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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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Thank you so muchhhhh
loris [4]

Answer:

I'm taking a wild guess at c

Explanation:

c. winter solstice

6 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
Read 2 more answers
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m

8 0
2 years ago
Plzpzlpzlzplzplzplzpz this one also<br> all questions ​
WINSTONCH [101]

Answer:

I didn't know these questions sorry

4 0
2 years ago
Un prisma de cemento pesa 2500 N y ejerce una presión de 125 Pa, ¿cuál es el valor del área en la cual se apoya?
Eduardwww [97]

Answer:

Area = 20 m²

Explanation:

Given the following data;

Force = 2500 N

Pressure = 125 Pa

To find the area on which it rest;

Mathematically, pressure is given by the formula;

Pressure = \frac {Force}{area}

Making area the subject of formula, we have;

Area = \frac {Force}{pressure}

Substituting into the formula, we have;

Area = \frac {2500}{125}

Area = 20 m²

3 0
3 years ago
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