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3 years ago

Describe the basis concept of the atwood’s machine

1 answer:

8 0

Answer:The Atwood Machine is a device that demonstrates the basic principles of acceleration and dynamics. You'll mostly see Atwood machines in Physics laboratories and classrooms. It consists of two objects with different masses that hang vertically from a frictionless pulley that has a very small, negligible mass.

Explanation:

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If 270 watts of power is used in 42 seconds, how much work was done

navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

Step 1: Find the Formula

P = W / T

Step 2: Make W the subject of the equation.

W = PT

Step 3: Given.

P = 270 Watts, T = 42 seconds

Step 4: Substitute these values into equation 2 .

W = 270(42)

Step 5: Simplify.

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0

3 years ago

Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector

garri49 [273]

Answer:

$5.2\ \text{m/s}$

$70^{\circ}$ south of east

Explanation:

$v_a$ = 3 m/s

$\theta_a$ = $20^{\circ}$ north of east

$v_b$ = 6 m/s

$\theta_b$ = $40^{\circ}$ south of east = $360-40=320^{\circ}$ north of east

x and y component of $v_a$

$v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}$

$v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}$

x and y component of $v_b$

$v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}$

$v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}$

$\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}$

Magnitude

$|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}$

Direction

$\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}$

The magnitude of the change in velocity vector is $5.2\ \text{m/s}$ and the direction is $70^{\circ}$ south of east.

6 0

3 years ago

An object is thrown with an initial speed v near the surface of Earth. Assume that air resistance is negligible and the gravitat

IgorLugansk [536]

Answer:

E. downward and constant

Explanation:

Freefall is a special case of motion with constant acceleration because the acceleration due to gravity is always constant and downward. This is true even when an object is thrown upward or has zero velocity.

For example, when a ball is thrown up in the air, the ball's velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration ggg, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.