Ask question

Ask question

All categories

3 years ago

12

If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled

and the distance is cut in half, what is the new value of the force of repulsion, F2? Express your answer in terms of F1.

1 answer:

Alisiya [41]3 years ago

6 0

Answer: $F_2=16\times F_1$

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

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

where $q_1$ and $q_2$ is the charges of particle

r=distance between charge particle

$=\frac{kq^2}{r^2}$ when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

$F_2=\frac{k(2q)^2}{(0.5r)^2}$

$F_2=\frac{kq^2}{r^2}\times 4\times 4$

$F_2=16\times F_1$

You might be interested in

A tractor ploughing a field accelerates at 2 m/s2

Helen [10]

Answer:

3 m/s

Explanation:

Given:

a = 2 m/s²

Δx = 10 m

v = 7 m/s

Find: v₀

v² = v₀² + 2a (x − x₀)

(7 m/s)² = v₀² + 2 (2 m/s²) (10 m)

v₀ = 3 m/s

7 0

3 years ago

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

3 years ago

Tema [17]

Answer:

B.

1-3hours

Explanation:

This is because a diabetic patients have increased blood sugar or glucose concentration. After eating a meal, the blood glucose concentration will be increased as it is been accumulated . Therefore it is best diabetic patient exercise at that hour to reduce it's blood glucose concentration.

6 0

2 years ago

What do we mean by the observable universe?

mote1985 [20]

The observable universe<span> is a spherical region of the </span>Universe, <span>comprising all matter that can be observed from Earth at the present time, because light and other signals from these objects have had time to reach Earth since the beginning of the cosmological expansion.

</span>

4 0

3 years ago

A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of 4° with the vertical. If g=10m/

nataly862011 [7]

Answer:

Train accaleration = 0.70 m/s^2

Explanation:

We have a pendulum (presumably simple in nature) in an accelerating train. As the train accelerates, the pendulum is going move in the opposite direction due to inertia. The force which causes this movement has the same accaleration as that of the train. This is the basis for the problem.

Start by setting up a free body diagram of all the forces in play: The gravitational force on the pendulum (mg), the force caused by the pendulum's inertial resistance to the train(F_i), and the resulting force of tension caused by the other two forces (F_r).

Next, set up your sum of forces equations/relationships. Note that the sum of vertical forces (y-direction) balance out and equal 0. While the horizontal forces add up to the total mass of the pendulum times it's accaleration; which, again, equals the train's accaleration.

After doing this, I would isolate the resulting force in the sum of vertical forces, substitute it into the horizontal force equation, and solve for the acceleration. The problem should reduce to show that the acceleration is proportional to the gravity times the tangent of the angle it makes.

I've attached my work, comment with any questions.

Side note: If you take this end result and solve for the angle, you'll see that no matter how fast the train accelerates, the pendulum will never reach a full 90°!

8 0

3 years ago