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2 years ago

What factors do NOT affect friction between two objects? Explain how you know this.

Physics

1 answer:

Black_prince [1.1K]2 years ago

4 0

The frictional force between two bodies depends mainly on three factors: (I) the adhesion between body surfaces

(ii) roughness of the surface

(iii) deformation of bodies.

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How do computer get input

Vitek1552 [10]

an input device is (a piece of computer hardware equipment) used to provide data and control signals to an information processing system such as acomputer or information appliance.

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2 years ago

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An oil gusher shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. Neglecting air resistance but not the

Mariana [72]

Answer:3764.282 KPa

Explanation:

Given gusher shoots oil at h=25 m

i.e. the velocity of jet is

v=\sqrt{2gh}[/tex]

v=22.147 m/s

Now the pressure loss in pipe is given by hagen poiseuille equation

$\Delta P=\frac{32L\mu v}{D^2}$

$\Delta P=\frac{32\times 50\times 22.147\times 1}{10^{-2}}$

$\Delta P=3543.557 KPa$

For 25 m head in terms of Pressure

$\Delta P_2=\rho \times g\times h=220.725 KPa$

Total Pressure= $\Delta P+\Delta P_2$ =3543.557+220.725=3764.282 KPa

4 0

3 years ago

What will the pressure of gas be if the temperature rises to 87°C

weeeeeb [17]

Answer: 0 degrees Celsius or 32 degrees Fahrehit

Explanation:

3 0

2 years ago

1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)

cupoosta [38]

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b) x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

5 0

2 years ago

A young parent is dragging a 65 kg (640 N) sled (this includes the mass of two kids) across some snow on flat ground, by means o

Delicious77 [7]

Answer:

b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247

Explanation:

a) In the attachment we can see the free body diagram of the system

b) Let's write Newton's second law on the y-axis

N + T_y -W = 0

N = W -T_y

let's use trigonometry for tension

sin θ = T_y / T

cos θ = Tₓ / T

T_y = T sin θ

Tₓ = T cos θ

we substitute

N = W - T sin 30

we calculate

N = 640 - 160 sin 30

N = 560 N

c) as the system goes at constant speed the acceleration is zero

X axis

Tₓ - fr = 0

Tₓ = fr

we substitute and calculate

fr = 160 cos 30

fr = 138.56 N

d) the friction force has the formula

fr = μ N

μ = fr / N

we calculate

μ = 138.56 / 560

μ = 0.247

4 0

2 years ago