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3 years ago

How is power defined?

Physics

2 answers:

erastova [34]3 years ago

6 0

d. the rate at which work is accomplished

kumpel [21]3 years ago

5 0

D is the answer to this question. Hope this helps!

You might be interested in

If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno

vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0

3 years ago

What is the angular displacement of the wheel between t = 5 s and t = 15 s?

Mkey [24]

The question seems incomplete. The complete text is:

a)What is the angular displacement of the wheel between t = 5 s and t = 15 s

b)What is the angular velocity of the wheel at 15 s

And it refers to the attached figure.

a) 25 rad

The graph shown represent the angular position of the wheel at different times.

Therefore, we can simply calculate the angular displacement between two times by calculating the difference between the angular position at t2 and the angular position at t1.

At $t_1 = 5 s$ , the angular position from the graph is $\theta_1 = 100 rad$

At $t_2 = 15 s$ , the angular position from the graph is $\theta_2 = 125 rad$

Therefore, the angular displacement is

$\Delta \theta= \theta_2 - \theta_1 = 125-100 = 25 rad$

2) -5.0 rad/s

For a angular displacement vs time graph, the angular velocity at any time is simply equal to the slope of the curve at that time.

Here we want to calculate the angular velocity at t = 15 s, so we have to calculate the slope at that time.

By noting that the slope is constant in the last part of the motion, we find that the slope between 10 s and 20 s is:

$\frac{\Delta \theta}{\Delta t}=\frac{100 rad - 150 rad}{20 s - 10 s}=-5.0 rad/s$

This slope is constant between 10 s and 20 s, so the angular velocity of the wheel at t = 15 s

$\omega = -5.0 rad/s$

5 0

3 years ago

If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object

777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0

3 years ago

What is another principle an artist can follow to create the illusion of depth on a flat surface?

Daniel [21]

It can be both flat or it can be when you have new eyeglasses on and you look down it makes you think the ground looks like that but its not

3 0

3 years ago

How does running involve science..? please dont copy off of something

Mila [183]

Running means your body must convert stored glucose into energy through glycolysis.

5 0

3 years ago

Read 2 more answers