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3 years ago

State the difference between viscosity and friction

Physics

2 answers:

zhuklara [117]3 years ago

8 0

The answer above me is correct

spin [16.1K]3 years ago

4 0

Answer:

The main difference between friction and viscosity is that friction is used to refer to forces that resist relative motion, in general, whereas viscosity refers specifically to resistive forces that occur between layers of a fluid when fluids attempt to flow.

Explanation:

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Salt is now added to the water in the bucket, increasing the density of the liquid. what happens to the tension in the string ?

AlexFokin [52]

The question involves a ping-pong ball that is held submerged in a bucket by a string attached to the bottom of the bucket.

The answer is the tension of the string will increase. This is because making the water salty increases its density, and consequently, increases its buoyancy. This is why sea water is more buoyant than fresh water. Therefore the ping pong is pushed more upwards by the water when salt is added than initially. This gives the string more tension.

5 0

3 years ago

Without fluid friction, all objects accelerate at?

lys-0071 [83]

Answer:

Acceleration of gravity= $9.8m/s/s$

Explanation:

Newton's Second Law-acceleration is proportional to the net force acting on an object.

All objects usually free fall at the same acceleration of $9.8m/s/s$ -this regardless of their mass. This acceleration is known as acceleration of gravity.

7 0

3 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

4 years ago

HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$