<u>Answer:</u>
The distance from earth to sun is 387.5 times greater than distance from earth to moon.
<u>Solution:</u>
Given, the distance from Earth to the sun is about ![9.3 \times 10^{7} \mathrm{miles}](https://tex.z-dn.net/?f=9.3%20%5Ctimes%2010%5E%7B7%7D%20%5Cmathrm%7Bmiles%7D)
The distance from Earth to the Moon is about ![2.4 \times 10^{5} \mathrm{miles}](https://tex.z-dn.net/?f=2.4%20%5Ctimes%2010%5E%7B5%7D%20%5Cmathrm%7Bmiles%7D)
We have to find how many times greater is the distance from Earth to the Sun than Earth to the Moon?
For that, we just have to divide the distance between earth and sun with distance between earth to moon.
Let the factor by which distance is greater be d.
![\text { Now, } \mathrm{d}=\frac{\text { distance between sun and earth }}{\text { distance between moon and earth }}=\frac{9.3 \times 10^{7}}{2.4 \times 10^{5}}=\frac{9.3}{2.4} \times 10^{7-5}\\\\=3.875 \times 10^{2}=387.5](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Now%2C%20%7D%20%5Cmathrm%7Bd%7D%3D%5Cfrac%7B%5Ctext%20%7B%20distance%20between%20sun%20and%20earth%20%7D%7D%7B%5Ctext%20%7B%20distance%20between%20moon%20and%20earth%20%7D%7D%3D%5Cfrac%7B9.3%20%5Ctimes%2010%5E%7B7%7D%7D%7B2.4%20%5Ctimes%2010%5E%7B5%7D%7D%3D%5Cfrac%7B9.3%7D%7B2.4%7D%20%5Ctimes%2010%5E%7B7-5%7D%5C%5C%5C%5C%3D3.875%20%5Ctimes%2010%5E%7B2%7D%3D387.5)
Hence, the distance from earth to sun is 387.5 times greater than distance from earth to moon.
For this, it is kind of like a triangle and having to find the diagonal. Let's see what the two legs of the triangle would be.
Leg one= 7 units
Leg two= 8 units
8²+7²= c²
64 + 49=113
√113≈10.6
The line is 10.6 units long.
Answer:
14 42 in thirds is 14 so if he had to pay a third he paid 14 bucks
Step-by-step explanation:
42/3=14
Vcylinder=hpir^2
Vsphere=(4/3)pir^3
Vcone=(1/3)hpir^2
Vcylinder=15*pi*5^2=375pi in^3
Vsphere=(4/3)*pi*6^3=288pi in^3
Vcone=(1/3)*15*pi*8^2=320pi in^3
greatest is Vcylinder at 375pi in^3
answer is A (cylinder)
Answer:
(a) Speed = 6.875 m/s
(b) Kinetic Energy = 1.70 kJ
(c) Average power = 1.06 kW
Step-by-step explanation:
Given:
d = 5.5 m
t = 1.6 s
W = 706 N
Part (a)
Using the equation of motion with the assumption of zero initial speed:
![d = \frac{1}{2} ({v-0})t](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28%7Bv-0%7D%29t)
![5.5= \frac{1}{2} (1.6v)](https://tex.z-dn.net/?f=5.5%3D%20%5Cfrac%7B1%7D%7B2%7D%20%281.6v%29)
The speed of the sprinter then become:
![v = 6.875 m/s](https://tex.z-dn.net/?f=v%20%3D%206.875%20m%2Fs)
Part (b)
Convert the weight of the sprinter to mass:
![m = \frac{w}{g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bw%7D%7Bg%7D)
![m = \frac{706}{9.8}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B706%7D%7B9.8%7D)
![m = 72 kg](https://tex.z-dn.net/?f=m%20%3D%2072%20kg)
KE (Kinetic Energy) can then be calculated as:
![KE = \frac{1}{2} mv^{2}](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D)
![KE = \frac{1}{2} (72)(6.875)^{2}](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2872%29%286.875%29%5E%7B2%7D)
![KE = 1701.56 J = 1.70 kJ](https://tex.z-dn.net/?f=KE%20%3D%201701.56%20J%20%3D%201.70%20kJ)
Part (c)
The change of the kinetic energy is equivalent to the work done by the sprinter. The average power P is the rate of the work done and can be calculated as
![P = \frac{KE}{t} \\\](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BKE%7D%7Bt%7D%20%5C%5C%5C)
![P = \frac{1701.56}{1.6}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B1701.56%7D%7B1.6%7D)
![P = 1063.48 W = 1.06 kW](https://tex.z-dn.net/?f=P%20%3D%201063.48%20W%20%3D%201.06%20kW)