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Nookie1986 [14]
3 years ago
13

A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +

6.5 m/s relative to the ground. What is the velocity of the girl relative to the lawnmower? A. 0 m/s B. +5.75 m/s C. +6.5 m/s D. +7.25 m/s
Physics
2 answers:
SashulF [63]3 years ago
8 0

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are V_a and V_b respectively, the relative velocity of B with respect to A is given by:

V_{ba}=V_b-V_a

Thus, the velocity of the girl relative to the lawnmower is:

V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}

ZanzabumX [31]3 years ago
4 0

I took the test and got B) +5.75 m/s correct

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Answer:

they stay shifted the same amount to the red

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The __________ on the left side of an equation must be balanced with the atoms on the right side of the equation.
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An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

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T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

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