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3 years ago

ADvantage of friction

Physics

1 answer:

kondor19780726 [428]3 years ago

7 0

Answer:

1. Friction enables us to walk freely.

2. It helps to support ladder against wall.

3. It becomes possible to transfer one form of energy to another.

4. Objects can be piled up without slipping.

You might be interested in

Power is the product of_____.

tia_tia [17]

Heya user☺☺

All options are wrong here.

The correct answer is..

Work/Time.

Hope this will help☺☺

3 0

3 years ago

Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?

Talja [164]

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

$\frac{10N}{50kg}=a$

A newton is just $\frac{kg·m}{s^{2}}$

$a=\frac{\frac{10kg·m}{s^{2}}}{50kg}$

The s^2 and 50 kg you multiply

$a=\frac{10kg·m}{50kg·s^{2}}$

The kg's cancel and 10/50 is 1/5

$\frac{1}{5}·\frac{m}{s^{2}}$

So the acceleration is 1/5 m/s^2

3 0

3 years ago

Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar

den301095 [7]

Answer:

electric field Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

a) Let's write the electric field for each charge and the total field

E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

Et = E1 + E2 + E3

E1 = k q / (x-a)²

E2 = k (-2q) / x²

E3 = k q / (x + a)²

Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ where x << 1, for this we take factor like x from all the equations

Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

(1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

(1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0

3 years ago

Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the

inna [77]

Answer:

The Internal energy of the gas did not change

Explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W ------ ( 1 )

Δ U = change in internal energy

Q = heat added

W = work done

since Q = W. the value of ΔU will be = zero i.e. No change

4 0

3 years ago

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

Oxana [17]

Answer:

A) $I_{total}$ = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = $I_{cm}$ + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

$I_{total}$ = $I_{body}$ + 2 $I_{arm}$

$I_{body}$ = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

$I_{arm}$ = M D²

Let's write the equation

$I_{total}$ = ½ M R² + 2 (m D²)

Let's calculate

$I_{total}$ = ½ 56 0.20² + 2 4 0.20²

$I_{total}$ = 1.12 + 0.32

$I_{total}$ = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0

3 years ago