Complete question:
use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is 2.2 x 10⁻⁵ km/s/Lyr
Answer:
The distance to the quasar is 1.02 x 10¹⁰ Lyr
Explanation:
Given;
speed of light, v = 300, 000 km/sec
Hubble's constant, H₀ = 2.2 x 10⁻⁵ km/s/Lyr
percentage of the quasar recession = 75% of speed of light
Hubble's Law is given by;
Therefore, the distance to the quasar is 1.02 x 10¹⁰ Lyr
Answer:
Explanation:
Given that
We know that
By using energy conservation
Heat lost by water = Heat gain by block
Therefore the specific heat of the block will be 0.501 kJ/kg.K
Pressure drop across the tube.
Viscosity of the fluid.
Diameter of the tube.
Answer:
-39.2m/s
Explanation:
Using the equation of motion;
v = u + at
Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g
v = u - gt
Since g = 9.8m/s²
t = 4.0s
u = 0m/s
v = 0 + (-9.8)(4)
v = 0 + (-9.8)(4)
v = -39.2m/s
Hence the speed of the ball before release is -39.2m/s
Evaporation - Heating
Condensation - Cooling
Freezing - Cooling
Melting - Heating
Sublimation - Heating
Deposition - Cooling