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Elenna [48]
3 years ago
5

Which of the following results when a crest and trough meet?

Physics
1 answer:
kifflom [539]3 years ago
6 0
D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.
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A 12 kg<br> mass is lifted to a height of 2 m. What is its potential energy<br> at this position?
Romashka-Z-Leto [24]

Answer:

Explanation:

Potential energy is the energy stored within an object, due to the object's position, arrangement or state

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2 years ago
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The conduction of heat from hot body to cold body is an example of what thermodynamics process?<br>​
Marizza181 [45]

Answer:

Heat flow

Explanation:

6 0
2 years ago
PLEASE HELP ME PLEASE
Afina-wow [57]

Answer:

For A

Displacement= 1/2*3*6= 9m

For B

Displacement= 1/2*4*4= 8m

5 0
3 years ago
A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi
tekilochka [14]

Answer:

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg

Explanation:

Hi there!

Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:

PE = EPE

m · g · h = 1/2 k · x²

Where:

m = mass.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression distance

The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:

EPE =1/2 k · x²

EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J

Then, using the equation of gravitational potential energy:

PE = m · g · h =  541.2 J

m =  541.2 J/ g · h

m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)

m = 16.6 kg

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.

6 0
3 years ago
A diver jumps off a cliff 50m high and needs to clear the rock that extend outward 5.0m from the base of the cliff. The diver ju
igor_vitrenko [27]

Answer:

He should run at least at 1.5 m/s

The diver will enter the water at an angle of 87° below the horizontal.

Explanation:

Hi there!

The position and velocity of the diver are given by the following vectors:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the  upward direction as positive)

v = velocity vector at time t

Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.

We know that, to clear the rocks, the position vector r final (see figure) should be:

r final = ( > 5.0 m, -50 m)

So let´s find first at which time the y-component of the vector r final is - 50 m:

y = y0 + v0y · t + 1/2 · g · t²

-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m

Solving the quadratic equation

t = 3.4 s

Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:

x = x0 + v0x · t      (x0 = 0)

5.0 m < v0x · 3.4 s

v0x > 5.0 m / 3.4 s

v0x > 1.5 m/s

The initial horizontal velocity should be greater than 1.5 m/s

To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:

vfx > 1.5 m/s

Now, let´s calculate vfy:

vfy = v0y + g · t

vfy = 2.1 m/s - 9.8 m/s² · 3.4 s

vfy = -31 m/s

Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:

|v| = \sqrt{(1.5 m/s)^{2} + (31m/s)^{2}} = 31 m/s

Then the final velocity of the diver will be greater or equal than 31 m/s.

To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:

cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s

θ = 87°

The diver will enter the water at an angle of 87° below the horizontal.

8 0
3 years ago
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