All categories

3 years ago

Which law of motion describes squeal and oppisit forces of action and reaction

1 answer:

4 0

These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, 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.

Hope this helps! :)

You might be interested in

Plz answer this asap first answer gets brainliest and 100 points

Ratling [72]

The answer is C. From the yolk to the white.

4 0

3 years ago

Read 2 more answers

Following are the different layers of the Sun's atmosphere. Rank them based on the order in which a probe would encounter them w

Ghella [55]

Answer:

Sequence of layers to encounter while travelling from Earth to the surface of the sun are:

Option (A) - Corona

Option (C) - Chromosphere

Option (B) - Photosphere

Explanation:

Corona- It represents the extreme outer region of the surface of the sun. It is normally not visible because of the light emitted from the sun. So it is difficult to be seen from the naked eye, but it can be seen during the time of total solar eclipse. The temperature in this region is about several million degrees.

Chromosphere- It is a layer that lies between the Corona and the Photosphere. This layer has a thickness of about 2000 kilometers and the temperature in this layer ranges from about 6000°C to 20,000°C.

Photosphere- This layer is the bright visible layer of the sun which is comprised of plasma and dark and cool sunspots, that forms on the sun due to the emerging of the magnetic field of the sun from its surface.

Thus, the correct sequences are mentioned above.

3 0

3 years ago

PHYSICS CIRCUIT QUESTION PLEASE HELP!! 20 Points!

dimulka [17.4K]

This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.

If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.

The energy dissipated by the circuit is the energy delivered by
the battery. We'd know what that is if we knew I₁ . Everything that
flows in this circuit has to go through R₁ , so let's find I₁ first.

-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
-- I₁ is 10volts/10Ω = 1 Ampere.

-- R1: 1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts.
    7 of the 10 appear across R₁ .
   So the other 3 volts appear across all the business at the bottom.

-- R₂: 3 volts across it = V₂.
   Current through it is I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4: 6Ω in the series combination
   3 volts across it
   Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere

-- Remember that the current is the same at every point in
a series circuit. I₃ and I₄ must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.

-- R₃: 1/2 Ampere through it = I₃ .
   1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄: 1/2 Ampere through it = I₄
   1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that I₂ is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches I₁ coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.

Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up. But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.

The power supplied by the battery = (voltage) · (current)

   = (10 volts) · (1 Amp) = 10 watts .

"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.

   (30 minutes) · (60 sec/minute) = 1,800 seconds

   (10 joules/second) · (1,800 seconds) = 18,000 joules in 30 min

The power (joules per second) dissipated by each individual resistor is

   P = V² / R
   or
   P = I² · R ,

whichever one you prefer. They're both true.

If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total.

They're not asking for that. But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.

7 0

3 years ago

True or False – All transformations fit the Law of Conservation of Energy.

Nataly [62]

Answer:

The correct answer is true.

5 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$