Answer:
v=0.04m/s
Explanation:
To solve this problem we have to take into account the expression
where v and r are the magnitudes of the velocity and position vectors.
By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that
the maximum relative velocity is 0.04m/s
hope this helps!!
Answer:
10.89 J.
Explanation:
The following data were obtained from the question:
Mass (m) = 12.5 kg
Velocity (v) = 1.32 m/s
Work done =?
To obtain the workdone, we shall determine the kinetic energy of the object since work and energy has the same unit of measurement. This is illustrated below:
Mass (m) = 12.5 kg
Velocity (v) = 1.32 m/s
Kinetic energy (K.E) =?
K.E = ½mv²
K.E = ½ × 12.5 × 1.32²
K.E = 6.25 × 1.7424
K.E = 10.89 J
The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.
de Broglie wavelength (λ) is given by the equation
λ = h/p
where h=Planck’s constant whose value is 6.62 x 10^(−34) joule-seconds and
p = momentum of the particle(here electron)
In terms of kinetic energy(E) momentum(p) can be written as,
p=(2mE)^1/2
where m=mass of the particle.
Hence λ becomes
1 λ = h(2mE)^-1/2
Given here, E = 13.6 eV = 13.6×1.6×10^-19 joule
m(mass of electron)= 9.1×10^-31 kg
Putting these values in equation (1) we get ,
λ =0.332×10^(-9) meter
=3.32×10^(-10) meter
=3.32 Å
Answer:
The position of the spring in terms of g, m & k is 
Explanation:
Stiffness of the spring = k
Mass = m
When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.
⇒ Force exerted on the spring F = k x
⇒ m g = k x
⇒
This is the position of the spring in terms of g, m & k.
