1) Length of the wire.
2) Thickness of the wire.
3) Temperature.
4) Type of metal.
Hope this helps!
-Payshence
Answer:
Explanation:
Given
mass of spring 
extension in spring 
downward velocity 
Position in undamped free vibration is given by

where 
also 



it is given


substituting values we get







Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.
However, photons are considered to have a rest mass zero.
However note the point carefully "rest mass". A body in motion cannot have mass to be zero.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>
Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s



While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

Substitute the given values to find "terminal speed"




The “terminal speed” of the ball bearing is 5.609 m/s
Answer:
t = 120.5 nm
Explanation:
given,
refractive index of the oil = 1.4
wavelength of the red light = 675 nm
minimum thickness of film = ?
formula used for the constructive interference

where n is the refractive index of oil
t is thickness of film
for minimum thickness
m = 0


t = 120.5 nm
hence, the thickness of the oil is t = 120.5 nm