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kenny6666 [7]
3 years ago
7

Two capacitors C1 and C2 are charged to 120V and 200V respectively. It is found that by connecting them together the potential o

n each one can be made zero. Then what is the relation between C1 and C2
Physics
1 answer:
Reil [10]3 years ago
6 0
V1=q1/C1. V2=q2/C2. If the potentials cancel then the charge, q, was equal. q1=q2. Therefore V1*C1=V2*C2, or
C1/C2=V2/V1=200/120=1.667
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Ear "popping" is an unpleasant phenomenon sometimes experienced when a change in pressure occurs, for example,in an airplane. If
marshall27 [118]

Answer:

H = 0.00058m = 0.58mm of Mercury

<em><u>h = 160m for popping at 8000m</u></em>

Explanation:

Air density decreases with altitude (increasing height).

At sea level air density = 1.22kg/m^3

while at 3000m above it is approximately 0.8 kg/m^3

ASSUMPTION:

we neglect the change in density across 100m vertically

density of mercury = 13600kg/m ^3

we equate the pressure for mercury and air

AIR(pgh) = Mercury(pgh)

0.8 * g * 100 = 13600 * g * H

H = 0.00058m = 0.58mm of Mercury

At even farther heights, the air density further drops

at 8000m above it is 0.52kg/ m^3  (source:https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html)

***for the ear to pop again, it has to experience same change in pressure as of at 3000m

Change in P for AIR at 3000m = Change in P for AIR at 8000m

ASSUMPTION : gravitational acceleration does not change with altitude

pgh = pgh

0.8 * g * 100 = .5 * g *<em><u>h</u></em>

<em><u>h = 160m</u></em>

3 0
4 years ago
A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k
Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

7 0
3 years ago
N 1800kg car has an<br> of 3.8m/s? What is it<br> on the car?<br> acceleration<br> force acting
nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

a=3.8 m/s^2

Substituting,

F=(1800)(3.8)=6840 N

7 0
3 years ago
Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . At a certain instant an object is
anastassius [24]

Answer:

After 2 seconds the velocity of the object is 0 m/s.

Explanation:

The velocity at 2 seconds later can be calculated as follows:

V_{f} = V_{0} - gt

Where:

V_{f}: is the velocity at 2 seconds

V_{0}: is the initial velocity = 20 m/s

g: is the gravity = 10 m/s²

t: is the time = 2 s

Hence, the final velocity is:

V_{f} = V_{0} - gt = 20 m/s - 10 m/s^{2}*2 s = 0 m/s

This value (0 m/s) means that the object has reached the maximum height.

Therefore, after 2 seconds the velocity of the object is 0 m/s.

I hope it helps you!

6 0
3 years ago
If you throw a ball up with a velocity of 7 m/s, how long will it take for the ball to reach the top of its path?
nirvana33 [79]

Answer:

c. 0.71 [s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -g*t

where:

Vf = final velocity = 0 (because when the ball reaches the top, there is no movement)

Vo = initial velocity = 7 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: The negative sign of the equation means that the movement is againts the direction of the gravity acceleration.

0 = 7 - (9.81*t)\\t = 0.713 [s]

4 0
3 years ago
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