What happens to the period of a spring-cart system motion when the system is in orbit inside the international space station?
1 answer:
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Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
Hence the velocity of the particle at time t is
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second
Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface,
Surface charge density of second surface,
The electric field at a point between the two surfaces is given by :
E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.