Can at transform plate boundaries, two plate move toward each other true or false

An engine flywheel initially rotates counterclockwise at 6.77 rotations/s. Then, during 23.9 s, its rotation rate changes to 3.5

olganol [36]

Answer:

-2.70 rad/s²

Explanation:

Given that

ω1 = initial angular velocity of the flywheel, which is 6.77 rev/s

If we convert it to rad/s, we have

(6.77 x 2π) rad/s = 13.54π rad/s

ω2 = final angular velocity of the flywheel = -3.51 rev/s,

On converting to rad/s also, we have

(-3.51 x 2π) rad/s = 7.02π rad/s

α = average angular acceleration of the flywheel = ?

Δt = elapsed time = 23.9 s

Now, using the formula, α = (ω2 - ω1)/Δt. On substituting, we have

α = (-7.02π rad/s - 13.54π rad/s)/23.9 s

α = -20.56π rad/s / 23.9 s

α = -64.59 rad/s / 23.9 s

α = -2.70 rad/s²

Therefore, the average angular acceleration of the flywheel is -2.70 rad/s²

7 0

3 years ago

Two cars, one in front of the other, are traveling down the highway at 25 m/s. the car behind sounds its horn, which has a frequ

netineya [11]

<span>You are given two cars, one in front of the other, that are traveling down the highway at 25 m/s. You are also given a frequency of 500 Hz of the car travelling behind it. You are asked what is the frequency heard by the driver of the lead car. This problem can be solved using the Doppler effect

sound frequency heard by the lead car = [(speed of sound + lead car velocity)/( speed of sound + behind car velocity)] * (sound of frequency of the behind car)
</span>sound frequency heard by the lead car = [(340 m/s + 25 m/s)/(340 m/s - 25 m/s)] * (500 Hz)
sound frequency heard by the lead car = 579 Hz

7 0

3 years ago

Gravity can be described as (2 points)

ladessa [460]

Answer:

"the force of attraction between two objects"

Explanation:

According to Newton's Universal Law of Gravitation, gravity is a force of attraction acting between objects that possess mass. The fact that we only observe gravitational attraction (as opposed to repulsion) makes gravity unique among the known forces.

6 0

3 years ago

A wave of amplitude 20mm has intensity Ix. Another wave of the same frequency but of amplitude 5mm has an intensity Iy.

Alexeev081 [22]

Answer:

(C) 16

Explanation:

Given:

The amplitude of first wave (s₁) = 20 mm

The amplitude of second wave (s₂) = 5 mm

Intensity of first wave = Iₓ

Intensity of second wave = $I_y$