Answer:
period of rotation is 9.9843 years
Explanation:
Given data
sides = 1.0×10^12 m
to find out
period of rotation
solution
first we use the gravititional formula i.e
F(g) = 2 F cos 30
here F = G × mass(sun)² / radius²
F = 6.67 ×
× (1.99×
)² / (1×
)²
so
F(g) = 2 × 6.67 ×
× (1.99×
)² / (1×
)² × (1.73/2)
F(g) = 4.57 ×
N
and
by the law of sines
r/sin30 = s/sin120
so r will be
r = 1.0×10^12 (0.5/.87)
r = 5.75 ×
m
we know that gravititional force = centripetal force
so here
centripetal force = mv² /r
centripetal force = 1.99×
v² / 5.75 ×
so
4.57 ×
= mv² /r
4.57 ×
= 1.99×
v² / 5.75 ×
v² = 4.57 ×
× 5.75 × ![10^{11} / 1.99× [tex]10^{30}](https://tex.z-dn.net/?f=10%5E%7B11%7D%20%2F%201.99%C3%97%20%5Btex%5D10%5E%7B30%7D)
and
v = 11480 m/sec
and we know period of rotation formula
t = d /v
t = 2
r /v
t = 2 ×
× 5.75 × [tex]10^{11} / 11480
t = 314 × [tex]10^{5} / 31536000 year
t = 9.9843 year
so period of rotation is 9.9843 years
Answer:
The correct answer is B.
The astronaut will know due to the light from the explosion.
Explanation:
Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.
Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.
Cheers!
Keller’s law of planetary motion is the law of planetary motion
Answer:
121550 J
Explanation:
Parameters given:
Mass, m = 0.34kg
Specific heat capacity, c = 14300 J/kgK
Change in temperature, ΔT = 25K
Heat gained/lost by an object is given as:
Q = mcΔT
Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:
Q = 0.34 * 14300 * 25
Q = 121550J or 121.55 kJ
Data:
wavelength (λ) = ?
wave speed (c) = 150 m/s
frequency (f) = 15 Hz (Hz = 1/s)
Formula
![\lambda = \frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%20%5Cfrac%7Bc%7D%7Bf%7D%20)
Solving:
![\lambda = \frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bc%7D%7Bf%7D%20)
![\lambda = \frac{150}{15}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B150%7D%7B15%7D%20)
![\boxed{\lambda= 10m}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Clambda%3D%2010m%7D)
Answer:
<span>
wavelength = 10 m</span>