Answer:
The answer is A and D but if you want one anwer i think the anwer is A
Explanation:
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 .
Answer:
The answer is C. thinking about the smallest particles of matter without experimenting.
Answer:
The magnitude of the impulse delivered to the baseball is 7.0 Ns
Explanation:
Given;
mass of the foul ball, m = 0.14 kg
initial velocity, u = 40 m/s
final velocity, v = 30 m/s in perpendicular direction
Impulse is given as change in momentum;
initial momentum in horizontal direction, Pi = mu
Pi = 0.14 x 40 = 5.6 Ns
final momentum in perpendicular direction, Pf = mv
Pf = 0.14 x 30
Pf = 4.2 Ns
The resultant impulse is given by;
J² = 5.6² + 4.2²
J² = 49
J = √49
J = 7.0 Ns
Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns