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Ludmilka [50]
3 years ago
13

The drawing on the right shows a man (mass = 70kg) and the child (mass = 35 kg) push each other apart. The man moves away with a

speed of 0.3 m/s relative to the ice.
Calculate the velocity of the child relative to the ice.


LET MOMENTUM EQUATION GUIDE YOUR THINKING...


SHOW YOUR WORK - INCLUDE UNITS

Physics
2 answers:
nordsb [41]3 years ago
7 0

Refer to the attachment..

Pepsi [2]3 years ago
3 0

Answer:

<h2>Hope it would help you.</h2>

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Describe the motion represented by a horizontal line on a distance-time graph.
denpristay [2]

Answer:

hi, this is the answer

Explanation:

A horizontal line on a distance-time graph shows no change in distance, therefore there is no motion.

The object is stationary. ...

Constant speed is motion that occurs with the same ratio of distance to time throughout the entire length of the motion.

pls mark this as the brainliest...

3 0
3 years ago
A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field  is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

\Delta A = \frac{1}{2} \Delta \theta R^2

The  formula for the induced emf is

E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

=B\frac{\Delta  A}{\Delta  t}

B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

so, the induce EMF will be

E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V

6 0
3 years ago
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