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nadezda [96]
3 years ago
14

What are the applications of pascal's principle​

Physics
1 answer:
Murrr4er [49]3 years ago
5 0

Explanation:

  • The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
  • Hydraulic jack- used in the braking system of cars.
  • use of a straw- to suck fluids, which goes because of air pressure.
<h3>The question simply asks, where pressure can be applied. There are many others, such as <em><u>l</u></em><em><u>i</u></em><em><u>f</u></em><em><u>t</u></em><em><u> </u></em><em><u>p</u></em><em><u>u</u></em><em><u>m</u></em><em><u>p</u></em><em><u>.</u></em></h3>
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The process of physically separating a single-phase mixture into components if the components have different boiling points is c
kobusy [5.1K]
This process of physically separating a single phase mixture into several components would be termed simple distillation, I believe.

As it is mentioned that one is separating them based on their different boiling points.
6 0
3 years ago
The original magnitude of the force on the Q charge was F; what is the magnitude of the force on the Q charge now?
Otrada [13]

Complete Question:

Two small objects each with a net charge of Q (where Q is a positive number) exert a force of magnitude "F" on each other. We

replace one of the objects with another whose net charge is 4Q. The original magnitude of the force on the Q charge was "F"; what is the magnitude of the force on the Q charge now?

Answer:

4 F₀

Explanation:

Assuming that we can treat to both objects as point charges, we can find the force "F" that one charge exerts upon the other applying Coulomb´s law, as follows:

F₀ = K*Q₀² / r₁₂²

If we replace one of the charges by one with a 4Q₀ charge, the new value of F will be as follows:

F₁ = K*Q₀*4Q₀ / r₁₂² =( K*Q₀² / r₁₂²)* 4 = 4* F₀

This value is reasonable, as the electrostatic force is a linear - type one, so it is possible to use the superposition principle (we can get the force exerted by one charge on another without considering the ones due to another charges)

3 0
3 years ago
A 15kg dog jumps out of a 40kg canoe. If the dogs velocity is 1.2m/s, what is the velocity of the canoe?
vivado [14]

Answer:

v = -0.45 m/s

Explanation:

Assuming the canoe was initially at rest with momentum L = 0

and that the dog's velocity is in the positive direction

conservation of momentum

0 = 15(1.2) + 40v

v = -0.45 m/s

6 0
3 years ago
A gyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 rev/min.
SIZIF [17.4K]

Answer:

Part a)

a_t = 0.423 m/s^2

Part b)

a_c = 2113 m/s^2

Part c)

d = 80 m

Explanation:

Part a)

as we know that angular acceleration of the wheel is given as

\alpha = 13.2 rad/s^2

now the radius of the wheel is given as

R = 3.21 cm

so the tangential acceleration is given as

a_t = R\alpha

a_t = (0.0321)(13.2)

a_t = 0.423 m/s^2

Part b)

frequency of the wheel at maximum speed is given as

f = 2450 rev/min

f = \frac{2450}{60} = 40.8 rev/s

now we know that

\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s

now radial acceleration is given as

a_c = \omega^2 r

a_c = (256.56)^2(0.0321) = 2113 m/s^2

Part c)

total angular displacement of the point on rim is given as

\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2

here we know that

\omega = \omega_0 + \alpha t

256.56 = 0 + 13.2 t

t = 19.4 s

now angular displacement will be

\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2

\Delta \theta = 2493.3 rad

now the distance moved by the point on the rim is given as

d = R\theta

d = (0.0321)(2493.3)

d = 80 m

7 0
3 years ago
7. A car stops at a red light. The light turns green
Readme [11.4K]

Answer:

Graph C

Explanation:

This is the answer because it is the only one that shows the vehicle accelerate to a constant speed before stopping and slowing down.

8 0
2 years ago
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