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2 years ago

What is static friction?

Physics

1 answer:

givi [52]2 years ago

7 0

Answer:

Plzzzzzzzzzzzzzzzz brainliest

Explanation:

In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.

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Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

Rina8888 [55]

<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>

8 0

3 years ago

Read 2 more answers

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

when energy is transferred from one part of a system to another, some of the energy is lost during the transfer and cannot be us

AlladinOne [14]

Sentences A and D describe examples of energy transformation.Heat is produced when a car's tires rub against the pavement and as electricity passes across power wires, they become hotter.

<h3>What is the law of conservation of energy?</h3>

According to the law of conservation of energy, the energy of an isolated system stays unchanged throughout time.it is said to be conserved.

Energy cannot be created nor destroyed and can be transferred from one form to the other form.

The complete question is

"When energy is moved from one component of a system to another, some of the energy is lost and cannot be used as planned.

Which two statements provide examples?

A. Friction between a car's tires and the road produces heat.

B. Sunlight strikes a solar panel, generating electricity.

c. Stereo speakers emit a sound when powered by electrical energy,

D. Wind moves a turbine, generating electricity.

I E. Power lines heat up as electricity flows through them."

Some of the energy wasted during the movement of energy from one section of a system to another is heat is produced by friction between a car's tires and the road and as electricity passes via power lines, they heat up.

Hence, sentences A and D describe examples of energy transformation.

To learn more about the law of conservation of energy refer to the link;

brainly.com/question/2137260

#SPJ1

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2 years ago

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$