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:
Option A
D = m/v
Explanation:
Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume
Therefore, option A is the right option
The cliff is 2042 ft away.
We know that the speed of sound in air is directly proportional to the absolute temperature.
First convert the Fahrenheit temperature to Celsius;
°C = 5/9(44.5 - 32)
°C = 6.9 °C
Applying the formula;
V1/V2 = √T1/T2
Where; V1 = velocity of sound in air at 0°C
V2 = Velocity of sound in air at 6.9 °C
1087/V2 = √273/279.9
V2= 1101 ft/s
Given that; V = 2s/t
Where s is the distance of the cliff
t is the time taken
1101 ft/s = 2s/3.71 s
s = 1101 ft/s × 3.71 s/2
s = 2042 ft
Learn more:brainly.com/question/15381147
Answer:
a. Decreases
b. Increases
c. Remains the same
d. Increases
Explanation:
a. Capacitance is given by c= Ak€/d
where A is conductivity plate with Area
K is a constant
€ is dielectric with permittivity.
d is the distance
b. Potential difference is given by
V = Ed, since, the electric field remains the
same, the potential diterence also increases with increase in distance.
Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.
c. Electric field remains the same because charge on the
plate remains the same.
d. since electric field remains the same and capacitance decreases, the energy increases.
E= 1/2c * Q^2