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4 years ago

What is capacitance?

Physics

1 answer:

Otrada [13]4 years ago

5 0

Answer:

A, the amount of charge stored per volt

Explanation:

The equation for capacitance is

C = Q/V

therefore, it is charge per volt

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If air resistance can be neglected, how does the acceleration of a ball that has been tossed straight upward compare with its ac

timama [110]

They are the same. If this is all happening on Earth, then the ball's acceleration is 9.8 m/s^2 in either case. That's the acceleration of gravity around here.

6 0

3 years ago

Convert 500,000 inches into miles

olga nikolaevna [1]

Answer: 7.89141414

Explanation:

3 0

3 years ago

Read 2 more answers

In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room." If the room radius is 5.5 m, and the r

Crank

Answer:

0.181

Explanation:

We can convert the 0.5 rps into standard angular velocity unit rad/s knowing that each revolution is 2π:

ω = 0.5 rps = 0.5*2π = 3.14 rad/s

From here we can calculate the centripetal acceleration

$a_c = \omega^2r = 3.14^2*5.5 = 54.3 m/s^2$

Using Newton 2nd law we can calculate the centripetal force that pressing on the rider, as well as the reactive normal force:

$F = N = a_cm = 54.3 m$

Also the friction force and friction acceleration

$F_f = N\mu = 54.3 m \mu N$

$a_f = F_f / m = 54.3 \mu$

For the rider to not slide down, friction acceleration must win over gravitational acceleration g = 9.81 m/s2:

$g = a_f = 54.3 \mu$

$9.81 = 54.3 \mu$

$\mu = 9.81 / 54.3 = 0.181$

6 0

3 years ago

Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit aro

Sergeu [11.5K]

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

T² = ( $\frac{4\pi ^2}{ G M_s}$ a³ = K_s a³

Let's apply this equation to our case

a = $\sqrt[3]{ \frac{T^2}{K_s} }$

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

10² = K_s $a_{w}^3$

a_w = $\sqrt[3]{ \frac{100}{ K_s} }$

a_w = $\frac{1}{ \sqrt[3]{K_s} }$ 4.64

Planet X

a_x = $\sqrt[3]{ \frac{640^3}{K_s} }$

a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

a_y = $\sqrt[3]{ \frac{80^2}{K_s} }$

a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

a_z = $\sqrt[3]{ \frac{270^2}{K_s} }$

a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

3 0

3 years ago

If the cd rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

Katen [24]

The solution for this problem is:

500 revolution per minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s

Angular velocity ω = 2π N

Angular acceleration α= (ω2 - ω1) /t

ω2 = 0

α = - ω1/t = -2π N /t

N = 500 rpm = 8.33 r p s.

α = -2π 8.33 /2.6 =- 20 rad/s^2

3 0

4 years ago