Answer:
The rate of transfer of energy is equal to 23.76W or 23.76J/s as may be required both forms are correct. The physical quantities needed to calculate the rate of energy transfer are the linear mass density or mass per unit length, tension force, amplitude, angular frequency( which is equal to 2pi •f )
Explanation:
The required quantity is the average power or average rate of energy transfer which differs from the maximum or instantaneous rate of energy transfer. The calculation steps to the answer above can be found in the attachment below. Should the requested quantity be the instantaneous quantity the answer will be 2 x Pav which equals 47.52W or 47.52J/s.
The Barometer: The barometer is a device meant for measuring the local atmospheric pressure. ...
Piezometer or Pressure Tube: ...
Manometers: ...
The Bourdon Gauge: ...
The Diaphragm Pressure Gauge: ...
Micro Manometer (U-Tube with Enlarged Ends):
There are three basic steps in a nuclear fission chain reaction are Initiation, Propagation, Termination.
<h3>
What are the three steps of nuclear fission ?</h3>
1. Initiation
The nuclear fission of 
is started by the absorption of a neutron; a single atom must react in order for the chain reaction to begin.
2. Propagation:-
With each stage producing additional product, this component of the process is repeated repeatedly. When 
splits apart, neutrons are released, which start the nuclear fission of more uranium atoms.
3. Termination:
The chain will come to an end eventually. Termination might take place if the reactant is exhausted or if the subsequent neutrons in the chain leave the sample without being absorbed by .
To learn more about nuclear fission with the given link
brainly.com/question/913303
#SPJ4
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Yes it is a force omg so smart
