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Lostsunrise [7]
3 years ago
11

When energy decreases, particle motion: A. stops B. increases C. decreases

Physics
2 answers:
DaniilM [7]3 years ago
7 0
C. decreases here you go
bezimeni [28]3 years ago
5 0
Hi lovely,

The answer you're looking for would be C. Decreases

Whenever energy decreases/increases; Particle motion will do the same.
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xxMikexx [17]

Let's use Newton's 2nd law of motion:

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2 years ago
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What type of motion does the galaxy go through
Mariana [72]

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8 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
What are some of the forces that are found in skateboarding?
Dominik [7]
Normal force, friction force, gravitational force
7 0
3 years ago
you push a sled of mass 15 kg across the snow with a force of 180 N for a distance of 2.5 m. There is no friction. if the sled s
Oxana [17]
<span>7.7 m/s First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so 180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2 Now determine how long you pushed it. For constant acceleration the equation is d = 0.5 A T^2 Substitute the known values getting, 2.5 m = 0.5 12 m/s^2 T^2 2.5 m = 6 m/s^2 T^2 Solve for T 2.5 m = 6 m/s^2 T^2 0.41667 s^2 = T^2 0.645497224 s = T Now to get the velocity, multiply the time by the acceleration, giving 0.645497224 s * 12 m/s^2 = 7.745966692 m/s After rounding to 2 significant figures, you get 7.7 m/s</span>
5 0
3 years ago
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