Answer:
The quantitative relationship between heat transfer and temperature change contains all three factors: Q = mcΔT, where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific heat c is a property of the substance; its SI unit is J/(kg ⋅ K) or J/(kg ⋅ ºC). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is kcal/(kg ⋅ ºC).
Explanation:
Answer:
I = I₀ + M(L/2)²
Explanation:
Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.
The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.
The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀
The distance between the two axes is L/2 (total length of the rod divided by 2
From the parallel axis theorem we have
I = I₀ + M(L/2)²
Answer:
V = 0.0806 m/s
Explanation:
given data
mass quarterback = 80 kg
mass football = 0.43 kg
velocity = 15 m/s
solution
we consider here momentum conservation is in horizontal direction.
so that here no initial momentum of the quarterback
so that final momentum of the system will be 0
so we can say
M(quarterback) × V = m(football) × v (football) ........................1
put here value we get
80 × V = 0.43 × 15
V = 0.0806 m/s
KE = (1/2)·(mass)·(speed)²
KE = (1/2)·(50 kg)·(18 m/s)²
KE = (25 kg)·(324 m²/s²)
KE = 8,100 kg-m²/s²
KE = 8,100 Joules
