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Makovka662 [10]
3 years ago
12

A 650 × 10–4 F capacitor stores 24 × 10–3 of charge. What is the potential difference between the plates?

Physics
1 answer:
olasank [31]3 years ago
8 0

Answer:

0.369 V

<u>Explanation:</u>

Given :

Capacitance ( c ) = 650 × 10–4 F

Charge ( q ) = 24 × 10–3 C

We are asked to find potential difference ( v )!

We know:

q = c v

= > v = q / c

Putting values here we get:

= > v = ( 24 × 10–3 ) / ( 650 × 10–4 ) V

= > v = 240 / 650 V

= > v = 24 / 65 V

= > v = 0.369 V

Therefore, potential difference between the plates is 0.369 V

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The motorbike reaches 100 km/h in 3.5 seconds

Explanation:

The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the motorbike in this problem,

u = 0 (it starts from rest)

v = 100 km/h = 27.8 m/s is the final velocity

a=7.9 m/s^2 is the acceleration

Solving for t, we find the time it takes for the bike to reach that velocity:

t=\frac{v-u}{a}=\frac{27.8-0}{7.9}=3.5 s

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Particle accelerators make it easy for scientists to study what?
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1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's
sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth  = 5.97 x 10²⁴kg

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Unknown:

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Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

   F = \frac{Gm_{1} m_{2}  }{r^{2} }

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is  the distance

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During the time interval from 0.0 to 10.0 s, the position vector of a car on a road is given by x(t) = a + bt + ct2, with a = 17
Juli2301 [7.4K]

The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

<h3>Average velocity of the car</h3>

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

<h3>Average velocity</h3>

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1  + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

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B. ice and d basalt............
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