Answer: D = 16m
Explanation: given values: a = 2 m/s2, v = 4 m/s
In this case we have to determine the diameter of the Ferris wheel.
Diameter of circle is given as: D = 2.r.
First we have to find radius of wheel. The best way to find that is using the centripetal acceleration equation: a = v2/r
Plug in values in above equation to find radius: 2 m/s2 = (4 m/s)2/r 2 m/s2 = (16 m2/s2)/r r = (16 m2/s2)/2 m/s2
r = 8.0m
Diameter of Ferris wheel is:
D = 2.r.
D = 2.8m
D = 16m
I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.
I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:
Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>
</span>
</span>
</span>
<span>
Distance at Perihelion
(</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>
Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.
</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>
1/2 (50%) of that is 43.9845 Earth days
The average of the aphelion and perihelion distances is
1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU
This also happens to be 1/2 of the major axis of the elliptical orbit.
Answer:
1.736m/s²
Explanation:
According to Newton's second law;
where;
Fm is the moving force = 70.0N
Ff is the frictional force acting on the body
is the coefficient of friction
m is the mass of the object
g is the acceleration due to gravity
a is the acceleration/deceleration
The equation becomes;
Substitute the given parameters
Hence the deceleration rate of the wagon as it is caught is 1.736m/s²
Answer:
Explanation:
Mass of ice m = 500g = .5 kg
Heat required to raise the temperature of ice by 10 degree
= mass of ice x specific heat of ice x change in temperature
= .5 x 2093 x 10 J
10465 J
Heat required to melt the ice
= mass of ice x latent heat
0.5 x 334 x 10³ J
167000 J
Heat required to raise its temperature to 18 degree
= mass x specific heat of water x rise in temperature
= .5 x 4182 x 18
=37638 J
Total heat
=10465 +167000+ 37638
=215103 J
It must be a virtual image, because this is the only kind of image it can produce.
