E = 1/2 m v^2
108000m /3600 = 30m/s
E = 1525 * 0.5 * 30 = 22875 J
Answer:
Part 1
20 N
Part 2
0.4 m/s²
Part 3
4 m/s
Explanation:
The force which pulls the sled right = 50 N
The friction force exterted towards left by the snow = -30 N
The mass of the sled = 50 kg
Part 1
The sum of the forces on the sled, F = 50 N + (-30) N = 20 N
Part 2
The acceleration of the sled is given as follows;
F = m·a
Where;
m = The mass of the sled
a = The accelertion
a = F/m
∴ a = (20 N)/(50 kg) = 0.4 m/s²
The acceleration of the sled, a = 0.4 m/s²
Part 3
The initial velocity of the sled, u = 2 m/s
The kinematic equation of motion to determine the speed of the sled is v = u + a·t
The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;
v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.
Answer:
Longest wavelength, lowest intensity
Explanation:
