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2 years ago

A machine produces 4,000 J of work in 5 seconds. how much power does the machine produce.

Physics

1 answer:

ad-work [718]2 years ago

5 0

Explanation:

Power is defined as the work done per unit time or

$P = \dfrac{\Delta W}{\Delta t}$

$\;\;\;\;= \dfrac{4000\:\text{J}}{5\:\text{s}} = 800\:\text{Watts}$

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A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well

Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

Explanation:

8 0

2 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Bond [772]

The y-component of the acceleration is $0.33 m/s^2$

Explanation:

The y-component of the acceleration is given by:

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

$u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s$

where

u = 2.25 m/s is the initial velocity

$\theta_1 = 50.0^{\circ}$ is the initial direction

$v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s$ , where

v = 4.65 m/s is the final velocity

$\theta_2 = 120.0^{\circ}$ is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

$a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2$

Learn more about acceleration:

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2 years ago

Consider a box sitting in the back of a pickup. The pickup accelerates to the right, and because the bed of the pickup is sticky

andreyandreev [35.5K]

The force would be coming from the right causing the box the lean/ slide to left, if it wasnt sticky.

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3 years ago

Solar energy leaves the core of the sun in the form of

olasank [31]

photons and convection - density differences makes bubbles of hot stuff float up. pretty sure

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2 years ago