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boyakko [2]
2 years ago
13

A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the

relative speed of the two ships as measured by a passenger on either one of the spaceships
Physics
1 answer:
vaieri [72.5K]2 years ago
5 0

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

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The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

\mathbf{v_f ^2 = v_i^2 + 2gh}

\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}

\mathbf{v_f ^2 =151 m/s}

\mathbf{v_f  =\sqrt{151 m/s}}

\mathbf{v_f  =12.29 \ m/s}  

From the equation of continuity:

\mathbf{A_iV_i = A_fV_f}

\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}

\mathbf{ r^2_iV_i =  r^2_fV_f}

\mathbf{ (\dfrac{10}{2})^2\times 2.0 =  r_f^2 \times 12.29}

\mathbf{ 50 = 12.29 \times r_f^2}

\mathbf{ r_f=  \sqrt{\dfrac{50}{12.29} }}

\mathbf{ V_f= 2.02 \ cm }

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

brainly.com/question/10822213

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