Answer: 
Explanation:
Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period 
of a planet is proportional to the cube of the semi-major axis 
of its orbit”:
(1)
Now, if is measured in years (Earth years), and is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
So, knowing 
and isolating from (2) we have:
![a=\sqrt[3]{T^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{(619.36 years)^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%28619.36%20years%29%5E%7B2%7D%7D)
(4)
Finally:
T
his is the distance between the dwarf planet and the Sun in astronomical units
Converting this to kilometers, we have:
Answer:
179.47m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute
7750(179)+72(230) = (7750+72)v
1,387,250+16560 = 7822v
1,403,810 = 7822v
v = 1,403,810/7822
v= 179.47m/s
Hence the final velocity of the probe is 179.47m/s
Answer:
1,3
Explanation:
As the acceleration is -10m/s^2 , that means deceleration is occurring. That means, the object is slowing down.
v=u-at
or, 0=80-10t
or, t=8 seconds
So, the object will stop in 8 seconds.
So, the correct answers are 1 and 3.
Hope, this helps you.
initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find
here we know that
a = 9.8 m/s^2
t = 1.5 s
now from above equation
so it will hit the floor with speed 14.7 m/s
Answer:
R = 2Ω
Explanation:
Potential difference (V) = current (I) * Resistance (R)
V = IR
I = 2.0A
V = 10v
R = ?
V = IR
R = V / I
R = 10 / 2
R = 2Ω
The resistance across the wire is 2Ω
