Answer:
c. Time period remains the same in all.
Explanation:
In order to answer this question, we need to analyze the parameters, upon which the time period of a pendulum depends. We know that the time of a pendulum is given by the following formula:
T = 2π√(L/g)
where,
T = Time period
L = Length of pendulum
g = acceleration due to gravity
The formula clearly shows that the time period of the pendulum depends only upon the length of pendulum and value of g. And the time period of a pendulum does not depend upon the mass of the bob. Hence, the time period for each of the three pendulums will remain same. So, the correct option will be:
<u>c. Time period remains the same in all.</u>
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
Answer:
Explanation:
We know that the frequency of the nth harmonic is given by 
, where 
is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:
Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):
Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):
So the tension is:
Which for our values is:
Answer:
-20,000N
Explanation:
Force (N) = mass (kg) x acceleration (m/s²)
So,
Force = 2000 x -10
= -20,000N (Newtons)
Answer:
3125 N
Explanation:
diameter /2 =radius
so r1 =14cm , r2 =35cm
f1/A1 =f2/A2.
f2 = f1 × A2 / A1
=500×1225 pi cm² / 96 pi cm²
f2 =3125N
