Lets se
And
So
If spring constant is doubled mass must be doubled
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
Answer:
Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,
Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 
So,
l = 0.024 m
Then for relativistic length contraction,
Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
Gravitational force = G · (mass₁) · (mass₂) / (distance)
(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)
G = 6.67 x 10⁻¹¹ n-m² / kg² (the "gravitational constant")
Distance² = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)
Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)
Distance² = (418.209 N-m²) / (39N)
Distance² = 10.72 m²
<em>Distance = 3.275 meters</em>
An absurd scenario, but that's by golly what the math says with the numbers provided. I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.
