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3 years ago

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Two tiny conducting spheres are identical and carry charges of -18.4 µC and +53.0 µC. They are separated by a distance of 2.73 c

m. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?

1 answer:

melomori [17]3 years ago

4 0

Answer:

-11776.36 N

This force is attractive since both charges are of opposite sign

Explanation:

Given that

$q_1=-18.4\mu C=-18.4\times 10^{-6}C\ and\\\ q_2=53\mu C=53\times 10^{-6}C$

Distance between the spheres = 2.73 cm =0.0273 m

where K is Coulomb's constant = $9.10^ 9 [N.m^2 /C^2]$

According to coulombs law we know that force between two charges is given by

$F = \frac{kq_1q_2}{r^2}$

$F = \frac{(9\times10^9)\times (-18.4 \times 10^-^6)(53 \times 10^-^6}{0.0273^2)} \\= -11776.36N$

This force is attractive since both charges are of opposite sign

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A 320-kg satellite experiences a gravitational force of 800 N. What is the radius of the satellite’s orbit? What is its altitude

Westkost [7]

Answer:

Radius of orbit = 3.992 × $10^{7}$ m

Altitude of Satellite =33541.9× m

Explanation:

Formula for gravitational force for a satellite of mass m moving in an orbit of radius r around a planet of mass M is given by;

$F= G\frac{m M}{r^{2} }$

Where G = Gravitational constant = 6.67408 × 10-11 $\frac{m^{3} }{Kg sec^{2} }$

We are given

F= 800 N

m = 320 Kg

M = 5.972 × $10^{24}$ Kg

G = 6.67408 × 10-11 $\frac{m^{3} }{Kg sec^{2} }$

We have to find radius r =?

putting values in formula;

==> 800 =6.67408 × $10^{-11}$ × 320 × 5.972 × $10^{24}$ / $r^{2}$

==> 800= 39.8576 × $10^{13}$ × 320 / $r^{2}$

==> 800 = 12754.43 × $10^{13}$ / $r^{2}$

==> $r^{2}$ = 12754.43 × $10^{13}$ /800

==> $r^{2}$ =15.94 × $10^{13}$

==> r = 3.992 × $10^{7}$ m

==> r = 39920× $10^{3}$ m

This is the distance of satellite from center of earth. To find altitude we need distance from surface of earth. So we will subtract radius of earth from this number to find altitude.

Radius of earth =6378.1 km = 6378.1 × $10^{3}$ m

Altitude = 39920× $10^{3}$ - 6378.1 ×

3 0

2 years ago

(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed

stiks02 [169]

a) 7 buses

b) 6.8 m

Explanation:

a)

The motion of the motorcycle is a projectile motion, so it consists of 2 independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion along the vertical direction

The initial components of the velocity of the motorcycle are:

$v_x = u cos(32^{\circ})=(40.0)(cos 32^{\circ})=33.9 m/s\\v_y = u sin(32^{\circ})=(40.0)(sin 32^{\circ})=21.2 m/s$

The equation for the vertical motion of the motorcycle is

$y=h+u_y t - \frac{1}{2}gt^2$

where

y is the altitude at time t

h is the initial height

$g=9.8 m/s^2$ is the acceleration due to gravity

The bus top is at the same height of the initial ramp, so we have

$y=h$

And therefore, we can solve the equation for t, to find the time of flight:

$0=u_y t - \frac{1}{2}gt^2\\t(u_y-\frac{1}{2}gt)=0\\t=\frac{2u_y}{g}=\frac{2(21.2)}{9.8}=4.33 s$

Now we find what is the horizontal distance covered by the motorcycle in its jump, which is given by:

$d=v_x t = (33.9)(4.33)=146.8 m$

And since each bus has a length of L = 20.0 m, the number of buses that the motorcycle can clear with its jump is:

$n=\frac{d}{L}=\frac{146.8}{20}=7.34$

So, 7 buses.

b)

In the previous problem, we saw that the total range of the motion of the motorcycle is

$d=146.8 m$

And we said that this corresponds to 7 buses.

Each bus has a length of

L = 20 m

So, the total length of 7 buses is

$L' = 7L=7(20)=140 m$

Therefore, the range of the motorcycle is greater than the length of the buses by:

$\Delta x = d-L'=146.8-140 = 6.8 m$

which means he will miss the last bus by 6.8 meters.

6 0

3 years ago

During a baseball game, a batter hits a popup to a fielder 84 m away. The acceleration of gravity is 9.8 m/s 2 . If the ball rem

Nezavi [6.7K]

The ball rises up to 30.625 m high.

<h3>What is gravity?</h3>

The force that pulls items toward the centre of a planet or other entity is called gravity. All of the planets are kept in orbit around the sun by gravity.

Given parameters:

Horizontal range of the ball; R = 84 m.

Time of flight of the ball: T = 5 s.

Acceleration due to gravity; g = 9.8 m/s².

We have to find, maximum height obtained by the ball; H = ?

In projectile motion, time of flight is; T = 2usinθ/g

Where, u = initial velocity of the ball and θ = angle of projection.

So, 2usinθ/g = 5

⇒ usinθ = 5g/2 = (5×9.8)/2m/s = 24.5 m/s.

Then maximum height obtained by the ball;

H = u²sin²θ/2g

= 24.5²/(2×9.8) = 30.625 m.

Hence, maximum height obtained by the ball is 30.625 m.

Learn more about gravity here:

brainly.com/question/13860566

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10 months ago

What determines the properties of elements?

likoan [24]

Their atomic number, and hence the number of electrons.

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2 years ago

Which of the following types of electromagnetic radiation has the longest wavelength?

mash [69]

Answer:

Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.

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