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creativ13 [48]
3 years ago
15

A power source of 6.0 V is attached to the ends of a capacitor. The charge is 12 C.

Physics
1 answer:
brilliants [131]3 years ago
5 0
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Select the correct answer from each drop-down menu.
kirill115 [55]

Answer:

A

Explanation:

I think but I am sure

8 0
3 years ago
_____ motion occurs when tectonic plates move apart from each other. A. Divergent B. Transform C. Convergent
Likurg_2 [28]
A. Divergent (pretty sure)
4 0
3 years ago
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After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position
KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

W_p = KE_f - KE_i

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

KE_f = \frac{1}{2}mv_1^2

now work done is given as

W_p = \frac{1}{2}mv_1^2 - 0

so we can say

W_p = \frac{1}{2}mv_1^2

so above is the work done on the box to slide it from x1 to x2

3 0
3 years ago
A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is
Maru [420]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q  = -5.02×10^{-9}C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8×10^{-2}m) (21.8×10^{-2}m)  = 4.75×10^{-4}m²

A  = 4.75×10^{-4}m²

Surface charge density = Q/A

Surface charge density = (-5.02×10^{-9}C)/(4.75×10^{-4}m²)

Surface charge density = -1.057×10^{-5} Cm^{-2}

3 0
4 years ago
The spring of a toy car is wound by pushing the car
Zepler [3.9K]

The elastic potential energy stored in the car's spring during the process is 3.75 J

<h3>Determination of the spring constant</h3>

From the question given above, the following data were obtained:

  • Force (F) = 15 N
  • Extention (e) = 0.5 m
  • Spring constant (K) =?

K = F/e

K = 15 / 0.5

K = 30 N/m

<h3>Determination of the potential energy</h3>
  • Spring constant (K) = 30 N/m
  • Extention (e) = 0.5 m
  • Potential energy (PE) =?

PE = ½Ke²

PE = ½ × 30 × 0.5²

PE = 15 × 0.25

PE = 3.75 J

Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J

Learn more about energy stored in spring:

brainly.com/question/4280346

7 0
2 years ago
Read 2 more answers
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