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Sergio039 [100]
3 years ago
12

The available source of charge that pushes a charge through a circuit is

Physics
1 answer:
sergey [27]3 years ago
8 0
     Voltage, because the voltage applied to a resistor generates a current, which is the load flow. 

Letter C

If you notice any mistake in my english, please let me know, because i am not native.
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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo
vaieri [72.5K]
Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement =  2m + (-3)m.               Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement  is 1m behind his original starting point.
4 0
3 years ago
Elements that easily transmit electricity and heat display the property know as ______
andre [41]
It is known as conductivity
7 0
3 years ago
Read 2 more answers
20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu
Anna71 [15]

Answer:

B. V_{f}= 10\,cubic\,inches

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

PV=nRT

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

P_{i}V_{i}=n_{i}RT_{i} (1)

P_{f}V_{f}=n_{f}RT_{f} (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because T_{i}=T_{f}, n_{i}=n_{f} and R is an universal constant:

P_{i}V_{i}= P_{f}V_{f}, solving for V_{f}

V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}

V_{f}= 10 cubic\,inches

6 0
3 years ago
A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 m. The particle experiences a constant
kirill [66]

Answer:

Part a)

F = 6 \times 10^{-3} N

Direction of force is along the motion of charge

Part b)

E = 4000 N/C

direction of electric field is along the direction of motion

Explanation:

Part a)

As we know that the change in electric potential energy is equal to the work done by electric field

W = EPE_A - EPE_B

W = 9.0 \times 10^{-4} J

now from the equation of work done we know that

W = F.d

(9.0 \times 10^{-4}) = F(0.15)

F = 6 \times 10^{-3} N

Direction of force is along the motion of charge

Part b)

As we know the relation between electrostatic force and electric field given as

F = qE

(6 \times 10^{-3}) = 1.5 \times 10^{-6} E

E = 4000 N/C

direction of electric field is along the direction of motion

8 0
3 years ago
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