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andrew11 [14]
3 years ago
14

A car travels in a straight line for 4.4 h at a constant speed of 81 km/h. What is its acceleration?

Physics
2 answers:
Fed [463]3 years ago
7 0

The speed and length of time don't matter.

Constant speed in a straight line is the DEFINITION of ZERO acceleration.

irga5000 [103]3 years ago
5 0

Answer:

a = 0m/s^2

Explanation:

we can use the equation:

v_f = a(t) + v_i

And plug in our numbers:

81km/h = am/s^2(4.4h) + 81hm/h

0 = am/s^2(4.4h)

0 = am/s^2

a = 0m/s^2

Because the speed was constant, there was no difference between the final and initial velocities, meaning that once you do the math you see there is no acceleration affecting the speed.

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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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3 years ago
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction betwe
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prohojiy [21]

Answer:

<em>They represent kinetic energy</em>

Explanation:

<u>Kinetic Energy </u>

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