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morpeh [17]
1 year ago
12

You are moving at a speed 2/3 c toward randy when randy shines a light toward you. at what speed do you see the light approachin

g you
Physics
1 answer:
yarga [219]1 year ago
3 0
I see the light moving exactly at speed equal to c.

In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
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A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
Mekhanik [1.2K]
The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
7 0
1 year ago
Substance X is placed in a container with substance Y. Both substances are fluids. Substance X initially sinks to the bottom of
Brut [27]

Answer: Option (A) is the correct answer.

Explanation:

Convection is a process in which heat transfers from a hotter substance to a colder substance.

As a result, the substance which is less dense will rise and the more denser substance will sink due to the influence of gravity.

Thus, we can conclude that in the given situation substance X will rise due to convection.

3 0
2 years ago
Read 2 more answers
A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri
liubo4ka [24]

Answer:

T = 7.64 kN

F_y = 0.52 kN(Downwards)

F_x = 3.23 kN (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

F(4.50 cos\theta) + mg(2cos\theta) = T \times 3

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T

T = 7.64 kN

Now force balance in vertical direction

F + mg = Tsin65 + F_y

5 + 1.40 = 7.64 sin65 + F_y

F_y = 0.52 kN(Downwards)

Force balance in horizontal direction

F_x = T cos65

F_x = 7.64 cos65

F_x = 3.23 kN (Towards Left)

7 0
2 years ago
A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i
lukranit [14]

Answer:

6n!!!!!!!!!!!!!!!!!!

Explanation:

nnnn

8 0
1 year ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
2 years ago
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