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Vedmedyk [2.9K]
3 years ago
5

Charge Q = +4.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the po

tential difference between the center of the sphere and the surface of the sphere?
Physics
1 answer:
sattari [20]3 years ago
8 0

Answer:

3.6 x 10^8 V

Explanation:

Q = 4 m C = 4 x 10^-3 C

r = 5 cm = 0.05 m

The formula for the potential at the surface is

Vs = K Q / r = (9 x 10^9 x 4 x 10^-3) / 0.05 = 7.2 x 10^8 V

The formula for the potential at the centre is

Vc = 3/2 Vs

Vc = 1.5 x 7.2 x 10^8 V = 10.8 x 10^8 V

The difference in potential is

V = Vc - Vs = 10.8 x 10^8 - 7.2 x 10^8 = 3.6 x 10^8 V

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
nikitadnepr [17]

The radius of curvature of the proton's path while in the field is 66.67  × 10^{-2}.

b) Let R = radius curvature of protons path. Then,

relation b/w B, R, and v is: -

B = mv/eR\\R=mv/eB

R=\frac{1.6*10^{-27} * 20*10^{6}}{1.6*10^{-19}*0.3 }

R =66.67× 10^{-2}

Hence, the radius of curvature of the proton's path while in the field is 66.67 × 10^{-2}.

<h3>What do you mean by Magnetic field?</h3>

The magnetic influence on moving electric charges, electric currents and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field.  The magnetic field of a permanent magnet pulls on ferromagnetic substances like iron and attracts or repels other magnets. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilized in electromagnets, and electric fields that change in time produce magnetic fields that surround magnetized things.

To know more about Magnetic Field visit:

brainly.com/question/14848188

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5 0
2 years ago
How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

___________________________________________

coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

8 0
3 years ago
A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose
dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m  { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0
3 years ago
A wave has the wavelength equal to 396 nm. what must be the frequency of this wave
Anvisha [2.4K]
This assumes that the wave has velocity c (is light).

7 0
3 years ago
Consider a balloon of mass 0.030kg being inflated with a gas of density 0.54kg/m. What will be the volume of the balloon when it
Nonamiya [84]

the weight of the balloon is .030 * 10 = 0.3 N

the weight of the gas of volume v is 0.54*10 N

The lifting force of a volume of v m³ of displaced air is 1.29v N

so, we need

1.29*10*v = 0.3 + 0.54*10*v

or

1.29v = 0.03+0.54v

7 0
1 year ago
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