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2 years ago

If it takes you 20 joules of work to move a couch 10 meters in 10 seconds, what is the power?

Physics

1 answer:

Debora [2.8K]2 years ago

4 0

Answer:

Power = 2 j/s

Explanation:

Power = Work / Time

= 20/10

= 2 j/s

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What units are used to measure mass and weight?

horrorfan [7]

Mass- Mass is measured in kilograms (kg).

Weight- Weight is measured in Newton’s.

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3 years ago

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3 years ago

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James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s

liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{James}=\sqrt{\frac{2h}{g}}$

Vertical motion of John : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{John}=\sqrt{\frac{2h}{g}}$

So both times are same.

Both of them reach the lake at the same time.

3 0

3 years ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago

If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1

Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

$I=\frac{Q}{t}$

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

$\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}$

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>

3 0

1 year ago