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2 years ago

8

A small, uncharged metal sphere is placed near a larger, negatively charged sphere. Which diagram best represents the charge dis

tribution on the smaller sphere

1 answer:

Rudik [331]2 years ago

3 0

Answer:

I'm sorry I don't know what is the answer

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*NEED THIS ANSWERED ASAP*

Musya8 [376]

Answer:

When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero

Explanation:

plato

4 0

2 years ago

A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?

ELEN [110]

Answer:

$\boxed {\boxed {\sf 50 \ kg}}$

Explanation:

We are asked to find the mass of a cabinet, given the force and acceleration. According to Newton's Second Law of Motion, force is the product of mass and acceleration. The formula for this is:

$F= m \times a$

The force is 200 Newtons, but we should convert the units to make unit cancellation easier. 1 Newton is equal to 1 kilogram meter per second squared, so the force of 200 Newtons is 200 kilogram meters per second squared.

The mass is unknown and the acceleration is 4 meters per second per second or 4 meters per second squared.

F= 200 kg*m/s²
a= 4 m/s²

Substitute the values into the formula.

$200 \ kg *m/s^2 = m \times 4 \ m/s^2$

We are solving for the mass, m, so we must isolate the variable. It is being multiplied by 4 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 4 m/s²

$\frac {200 \ kg *m/s^2}{4 \ m/s^2}= \frac{m \times 4\ m/s^2}{4 \ m/s^2}$

$\frac {200 \ kg *m/s^2}{4 \ m/s^2} =m$

The units of meters per second squared cancel.

$\frac {200 \ kg }{4 }=m$

$50 \ kg =m$

The mass of the cabinet is <u>50 kilograms.</u>

5 0

2 years ago

Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle

Vinil7 [7]

Answer:

1/9 of that just outside the smaller sphere

Explanation:

The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is

$E_0=k\frac{Q}{R^2}$

The second sphere has a radius which is 3 times that of the smaller sphere:

$R'=3R$

So, the electric field just outside the second sphere is

$E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E_0}{9}$

So, the correct answer is 1/9.

7 0

2 years ago

1. Calculate the total heat content of 10 kg of ice at -23°C.?

vesna_86 [32]

To solve the problem, we must know the heat capacity of ice and water.

For Cp = 2090 J/kg C

H = mCpT

H = (10 kg) ( 2090 J/ Kg C) ( -23 C)

H = - 480700 J

For water Cp = 4180 j/kg C

H = (100 kg) ( 4180 J/kg C) ( 60 C)

<span>H = 2508000 J</span>

8 0

2 years ago

In one hour, coal supplies 500 000 J of energy. The energy amounts to 380 000 J. How much useful energy is produced in one hour?

Debora [2.8K]

Answer:

120,000J

Corrected question;

In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?

Explanation:

Given;

Total energy Et = 500,000 J

Wasted Energy Ew = 380,000J

The amount useful energy is the amount of energy that is available for supply.

This can be derived by subtracting the wasted energy from the total energy.

Useful energy = Total Energy - wasted energy

Eu = Et - Ew

Substituting the given values;

Eu = 500,000J - 380,000

Eu = 120,000 J

The amount of useful energy produced in one hour is 120,000 J

5 0

2 years ago