Answer:
1.16 Hz
Explanation:
frequency, basically, is the number of wave on 1 second
so, in math we write like this
f = n/t
n = number of waves
t = time to do that (in sec)
f = 140/120 = 7/6 Hz
f = 1.16 Hz
1) Hubble Space Telescope- Visible and near-ultraviolet
2) Compton Gamma Ray Observatory- Gamma Rays
3) Chandra X-ray Observatory- X-rays
4) Spitzer Space Telescope- infrared
Answer and Explanation:
This can be explained as in Rutherford's model of atom the electrons orbits the nucleus which means that they will travel around the nucleus with some velocity and hence radiate electromagnetic waves which results in the loss of energy due to which the electron keeps coming closer and eventually falls into the nucleus.
But Bohr came up with a better explanation as according to the Bohr's atomic model, electrons stay fixed in orbit with certain energy in different shells around the nucleus and can only jump from an energy level to another if that specific amount of energy is supplied to it.
This model is based on the quantization of energy thus giving an explanation why electrons do not fall into the nucleus of an atom.
Answer:
The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Explanation:
Given that,
Initial pressure = 10 bar
Temperature = 22°C
Final pressure = 2.0 bar
We using the value of h
The refrigerant during expansion undergoes a throttling process
Therefore, 
We need to calculate the quality of the refrigerant at the exit of the expansion valve
At 2.0 bar,
The property of ammonia
Using formula
Put the value into the formula
Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
