Answer
given,
mass of the rod = 1.50 Kg
length of rod = 0.85 m
rotational velocity = 5060 rev/min
now calculating the rotational inertia of the system.
where L is the length of road, we will take whole length of rod because mass is at the end of it.
I = 1.084 kg.m²
hence, the rotational inertia the system is equal to I = 1.084 kg.m²
Answer:
310 meters
Explanation:
Given:
v₀ = 0 m/s
t = 8.0 s
a = -9.8 m/s²
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²
Δy = -313.6
Rounded to two significant figures, the object fell 310 meters.
To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as
F = ma
Where,
m= Mass
a = Acceleration
At the same time the frictional force can be defined as,

Where,
Frictional coefficient
N = Normal force (mass*gravity)
Our values are given as,

By condition of Balance the friction force must be equal to the total net force, that is to say



Re-arrange to find acceleration,



Therefore the acceleration the horse can give is 
Answer:
51 Ω.
Explanation:
We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:
Resistor 1 (R₁) = 40 Ω
Resistor 3 (R₃) = 70.8 Ω
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?
Since the two resistors are in parallel connection, their equivalent can be obtained as follow:
R₁ₙ₃ = R₁ × R₃ / R₁ + R₃
R₁ₙ₃ = 40 × 70.8 / 40 + 70.8
R₁ₙ₃ = 2832 / 110.8
R₁ₙ₃ = 25.6 Ω
Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω
Resistor 2 (R₂) = 25.4 Ω
Equivalent Resistance (Rₑq) =?
Rₑq = R₁ₙ₃ + R₂ (series connection)
Rₑq = 25.6 + 25.4
Rₑq = 51 Ω
Therefore, the equivalent resistance of the group is 51 Ω.
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds