Answer: Find the answer in the explanation
Explanation:
Bohr postulated that :
1.) Electrons revolve in specific circular orbits around the nucleus and these orbits can be represented by the letters K, L, M, N.
2.) Electrons revolving in a particular orbit are in a stationary state. They can neither gain energy nor loose energy. 3.) But if an electron jumps from an higher orbit to a lower orbit it will emit energy in the form of radiations. Also if an electron jumps from lower orbit to a higher orbit it excites and absorbs energy in the form of radiations.
4.) Neil Bohr experimental findings about angular momentum of an electron shows more light on the stability of atom and the line spectrum of hydrogen atom
Take 771.9 Kg M/s because that's how fast she is moving and divide it by 93 kg because that is her force pushing down
771.9/93= 8.3
Answer: A. 8.3 M/s
P.s I changed my answer on the last question you asked so change that.
We could use the change of pressure to calculate for the height climbed by the mountain hiker. The change of pressure is given by
p = rho * g * h, where p is the change of pressure, rho is the air density, g is the acceleration due to gravity, and h is the height.
Using the conversion 1 mbar = 100 Pa,
(930 - 780)(100) = (1.20)(9.80)h
15000 = 1.20*9.80*h
h = 1.28 km
Answer:
At 400 m the potential energy of the mountain climber doubled the initial value.
Explanation:
Given;
initial height of the mountain climber = 200 m
final height of the mountain climber, = 400 m
The potential energy of the mountain climber is calculated as;
Potential energy, P.E = mgh
At 200 m, P.E₁ = mg x 200 = 200mg
At 400 m, P.E₂ = mg x 400 = 400mg
Then, at 400 m, P.E₂ = 2 x 200mg = 2 x P.E₁
Therefore, at 400 m the potential energy of the mountain climber doubled the initial value.
Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart

v is the velocity of jet
Q = A V


v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,

F n = 2295 N
now force along x direction



Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW