By definition we have that
force=dP/dt,
where
p is momentum
so
<span>momentum is force*time
p= 15*3 = 45 Ns , west.
</span><span>the change in momentum of the object is 45 N.s</span>
Answer:
Power output = 96.506 watts
Explanation:
Drag coefficient (Cd) = 0.9
V = 7.3 m/s
Air density (ρ) = 1.225 kg/m^(3)
Area (A) = 0.45 m^2
Let's find the drag force ;
Fd=(1/2)(Cd)(ρ)(A)(v^(2))
So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N
Drag power = Drag Force x Drag velocity.
Thus drag power, = 13.22 x 7.3 = 96.506 watts
Complete question:
(b) How much energy must be supplied to boil 2kg of water? providing that the specific latent heat of vaporization of water is 330 kJ/kg. The initial temperature of the water is 20 ⁰C
Answer:
The energy that must be supplied to boil the given mass of the water is 672,000 J
Explanation:
Given;
mass of water, m = 2 kg
heat of vaporization of water, L = 330 kJ/kg
initial temperature of water, t = 20 ⁰C
specific heat capacity of water, c = 4200 J/kg⁰C
Assuming no mass of the water is lost through vaporization, the energy needed to boil the given water is calculated as;
Q = mc(100 - 20)
Q = 2 x 4200 x (80)
Q = 672,000 J
Q = 672,000 J
Q = 672,000 J
Therefore, the energy that must be supplied to boil the given mass of the water is 672,000 J
Answer: R = 0.131 m
Explanation: The formulae for the distance between a fringe and the first (center) is given by
y = R×mλ/d
Where y = distance between first and nth fringe = 4mm = 4×10^-3m
λ = wavelength of light = 441.1nm = 441.1×10^-9m
R = distance between slits and screen =?
d = distance between slits = 0.29mm = 2.9×10^-4m
4×10^-3 = R ×2 ×441.4×10^-9/ 2.9×10^-5
Hence R = (4×10^-3) ×(2.9×10^-5)/2×441.4×10^-9
R = 1.16 × 10^-7/8.828×10^-7
R = 1.16/8.828
R = 0.131 m
Answer:
#1 ( Balanced ) ( Rest)
#2 ( Unbalanced) ( Accelerating)
Explanation:
