Answer:
Extraneous
Explanation:
Extraneous variables are any variables that you are not intentionally studying in your experiment or test
The central force acting on the electron as it revolves in a circular orbit is
.
The given parameters;
- <em>speed of electron, v = 2.2 x 10⁶ m/s</em>
- <em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>
<em />
The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

where;
is mass of electron = 9.11 x 10⁻³¹ kg

Thus, the central force acting on the electron as it revolves in a circular orbit is
.
Learn more about centripetal force here:brainly.com/question/20905151
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Answer:
The amount of work done required to stretch spring by additional 4 cm is 64 J.
Explanation:
The energy used for stretching spring is given by the relation :
.......(1)
Here k is spring constant and x is the displacement of spring from its equilibrium position.
For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

k = 4 x 10⁴ N/m
Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).
Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = 72 J
But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :
E = ( 72 - 8 ) J = 64 J
Kinetic energy is a result of mass in motion at a certain velocity.
<span>1 Joule = 1 kg • (m/s)<span>2
</span></span>the force as a function of mass of the object.