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3 years ago

What is transferred by all types of waves? A. energy B. electrons C. heat D. matter

Physics

2 answers:

erik [133]3 years ago

5 0

Energy, hope this helps you.

djyliett [7]3 years ago

3 0

<span>The question is what do all types of waves transfer. Waves, both mechanical and electromagnetic are disturbances that transfer energy without transporting matter. It doesnt matter wether they are propagating trough a medium or trough vacuum. Therefore the correct answer is A. energy.</span>

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The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di

nlexa [21]

Answer and Explanation:

Let:

$x(t)=Acos(\omega t+ \phi)$

The equation representing a simple harmonic motion, where:

$x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time$

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

$v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)$

Also, you may know these fundamental formulas:

$f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}$

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

$x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m$

(b)

$v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s$

(c)

$a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2$

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

$\phi = \frac{\pi}{5}$

(e)

$f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz$

(f)

$T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s$

8 0

3 years ago

Read 2 more answers

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car’s moti

Nitella [24]

The car is staying still. either getting petrol or stopped at a light .

3 0

3 years ago

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Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a

Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

Vf = √( 2 g X)

Vf = √(2 9.8 1)

Vf = 4.43 m/s

This is the speed with which it reaches the ground

Having the final speed we can find the time

Vf = Vo + g t

t = Vf / g

t = 4.43 / 9.8

t = 0.45 s

This is the time of fall of the body to touch the ground

3 0

3 years ago

Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,

crimeas [40]

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

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3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca

4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t + 1/2 a t²

s = 1/2 a t² ----------- (2)

compare equation (1) and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0

3 years ago