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andrezito [222]
3 years ago
13

Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

ectrostati, propulsion that equals 1.00% of the-gravitational atraction between twe, bodies
Physics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

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The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two
Maurinko [17]

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. \omega _0=0 rad/s

after t=10 s

\omega =10 rad/s

using \omega =\omega _0+\alpha t

10=0+\alpha \cdot 10

\alpha =1 rad/s^2

Revolutions turned in 2 s

\theta =\omega _0t+\frac{\alpha t^2}{2}

\theta =0+\frac{1\times 2^2}{2}

\theta =2 rad

To get revolution \frac{\theta }{2\pi }

=\frac{2}{2\pi}=0.318\ revolutions

3 0
3 years ago
Read 2 more answers
The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


4 0
3 years ago
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A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog
Hunter-Best [27]

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

6 0
3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0
3 years ago
A swimming pool is 4.0 m in depth; a swimmer at this depth feels discomfort in the ear. Calculate the net force on a 0.50-cm-dia
Mashcka [7]

The net force on a 0.50-cm-diameter eardrum is mathematically given as

F= 0.76969 N

<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>

Generally, the equation for Pressure is  mathematically given as

P = ρgh

Therefore

P= 1000*9.8*4

P= 39200 Pa

Where

A= pi*(0.005/2)^2

Generally, the equation for Net force is  mathematically given as

F = PA

F= 39200 *( pi*(0.005/2)^2)

F= 0.76969 N

In conclusion, The net force is

F= 0.76969 N

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5 0
1 year ago
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