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liraira [26]
3 years ago
11

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

distance between them, what happens to the magnitude of the force between the objects?
a. the force between them quadruples
b. the between them doubles
c. the between them is halved
d. the force between them is quartered
e. the force between them is unchanged
Physics
1 answer:
beks73 [17]3 years ago
4 0

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

q_1' = 2q_1\\q_2' = 2q_2

While the distance does not change, so the new force will be

F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F

so, the force will quadruple.

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(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh
makvit [3.9K]

Answers:

(a) 2509.98 m/s

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is 700 km  and whose gravitational acceleration at the surface is a_{g}=4.5 m/s^{2}

Knowing this, let's begin:

a) In this part we need to find the escape speed V_{e} on the asteroid:

V_{e}=\sqrt{\frac{2GM}{R}} (1)

Where:

G is the universal gravitational constant

M is the mass of the asteroid

R=700 km=700(10)^{3} m is the radius of the asteroid

On the other hand we know the gravitational acceleration is a_{g}=4.5 m/s^{2}, which is given by:

a_{g}=\frac{GM}{R^{2}} (2)

Isolating GM:

GM=a_{g}R^{2} (3)

Substituting (3) in (1):

V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R} (4)

V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)} (5)

V_{e}=2509.98 m/s (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

E_{o}=E_{f} (7)

Being:

E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R} (8)

E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h} (9)

Where:

E_{o} is the initial mechanical energy

E_{f} is the final mechanical energy

K_{o} is the initial kinetic energy

K_{f}=0 is the final kinetic energy

U_{o} is the initial gravitational potential energy

U_{f} is the final gravitational potential energy

m is the mass of the object

V=1510 m/s is the radial speed of the object

h is the distance above the surface of the object

Then:

\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h} (10)

Isolating h:

h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R (11)

h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m (11)

h=397042.215 m (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance h=981.8 km=981.8(10)^{3} m. This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2} (13)

Isolating V:

V=\sqrt{2a_{g} R(1-\frac{R}{R+h})} (14)

V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})} (15)

Finally:

V=1917.76 m/s

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Answer:

Part a)

KE = 77.95 J

Part b)

L = 3.16 m

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2

KE = \frac{7}{10} mv^2

KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)

KE = 77.95 J

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

mgLsin\theta = KE

\frac{42}{9.81}(9.81)L sin36 = 77.95

L = 3.16 m

Part c)

by equation of energy conservation we know that

\frac{7}{10}mv^2 = mgL sin\theta

so here we can see that distance L is independent of the mass of the sphere

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3 years ago
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