What term is used to describe how well liquids flow

Answers

Answer 1
Answer: viscosity is one term you could use
Answer 2
Answer: well water or any  liquid  actually takes the form of the cup of bowl or whatever you put the liquid in
REMEMBER- Liquid has no shape.
hope this helps!!!!!!!!

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When the two pistons have different surface areas, how will the pressure exerted on them compare?This is in a hydraulic system. Pascals principle
will the pressure be the same on both or will one be greater

Answers


Pascal's principle says that the pressure is the same
everywhere in a confined fluid, no matter how much
or how little area you're looking at.

THAT's why hydraulic systems work ... because the
SAME pressure over MORE area adds up to MORE force.

A weightlifter lifts a 1400 N barbell 2.5 meters. Calculate the work done during the lift

Answers

Answer: W = 3500 J

Explanation: Work is the product of the amount of force and the perpendicular distance or W = Fd.

W = Fd

   = 1400 N ( 2.5 m )

   = 3500 J

Answer:

Force = 1400N

Displacement = 2.5 m

work done = F×D

=1400×2.5

= 3500 joule

1. The force between a pair of charges is 100 newtons. The distance between the charges is 0.01 meter. If one of the charges is 2 u 10-10 C, what is the strength of the other charge?2. The force between two charges is 2 newtons. The distance between the charges is 2 u 10-4 m. If one of the charges is 3 u 10-6 C, what is the strength of the other charge?

Answers

Answer:

1. Q = 5.56 * 10^(-3) C

2.Q = 2.97 * 10^(-12) C

Explanation:

1. Electrostatic force is given as:

F = (kqQ)/(r^2)

where k = Coulombs constant

q = charge of first charge

Q = charge of the second charge

r = distance between them

From the question:

F = 100 N

q  = 2 * 10^(-10) C

r = 0.01 m

We need to find Q.

From the formula of force, we have that Q is:

Q = (Fr^2)/(kq)

Q = (100 * 0.01^2)/(8.99 * 10^9 *2 * 10^(-10)) \n\n\nQ = (0.01)/(1.798)\n \n\nQ = 0.00556 C = 5.56 * 10^(-3) C

This is the charge, Q, of the second charge.

2. From the question:

F = 2 N

q  = 3 * 10^(-6) C

r = 2 *10^(-4) m

We need to find Q.

Using the same formula for Q as in 1. above, we have that:

Q = (2 * (2 *10^(-4))^2)/(8.99 * 10^9 *3 * 10^(-6)) \n\n\nQ = (8 * 10^(-8))/(26970)\n \n\nQ = 2.97 * 10^(-12) C

This is the charge, Q, of the second charge.

Answer:

(1)  4.4835*10^(17)C

(2)2.9748*10^(-12)C

Explanation:

Force Between two charges is give by.

F=k(Q_1 *Q_2)/(r^2) , here k is called coulomb constant and has value = 8.967*10^9Nm^2/C.

(1) case, F =100N, r = 0.1m and Q_1=2*10^(-10)C substituting these values in above equation and solving for unknown gives us.

Q_2 = 4.4835*10^(17)C.

(2) Case, F = 2N, r = 2*10^(-4)m and Q_1=3*10^(-6)C.

again by substituting these in above equation and solving for unknown gives us.

Q_2=2.9748*10^(-12)C.

The gravity on earth is 6 times greater than the gravity on the moon. An object weighing 2 lbs. on the moon will weigh:

Answers

The answer would be 6 • 2= 12

Which describes an image that can be produced by a concave lens?The image is real and smaller than the object.
The image is virtual and larger than the object.
The image is virtual and smaller than the object.
The image is real and larger than the object.

Answers

Answer:

The image is virtual and smaller than the object.

Explanation:

A concave lens is a diverging lens, meaning that it spreads out light rays that have been refracted through it. A concave lens is thinner at its center than at its edges.

It always produces virtual images because the image appears to form on the same side of the lens where the object is placed.

The image is always smaller than the object. Far the object, smaller will be its image.

A light ray diagram of a concave lens is attached for a better understanding.

Answer: The image is virtual and smaller than the object.

Explanation:

A student pushes a 40-N block across the floor for a distance of 10 m. How much work was applied to move the block? A.
4 J

B.
40 J

C.
400 J

D.
4000 J

Answers

The correct answer to the question is C) 400 J.

CALCULATION:

As per the question, the force applied on the block F = 40 N.

The distance travelled by the block S = 10 m.

We are asked to calculate the work done by the block.

The work done by a body under a force F is calculated as -

             Work done W = \vec F.\vec S

                                     = FScos\theta

Here, \theta is the angle between force and displacement.

If \theta =\ 0^(0), then work done W = FScos0

                                                   = FS.     [ cos0 = 1]

Hence, the work done by the block W = FS

                                                                = 40 N × 10 m

                                                                =  400 Joule.         [ans]

Here, Joule is the unit of work.

Hence, the work done by the block s 400 Joule.