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Zielflug [23.3K]
3 years ago
9

You are in your car at rest when the traffic light turns green. You place your coffee cup on the horizontal dash and hit the gas

. The coffee cup does not slide as you accelerate forward. The work done by friction on your coffee cup is
a. equal to zero.

b. positive.

c. equal to the nonconservative work done by the car's engine.

d. negative.

e. equal in magnitude to the work done by the road on the car.
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

(d) Negative.

Explanation:

let's test each at a time.

(a) It can't be 0, because cup would slide back other wise.

(b) Positive, well if forward is positive, than the work done against the forward acceleration must be negative , so it can't be positive.

(c) Equal to non-conservative work done by the car's engine.

well no, because work done by car's engine dosen't go all of it into getting car to move, so it can't be that.

(d) negative, this look like it, because work that friction does must be nagative to counteract positive thrust of car which is positive and in forward direction.

(d) this can't be true.

So the answer is (d) negative.

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3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

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Answer:

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2 years ago
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Answer:

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Answer:

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