Answer:
76
At what angle height and range of projectile are equal?
It means that the range of the projectile with a given initial velocity is same for a pair of projection angles θ and 90° – θ.
The dimension of K is M/ T^2
according to the question T=2π square root ofm/k here 2 pi is constant so
T= root of m /k and root of k = root of m/ T now by squaring on both the sides we get the answer k= M/ T^2
complete question :
A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T=√2πm/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?
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Answer:
Explanation:
for rolling motion down the plane acceleration is given by the following expression
a = g sinθ / (1 + k² / R²)
here k is radius of gyration and R is radius of the object rolling down .
for cylinder I = 1/2 m R²
so k² = R² / 2
k² / R² = 1/2
a = g sinθ /( 1 + 1 / 2 )
= 2 / 3 x g sinθ
v = √ 2 a s
= √ (2 x 2 / 3 x g sinθ s )
= √ (4 / 3 x g h )
= √ (4 / 3 x g x .5 )
= √ 2g / 3
for sphere I = 2/5 m R²
so k² = 2/5 R²
k² / R² = 2 / 5
a = g sinθ / (1 + 2 / 5)
= 5 / 7 x g sinθ
v = √ 2 a s
= √ (2 x 5 / 7 x g sinθ s )
= √ (10/7 x g h )
Given
√ (10/7 x g h ) = √ 2g / 3
10/7 x g h = 2g / 3
h = 14 / 30 m
= .47 m .
