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lorasvet [3.4K]

3 years ago

5

7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca

pacitors?

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

q = C V charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor

N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors

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What is an non-example of a switch

loris [4]

Dont really understand what youre asking but switches can be used to turn things on and off.

3 0

3 years ago

An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

7 0

3 years ago

A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient

Basile [38]

Answer:

v_o = 4.54 m/s

Explanation:

<u>Knowns </u>

From equation, the work done on an object by a constant force F is given by:

W = (F cos Ф)S (1)

Where S is the displacement and Ф is the angle between the force and the displacement.

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:

K.E=1/2m*v^2 (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:

W = K.E_f-K.E_o (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:

∑F_y=N-mg

N=mg

Thus, the kinetic friction force is:

f_k = μ_k*N

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:

W_f=(f_k*cos(180) s

=-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:

W= -K.E_o

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force

W_f= -K.E_o

From equation (2), the work done by the friction force in terms of the initial speed is:

W_f=-1/2m*v^2

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:

-μ_k*mg*s = -1/2m*v^2

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s

v_o = 4.54 m/s

6 0

3 years ago

Read 2 more answers

Help with speed problems #1-3

-Dominant- [34]

1. Each plot represents the meters traveled by both the Hare and the Tortoise over a certain period of time (minutes).

2. The Tortoise lines show it lines is steadily increasing over a period of time. So as more time elapses the faster the tortoise becomes it travels more meters. The Tortoise line shows steady acceleration.

3. The Hare in the first 5 minutes had a rapid fast advancement up to 40 meters. But for the 5-20 mins. period the Hare did not move at all. Its speed stayed at the same place. But towards the end 20-25 mins. marks the Hare started moving again. At the end the Hare at first had a rapid acceleration but stopped for a long time then it sped up briefly.

8 0

3 years ago

In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

Aleonysh [2.5K]

Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light $I$ = 2500 W/m^2

speed of light $c$ <u> </u>= 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

$P_{abs}$ = $I/c$ = 2500/(3 x 10^8) = <em>8.33 x 10^-6 Pa</em>

b) 1 atm = 101325 Pa

$P_{abs}$ = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

$P_{ref}$ = $2I/c$ = twice the value for totally absorbing

$P_{ref}$ = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d) 1 atm = 101325 Pa

$P_{ref}$ = 2 x 8.23 x 10^-11 = <em>1.65 x 10^-10 atm</em>

8 0

3 years ago