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Snezhnost [94]
3 years ago
14

Strength of the electric force imagine two 1.0-g bags of protons, one at the earth's north pole and the other at the south pole.

Physics
1 answer:
Vladimir79 [104]3 years ago
4 0
(a) The mass of one proton is m_p = 1.67 \cdot 10^{-27} kg. Each bag has a mass of M=1.0 g=0.001 kg, so the number of protons in each bag is
N= \frac{M}{m_p}= \frac{0.001 kg}{1.67 \cdot 10^{-27}kg}=6.0 \cdot 10^{23}

(b) The distance between the two bags is twice the Earth's radius:
r=2 \cdot 6.37 \cdot 10^6 m=1.27 \cdot 10^7 m

The gravitational attraction between the two bags is given by
F_g = G  \frac{M_1 M_2}{r^2}=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(0.001 kg)^2}{(1.27 \cdot 10^7 m)^2}=4.1 \cdot 10^{-31}N

To calculate the electrostatic repulsion, we should first calculate the total charge of each bag, which is the charge of one proton times the number of protons:
Q=eN=(1.6 \cdot 10^{-19} C)(6.0 \cdot 10^{23})=9.6 \cdot 10^4 C

And now we can calculate the electrostatic repulsion between the two bags:
F_e = k_e  \frac{Q_1 Q_2}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(9.6 \cdot 10^4 C)^2}{(1.27 \cdot 10^7 m)^2}=5.14 \cdot 10^5 N

(c) we can easily see from point (b) that the gravitational attraction between the two bags is very tiny, so it cannot be felt. Instead, the electrostatic repulsion between the two bags is a very large number, so this can be felt.
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Explanation:

Conservation of momentum

Initial momentum is zero

3(15) + 2(v) = 0

v = - 22.5 m/s

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3 0
3 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
2 years ago
A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of work during this pull.
Wittaler [7]

Answer:

<h2>602.08 N</h2>

Explanation:

The force supplied by the train can be found by using the formula

f =   \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{578000}{960}  \\  = 602.083333...

We have the final answer as

<h3>602.08 N</h3>

Hope this helps you

7 0
3 years ago
The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e
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The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m


g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2

 
8 0
3 years ago
A student is performing experiments on a particular substance. Which
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Answer: A.

Explanation:

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