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Scorpion4ik [409]
3 years ago
11

Which statement best describes how waves carry energy?

Physics
2 answers:
ratelena [41]3 years ago
7 0

Correct answer choice is :


A) Energy moves from one place to another through the wave.


Explanation:


Waves can carry energy over a range without moving matter the complete distance. For instance, an ocean wave can move many kilometers without the water itself traveling many kilometers. The water moves up and down a motion recognized as a disturbance. A wave brings its energy without carrying matter. Waves are observed to travel through an ocean or lake, yet the water perpetually reverts to its rest state. Energy is carried through the medium, yet the water particles are not carried.


nignag [31]3 years ago
5 0
A.Energy moves from one place to another through the wave.
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A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi
Rus_ich [418]
A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² =  1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² =  1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s


b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s² 
h = ?

1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
3 0
2 years ago
A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s . assuming constant acc
mars1129 [50]

To solve this problem, we use the equation:

<span>d = (v^2  - v0^2) / 2a</span>

 

where,

d = distance of collapse

v0 = initial velocity = 101 km / h = 28.06 m / s

v = final velocity = 0

a = acceleration = - 300 m / s^2

 

d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)

<span>d = 1.31 m</span>

3 0
2 years ago
Difference between pulling and pushing force​
Mrac [35]

Answer:

Push and pull both are forces , but the difference is in their direction at which it is applied . If the force applied in the direction of motion of the particle then we call it as push . If that force applied in the direction OPPOSITE to the motion of particle then it is termed as pull

7 0
2 years ago
A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at
Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

4 0
2 years ago
Dont know whether its b or c. (cuz for sure it transports energy)
masha68 [24]

Sound waves are a type of classical waves and so they transport only energy without transporting matter through the medium.

6 0
2 years ago
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