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IrinaVladis [17]
3 years ago
11

What is newton's third law of physics?

Physics
1 answer:
kobusy [5.1K]3 years ago
7 0
Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
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A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en
Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient) 

W= f x d = uNd ; N=mg 
Delta E = 1/2 mV^2 -1/2mVi^2 

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term) 

This will then give us: 

1/2Vi^2-ugd = 1/2V^2 

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s 

 

6 0
3 years ago
If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice
Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

8 0
3 years ago
Why is the coefficient of friction independent of the mass?
Svet_ta [14]
Because the coefficient of friction depends on the surface
3 0
3 years ago
MATCH THESE ^-^ Match Newton's law with the correct statement.
ikadub [295]
1 and A
2 and B
3 and D
4 and C

4 0
3 years ago
Read 2 more answers
The y-position of a damped oscillator as a function of time is shown in the figure.
NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

<h3>at time, t = 0, y = 3.5</h3>

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

<h3>at time, t = 1 cm, y = - 3cm</h3>

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

<h3>at time, t = 2 cm, y = - 2cm</h3>

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0
2 years ago
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