Answer:

Explanation:
Given that,
The mass of a golf ball, m = 40 g = 0.04 kg
Its angular velocity, 
The radius of the sphere is 2.5 cm or 0.025 m
We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

Where I is moment of inertia
For sphere, 

So, the magnitude of the angular momentum of the sphere is
.
Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
Answer:
Explanation:
Given
Minute hand length =16 cm
Time at a quarter after the hour to half past i.e. 1 hr 45 min
Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270


(c)For the next half hour
Effectively it has covered 2 revolution and a quarter

angle turned 
(f)Hour after that
After an hour it again comes back to its original position thus displacement is same =25.136
Angle turned will also be same i.e. 