-1.6t+2.60=0$

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the same point.

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of car A:

$v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t$

The velocity of car B is given by the derivative of the position of car B:

$v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2$

Therefore, the distance between the two cars is not changing when the two velocities are equal:

$v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0$

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

The acceleration of each car is given by the derivative of the velocity of the car A.

The acceleration of car A is:

$a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40$

While the acceleration of car B is:

$a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t$

So, the two cars have same acceleration when

$a_A(t)=a_B(t)$

And solving the equation, we find:

$2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s$

So, the two cars have same acceleration at t = 2.67 s.

Learn more about accelerated motion:

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3 0

3 years ago

A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?

Dmitriy789 [7]

Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:

1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

Therefore, the Workdone is 592000 J.

6 0

3 years ago

1.Wave motion that is parallel to the wave direction is described

Dafna1 [17]

Answer:

Explanation:

A wave is a phenomenon that travels through a material medium without any permanent effect on the medium. It can be classified as mechanical or electromagnetic waves. And the two major types are transverse and longitudinal.

From the given question;

1. Longitudinal waves e.g sound waves, waves in a spring'

2. Frequency (number of cycles per second).

3. Wavelength of the wave. Measured in meters.

4. Transverse waves e.g light waves, water waves.

5. Wave speed.

6. Electromagnetic waves e.g ultraviolet waves, X-rays etc.

7. Amplitude.

8. Hertz.

9. Rarefaction

10. Compression

7 0

3 years ago

A satellite with a mass of 120 kg fires its rocket thrusters, which give an impulse of 7440 kg-m/s. What was the total change in

Marat540 [252]

From the calculation and the momentum of the body, the velocity is 62 m/s

<h3>What is momentum?</h3>

The term momentum refers to the product of mass and velocity. Now recall that the rate of change of momentum is equal to the impressed force.

Hence;

7440 kg-m/s = 120 kg * v

v= 7440 kg-m/s/120 kg

v = 62 m/s

Learn more about momentum:brainly.com/question/24030570

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3 0

2 years ago

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa

Anna [14]

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature $T_{1}$ = 68 °F = 20° c

Final temperature $T_{2}$ = 212 °F = 100° c

Specific heat of water $C = 4.186 \frac{KJ}{kg c}$

Now heat transfer Q = m × C × ( $T_{2}$ - $T_{1}$ )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

4 0

3 years ago