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koban [17]
3 years ago
12

Can someone please help thank u (. ^ ᴗ ^. )​

Physics
2 answers:
Mila [183]3 years ago
8 0

Answer:

3) The answer is (b) -8 m/s (pointing down)

4) The answer is (c) -9.8 m/s^2 (acceleration of a freely falling object is always due to gravity, -9.8 m/s^2, regardless o

Explanation:

kakasveta [241]3 years ago
6 0

3) The answer is (b) -8 m/s (pointing down)

4) The answer is (c) -9.8 m/s^2 (acceleration of a freely falling object is always due to gravity, -9.8 m/s^2, regardless of direction)

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The resistance of a conductor is given by
R= \frac{\rho L}{A}
where L is the length of the wire, \rho the resistivity of the material and A the cross-sectional area.

We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length: L'=4 L, then the new resistance is also 4 times the original value:
R'= \frac{\rho L'}{A}= \frac{\rho 4 L}{A}=4  \frac{\rho L}{A}=4 R
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In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v
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Answer:

True or False

Explanation:

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3 years ago
A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with
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Momentum, p = m.v
m of the girl = 60.0 kg
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B) momentum of the raft = - momentum of the girl = -24.0 N/s

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A miner develops cancer of the esophagus. Ten years before, he had been exposed to radiation when he worked for a year in a mine
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You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55
ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

v=v_0+at\\\\v^2=v_0^2+2ad

Dividing the second equation by the first one, we obtain:

v=\frac{v_0^2+2ad}{v_0+at}

And, since v_0=0, then:

v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2

So the acceleration is 3.30m/s^2.

7 0
4 years ago
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