Answer:
C)T
Explanation:
The period of a mass-spring system is:
As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.
Initially, the velocity vector is 
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 
, so the velocity is 
.
Converting back to direction and magnitude, we get 
Answer:
The train's displacement is zero.
Explanation:
Given data,
The time taken by the train from NY to Washington and back is, t = 6 h 5 min
The distance between the two stations is, d = 363 km
Therefore, the total distance the train traveled is, d' = 726 km
The displacement is defined as the change in position coordinates with respect to its original position.
If the train travels from one point and returns back to the same point after some time, there is no change in the position coordinates with respect to its original position.
Hence, the train's displacement is zero.
Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
Answer:
f = 931.1 Hz
Explanation:
Given,
Mass of the wire, m = 0.325 g
Length of the stretch, L = 57.7 cm = 0.577 m
Tension in the wire, T = 650 N
Frequency for the first harmonic = ?
we know,
μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,
v = 1074.49 m/s
The wire is fixed at both ends. Nodes occur at fixed ends.
For First harmonic when there is a node at each end and the longest possible wavelength will have condition
λ=2 L
λ=2 x 0.577 = 1.154 m
we now,
v = f λ
f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz
