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2 years ago

Which of the following is an example of an electric force?

1 answer:

8 0

The example of electric force in the given question is a Plastic wrap sticks to glass bowl containing leftovers. Option A is the right answer.

ELECTRIC FORCE

According to Coulomb's law, the electric force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

The electric force can be:

Force of attraction

Force of repulsion

The electric force can be exhibited in both static and current electricity. Example of static electricity are;

a glass rod rub with fur

an ebonite rod rub with silk.

e.t.c

The example of electric force in the given question is a Plastic wrap sticks to glass bowl containing leftovers.

The plastic wrap is able to stick to glass bowl container because of the force of attraction between the plastic wrap and the glass bowl container.

Therefore, the correct answer is the option A

Learn more about Electric Force here: brainly.com/question/4053816

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.

Contact [7]

The center of mass is given with this formula:
$x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}$
Velocity is:
$v=\frac{dv}{dt}$
So, for the velocity of the center of mass we have:
$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\$
In our case it is:
$v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}$

5 0

3 years ago

A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with

Naddika [18.5K]

The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

$F = \frac{mv}{t}$

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

$F = \frac{0.07 \times 56}{0.04}$

$F = \frac{3.92}{0.04}$

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: brainly.com/question/13590154

4 0

2 years ago

Read 2 more answers

16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d

dsp73

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t = power X time

W / t = P power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

5 0

2 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

1 year ago