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krok68 [10]
3 years ago
7

what does the term equilibrium refer to? a. the resting position of the wave b. the highest point in the wave c. the lowest poin

t of the wave d. the next subsequent wave e. the period of the wave
Physics
2 answers:
Brums [2.3K]3 years ago
7 0
A. the resting position of a wave
djverab [1.8K]3 years ago
6 0

Answer:

a. the resting position of the wave

Explanation:

The equilibrium refers to the resting position of the wave or the position it would be if it was undisturbed. And this is normally exactly midway between the highest point of the wave and the lowest point of the wave.

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Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba
Allushta [10]
The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
p_i = p_1-p_2 = mv-mv =0
So, the final momentum must be zero as well:
p_f = 0
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
p_f = mv_1 + mv_2 = 0
From which
v_1 = -v_2

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is \frac{1}{2} mv^2. Therefore, the total kinetic energy before the collision is
K_i = \frac{1}{2} mv^2 +  \frac{1}{2} mv^2 = mv^2
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
K_f = K_i = mv^2 (1)
But the final kinetic energy, Kf, is also
K_f =  \frac{1}{2} mv_1^2 +  \frac{1}{2}mv_2^2
Substituting v_1 = -v_2 as we found in the conservation of momentum, this becomes
K_f = mv_2 ^2
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write 
mv_2^2 = mv^2

Therefore, the two final speeds of the balls are
v_2 = v
v_1 = -v_2 = -v

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0
3 years ago
A 5000kg car traveling at 40m/s crashes into a wall and comes to a complete stop in 10ms. What was the force on the wall?​
Kisachek [45]

Answer:

20,000,000 N

Explanation:

First find the acceleration:

a = Δv / Δt

a = (0 − 40 m/s) / 0.010 s

a = -4000 m/s²

Next use Newton's second law to find the force on the car:

F = ma

F = (5000 kg) (-4000 m/s²)

F = -20,000,000 N

According to Newton's third law, the force on the wall is equal and opposite the force on the car.

F = 20,000,000 N

7 0
3 years ago
Please help I’ll give brainliest
marta [7]

Answer:

Repel

Unlike

Atrract

Fur

Balloon

Positivley charged

negative

postive

neutral

Explanation:

It goes from top to bottom

7 0
3 years ago
A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. how far will a point on the edge of the wheel
melisa1 [442]

First we find the angular acceleration \alpha. The angular velocities are \frac{240}{60} =4 \;rps and \frac{360}{60} =6 \;rps The angular velocities and \alpha are related as

\omega=\omega_0+\alpha t\\ \alpha =\frac{\omega - \omega_0}{t} \\ \alpha =\frac{6-4}{(6.5)} \\ \alpha =0.3077 \;rps^2\\ \alpha =(0.3077 )2\pi =1.933 \;rad/s^2

Angle turned in 6.5 seconds is

2 \alpha\theta =   \omega^2-\omega_0^2\\ \theta =  \frac{ \omega^2-\omega_0^2}{2 \alpha}\\  \theta =  \frac{ 6^2-4^2}{2 (1.933)}\\  \theta = 5.173 \; rad

The distance traveled by a point on the edge of the wheel is r \theta = \frac{33}{2}(5.173)= 85.35 \;cm

7 0
4 years ago
Which of the following is an inference? A) the flower is pink. B) vinegar has a pungent aroma. C) a dog barks. D) a cup with ste
velikii [3]

I'm like 99% sure that it's d.

8 0
3 years ago
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