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4 years ago

What are the three types of friction and when does each apply?

Physics

1 answer:

Anestetic [448]4 years ago

4 0

The force that opposes the motion of an object is called friction.

There are two kind of friction namely kinetic and static friction forces.

Kinetic friction is the force that is exerted on one surface by another when the two surfaces rub against each other because one or both surfaces are moving.

Static friction is the opposing force that acts against a secondary force that is trying to cause a stationery object to start moving. If there is no such force acting on the subject, the static friction is nil or zero.

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A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m

Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

$I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2$

The moment of inertia of the rod about its center is calculated as follows;

$I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2$

The initial rotational kinetic energy of the disk and rod;

$K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J$

The final rotational kinetic energy of the disk-rod system is calculated as follows;

$K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J$

The loss in rotational kinetic energy due to the collision is calculated as follows;

$\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J$

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

8 0

3 years ago

Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and the other has a mass of 52.0 kg. What is the gravit

algol [13]

Answer:

The gravitational force between them is $1.079\times10^{-7}\ N$ .

Explanation:

Given that,

Distance = 1.50 m

Mass of one student = 70.0 kg

Mass of other student = 52.0 kg

We need to calculate the gravitational force

Using formula of gravitational force

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁ = mass of one student

m₂ = mass of other studen

r = distance between them

Put the value into the formula

$F=\dfrac{6.67\times10^{-11}\times70.0\times52.0}{1.50^2}$

$F=1.079\times10^{-7}\ N$

Hence, The gravitational force between them is $1.079\times10^{-7}\ N$ .

5 0

4 years ago

A 25.0 kg object is held 8.50 m above the ground. Calculate its PE

pantera1 [17]

Answer:=mgh

Explanation:

6 0

3 years ago

A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

7 0

4 years ago

An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres

Paha777 [63]

A) The biggest astronomical object is the Universe, which contains billions of galaxies among which there is the Milky Way.
The Milky Way contains thousands of planetary systems, among which the Solar System.
The Solar System contains many <span>planets <span>(but only one star, the Sun)</span>,</span> among which there is Earth.
Therefore you can label:
A = Universe, B = Milky Way, C = Solar system, D = Earth

b) Given what we said before, you could label D also any other planet in the Solar System, therefore you can choose among Mercury, Venus, Mars, Jupiter, Saturn, Uranus, and Neptune.

8 0

4 years ago