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ohaa [14]
3 years ago
14

Both the electrical force and the gravitational force between two objects share which relationship?

Physics
2 answers:
love history [14]3 years ago
8 0

Answer : They are inversely proportional to the square of the distance.

Explanation :

We know that the electrical or electrostatic force is given by :

F=\dfrac{1}{4\pi \epsilon_0}\dfrac{q_1q_2}{r^2}.........(1)

Where,

\dfrac{1}{4\pi \epsilon_0} is constant

q_1 q_2 are the electric charges,

r is the distance between two charges.

Similarly, the gravitational force is given by :

F=G\dfrac{m_1m_2}{r^2}..........(2)

Where

G is the universal gravitational constant

m_1 and m_1  are the masses

r is the distance between them

From equation (1) and (2), it is clear that the both electrical as well as gravitational forces are inversely proportional to the square of the distance between them.    

Hence, the correct option is (b) " They are inversely proportional to the square of the distance" .

kolbaska11 [484]3 years ago
7 0

They are inversely proportional to the square of the distance.

Explanation:

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the two objects.

The magnitude of the electrical force between two objects is given by

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

where k is the Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between the two objects.

In both cases, we see that the magnitude of the force is iinversely proportional to the square of the distance, \frac{1}{r^2}, so the correct option is

They are inversely proportional to the square of the distance.

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The total average velocity v=+1.41 m/s (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion (S_1=6.30 km=6300 m, due west) and of the displacement in the second part of the motion (S_2, due east).
-The total time taken t is the time taken for the first part of the motion, t_1, and the time taken for the second part of the motion, t_2. t_1 can be found by using the average velocity and the displacement of the first part:
t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s

t_2, instead, can be written as \frac{S_2}{v_2}, where v_2=-0.630 m/s is the average velocity of the second part of the motion (with a negative sign, since it is due east). 

Therefore, we can rewrite the initial equation as:
v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
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A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at
VARVARA [1.3K]

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

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mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

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K.E₂ = ¹/₂v²(m₁ + m₂)

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K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

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                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

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