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Assume we are given an electric field set up by an unknown charge distribution. U0 is the amount of work needed to bring a point charge of charge q0 in from infinity to a point P. If the charge q0 is returned to infinity, how much work would it take to bring a new charge of 4 q0 from infinity to point P?

Answers

Answer 1

Answer:

Answer:

$4U_0$

Explanation:

We are given that

Amount of work needed to bring a point charge q0 from infinity to a point P= $U_0$

We know that potential at point P= $V=(U)/(q)$

$U=Vq$

Where U=Amount of work needed to bring a point charge q from infinity to a point P

Initially , $V=(U_0)/(q_0)$

New charge, q= $4q_0$

Then, work done,U= $(U_0)/(q_0)* (4q_0)=4U_0$

Hence, the amount of work needed to bring a new charge 4q0 from infinity to point P= $4U_0$

Answer 2

Answer:

Answer:

$4U_(0)$

Explanation:

V = u / q,

Work = P = V

1U / 1/4 = 4U

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1. A bicyclist starts at 2.5 m/s and accelerates along a straight path to a speed of 12.5 m/s ina time of 4.5 seconds. What is the bicyclist's acceleration to the nearest tenth of a m/s??

A girl weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm under her weight. Find the spring constant and the total work done on it during the compression.

Answers

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that $v = f * d$ where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?

Answers

Answer:

Δ KE = - 8.75 x 10⁻⁴ J

Explanation:

given,

mass of applesauce = 7 g = 0.007 Kg

initial velocity, u = 0.5 m/s

final velocity, v = 0 m/s

Decrease in kinetic energy = ?

initial kinetic energy

$KE_1=(1)/(2)mu^2$

$KE_1=(1)/(2)* 0.007 * 0.5^2$

KE₁ = 8.75 x 10⁻⁴ J

final kinetic energy

$KE_2=(1)/(2)mv^2$

$KE_2=(1)/(2)* 0.007 * 0^2$

KE₂ =0 J

Decrease in kinetic energy

Δ KE = KE₂ - KE₁

Δ KE = 0 - 8.75 x 10⁻⁴

Δ KE = - 8.75 x 10⁻⁴ J

decrease in kinetic energy of the applesauce is equal to 8.75 x 10⁻⁴ J

Final answer:

The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.

Explanation:

This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.

When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).

Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.

Learn more about Kinetic Energy here:

brainly.com/question/33783036

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The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?

Answers

Answer:

0.00098 N

Explanation:

The weight of an object is given by:

$W=mg$

where

m is the mass of the object

g is the gravitational acceleration on the planet

In this problem, we have:

$m=0.0001 kg$ is the mass of the honeybee

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting into the equation, we find:

$W=mg=(0.0001 kg)(9.8 m/s^2)=0.00098 N$

weight = mg
here m = 0.000100 g = 9.80
hence weight = 0.00980 kgm/s2

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta$

Where:

$\omega_(f)$ : is the final angular velocity = 0

$\omega_(0)$ : is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = (\tau)/(I)$

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

$I = (mr^(2))/(2)$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_(0)^(2) + 2\alpha*\theta$

$\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns$

I hope it helps you!

A piston above a liquid in a closed container has an area of 0.75m^2, and the piston carries a load of 200kg. What will be the external pressure on the upper surface of the liquid?

Answers

Answer:

2613.3 pa

Explanation:

p=F/A

p=ma/A

p=200×9.8/0.75

p=2613.3

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A screw can be considered a type of