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2 years ago

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In an adiabatic process, the greater the compression of a gas, the greater the pressure of that gas. Based on this information,

which graph best shows the relationship between the pressure and temperature of a gas?

2 answers:

11Alexandr11 [23.1K]2 years ago

8 0

the answer is A the diagram that is

Maslowich2 years ago

6 0

Answer:

A.)

Explanation:

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Sphere A has a charge of +2 x 10 to the power of -6 coulomb and is brought into contact with a similar sphere, B which has a cha

Setler [38]

When sphere A and B are brought in contact and separated, charge on each sphere becomes [2x10^-6 + (-4x10^-6)]/ 2 = -1x10^-6 C.

That is, charge is equally separated and is the average of charges on both spheres. The reason behind equal charge on both spheres after separation is, when they are kept in contact, their potential difference becomes same.

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2 years ago

Which statement best summarizes the central idea of “Applications of Newton’s Laws”?

Shkiper50 [21]

Newton's three forces, normal, tension and friction, are present in a surprising number of physical situations

Newton's Laws, that describe the relationship between an obejct and the forces acting upon it, apply in almost every physical situation, from quantum mechanics to electricity.

The correct answer is:

Newton’s laws can explain the forces that occur between objects every day

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2 years ago

Read 2 more answers

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0

2 years ago

A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?

Lena [83]

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω = 11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

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2 years ago

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0

2 years ago