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Nat2105 [25]
2 years ago
11

At a given location from a source charge, the electric field______________.a. strength is dependent on the source charge and the

charge that is at the given location. points away from the source charge, regardless of the sign of the source b. points away from the source charge, if the source charge is positive. c. strength is only dependent on the source charge.d. points toward the source charge, if the source charge is negative.
Physics
1 answer:
noname [10]2 years ago
3 0

Answer:

b. The electric field points away from the source charge, if the source charge is positive

d. The electric field points toward the source charge, if the source charge is negative.

Explanation:

A positive source charge would create an electric field that would exert a repulsive effect upon a positive test charge. Thus, the electric field vector would always be directed away from positively charged objects. On the other hand, a positive test charge would be attracted to a negative source charge. Therefore, electric field vectors are always directed towards negatively charged object.

Also electric field strength depends only on test charge

The correct options include b and d

The electric field points away from the source charge, if the source charge is positive.

Also, the electric field points toward the source charge, if the source charge is negative.

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What happens during the process of deposition
frez [133]
Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .
8 0
2 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5
Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0
3 years ago
How close does the proton get to the line of charge?
loris [4]

Answer:

12 cm

Explanation:

3 0
2 years ago
1. Calculate the momentum of each car before the collision: SHOW YOUR WORK!
a_sh-v [17]

Answer:

Momentum of red car = 5kgm/s

Momentum of blue car = 0kgm/s

Explanation:

Momentum = mass × velocity

For the red car

Mass = 1kg

Velocity = 5m/s

Momentum of the red car = 1kg × 5m/s

Momentum of the red car = 5kgm/s

For the blue car.

Mass = 1kg

Velocity = 0m/s(shows that the blue car is stationery)

Momentum = 1kg ×0m/s

Momentum of the blue car = 0kgm/s

3 0
2 years ago
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