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Vera_Pavlovna [14]
3 years ago
5

Wave energy can only be transmitted through a material medium. wave energy can only be transmitted through a material medium.

Physics
1 answer:
S_A_V [24]3 years ago
3 0
The correct answer is False.
In fact, electromagnetic waves (which carry energy), do not need a material medium to travel, because they can travel in vacuum as well. So, wave energy can be transmitted also through vacuum.
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What requirement must a force acting on a object satisfy in order for the object to undergo simple harmonic motion?
viktelen [127]

Answer:

Simple harmonic motion is the movement of a body or an object to and from an equilibrium position. In a simple harmonic motion, the maximum displacement (also called the amplitude) on one side of the equilibrium position is equal to the maximum displacement.

The force acting on an object must satisfy Hooke's law for the object to undergo simple harmonic motion. The law states that the force must be directed always towards the equilibrium position and also directly proportional to the distance from this position.

6 0
3 years ago
. What is the relationship between potential energy, kinetic energy, and speed as the skater moves down and up the U-shaped ramp
kykrilka [37]
Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest

Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest

THEREFORE, PE and KE have an INVERSE RELATIONSHIP.
6 0
3 years ago
Read 2 more answers
A dog is walking at 2m/s and then begins to run at a speed of 6m/s. What is his acceleration if his total travel time is 2 secon
fiasKO [112]
The formula for velocity vf = vi + at

First list your given information

2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)

Since you want the a for acceleration get a by itself

a = (vf-vi)/t

So a= (6-2)/2

a= 4/2

a=2

Now units

the units for acceleration are m/sx^{2}

2m/sx^{2}
7 0
3 years ago
What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri
Yakvenalex [24]

Answer:

The force is -1.67\times10^{5}\ N

Explanation:

Given that,

Mass of car = 64 kg

Suppose,  a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

v^2-u^2=2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}

a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}

a=-2623.45\ m/s²

We need to calculate the force

Using formula of force

F=ma

F=64\times(-2623.45)

F=-1.67\times10^{5}\ N

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is -1.67\times10^{5}\ N

5 0
3 years ago
two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity o
Tems11 [23]

Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

7 0
3 years ago
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