Answer:
The force constant of the spring is 317.8 N/m.
Explanation:
Given that,
Frequency 
We need to calculate the reduced mass
Using formula of reduced mass
Where, 
= atomic mass of H
= atomic mass of I
Put the value into the formula
We need to calculate the force constant of the spring
Using formula of frequency
Put the value into the formula
Hence, The force constant of the spring is 317.8 N/m.
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
The false statement is 'Electric field lines form closed loops'.
Explanation:
- Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
- These lines when close together represents high intensity and when far apart shows low intensity of the field.
- These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
- These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.
Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.
Explanation:
Measure unstretched length of spring, L. E.g. L = 0.60m.
Set mass to a convenient value (e.g. m = 0.5kg).
Hang mass.
Measure new spring length, L'. E.g. L' = 0.70m.
Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m
Use mg = ke (in equilibrium weight = tension)
k = mg/e
Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).
k = 0.5*10/0.10 = 50 N/m
Repeat for a few different masses. (L always stays the same.)
Take the average of your k values.
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
