Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
Complete Question
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
Answer:
The electric field vector of the satellite broadcast as measured at the surface of the earth is
Explanation:
From the question we are told that
The height of the satellite is
The power output of the satellite is
Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is mathematically represented as
substituting values
This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be mathematically represented as
Where is the amplitude of the electric field vector of the satellite broadcast so
substituting values
Answer:
24,187.04 J ≈ 24,200 J
Explanation:
mass (m) = 544 kg
initial speed (u) = 6.75 m/s
final speed (v) = 15.2 m/s
change in height (Δh) = -14 m (negative sign is because there is a decrease in height )
acceleration due to gravity (g) = 9.8 m/s^{2}
How much work was done on the raft by non conservative forces?
work done = change in energy of the system = change in kinetic energy + change in potential energy
work done = () + (mgΔh)
work done = () + (544 x 9.8 x (-14))
work done = 50449.76 - 74,636.8
work done = 24,187.04 J ≈ 24,200 J