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Soloha48 [4]
3 years ago
8

Which has more momentum: a 30 kg bicycle moving at 10 m/s or a 300 kg car at rest?

Physics
1 answer:
Hatshy [7]3 years ago
5 0
<h2>Answer:</h2>

Momentum = mass × velocity

<h3>Case I </h3>

Momentum of bicycle = 30 × 10 = 300 kgms¯¹

<h3>Case II </h3>

Momentum of car = 300 × 0 = 0 kgms¯¹

<h3>Therefore, 300 > 0 so momentum of bicycle is more. Hence, option A is correct. </h3>
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Digiron [165]

Answer:

Incident ray

Explanation:

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.
8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

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U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

4 0
3 years ago
4) For the situation pictured below, F1 = 20.0 N east and F2 = 30.0 N west ,
maksim [4K]

Answer:

10.0 N West

Explanation:

Imagine East is positive and West is negative. Essentially now we have 2 forces pulling opposite each other on the same plane, so we can add them up.

F1=20N

F2=-30N

F1+F2=Net Force

20+(-30)=-10

Therefore the Net Force is 10N West.

8 0
3 years ago
A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total
Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

v_2 = v - 1

now we know that total time of the motion is 19 hr

so we will have

t_1 = \frac{90}{v}

t_2 = \frac{90}{v - 1}

so we will have

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\frac{90}{v} + \frac{90}{v-1} = 19

90(2v - 1) = 19(v^2 - v)

19 v^2 - 199 v + 90 = 0

by solving above equation we have

v = 10 mph

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

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3 years ago
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