A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Answer:
The frequency is 302.05 Hz.
Explanation:
Given that,
Speed = 18.0 m/s
Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .
We need to calculate the frequency
Using formula of frequency
Where, f = frequency
v = speed of sound
= speed of passenger
= speed of source
Put the value into the formula
Hence, The frequency is 302.05 Hz.
Answer:
Kepler's First Law: each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the orbital ellipse. The Sun is at one focus. The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit.
Answer:
6.67×10⁻⁸ cm³/g/s²
Explanation:
6.67×10⁻¹¹ Nm²/kg²
= 6.67×10⁻¹¹ (kg m/s²) m²/kg²
= 6.67×10⁻¹¹ m³/kg/s²
= 6.67×10⁻¹¹ m³/kg/s² × (100 cm/m)³ × (1 kg / 1000 g)
= 6.67×10⁻⁸ cm³/g/s²
