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miskamm [114]
4 years ago
14

Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of sp

aceship 1 relative to spaceship 2?
Physics
1 answer:
Studentka2010 [4]4 years ago
8 0

Answer:

The relative speed of 1 relative to 2 is 0.88c

Explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

v_{ab}=\frac{v_{a}+v_{b}}{1+\frac{v_{a}v_{b}}{c^{2}}}

Since it is given that

v_{a}=0.6c\\\\v_{b}=0.6c

Applying values in the formula we get

v_{ab}=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}\\\\v_{ab}=0.88c

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What are (a) the kinetic energy, (b) the rest energy, and (c) the total energy of a 1.50 g particle with a speed of 0.600 c ?
MAVERICK [17]
Kinetic energy = 1/2 m v^2 = 1/2 x1.5 x10^-3 x 0.36 
5 0
3 years ago
13
gladu [14]

v = u + at

50 = 0 + a*10

50 = 10a

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4 0
2 years ago
What is the new volume of the gas if the pressure on 350 L of oxygen
Iteru [2.4K]

Answer:

420 L

Explanation:

Applying Boyle's Law,

PV = P'V'.................... Equation 1

Where P = Initial pressure, P' = Final pressure, V = Initial volume, V' = Final volume.

make V' the subject of the equation

V' = PV/P'.................... Equation 2

From the question,

Given: P = 720 mmHg, V = 350 L, P' = 600 mmHg

Substitute these values into equation 2

V' = (720×350)/600

V' = 252000/600

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3 years ago
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

P_{2} > P_{1}.

6 0
4 years ago
A soccer ball is rolling past Ramon. He kicks it in the same direction that it IS rolling. What will MOST LIKELY happen?
vovikov84 [41]

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6 0
3 years ago
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