Answer:
I₁ =1250 kg.m²
Explanation:
Given that
Angular speed of Merry ,ω₁= 0.2 rad/s
Angular speed of technician ,ω₂= 0.04 rad/s
Moment of the inertia of the technician ,I₂= 5000 kg.m²
Lets assume that
Moment of the inertia of merry with respected to the ground=I₁
There is no any external torque ,that is why angular momentum of the system will be conserve.
Now by conserving angular momentum
I₁ ω₁=(I₁+I₂)ω₂
I₁ x 0.2 = (I₁ +5000 ) x 0.04
I₁ (0.2-0.04) = 5000 x 0.04

I₁ =1250 kg.m²
Answer:
Before start of slide velocity will be 14.81 m/sec
Explanation:
We have given coefficient of static friction 
Angle of inclination is equal to 


Radius is given r = 28 m
Acceleration due to gravity 
We know that 



So before start of slide velocity will be 14.81 m/sec
Use pythagorean's theorem for this, with 7 as a and 5 as b. pythagorean's theorem says that a^2 + b^2 = c^2, so 7^2 * 5^2 = c^2. this gives you 49 + 25 = c^2, so 74 = c^2. c = sqrt 74, which is approximately 8.60 km
Answer:
B: 1530 N
Explanation:
We know ghat formula for force of gravity is;
F_grav = G•m1•m2/d²
We are told that two clay masses produce a gravitational force of 340N.
Thus;
G•m1•m2/d² = 340
Now, if the distance is divided by 3 and the mass of one is divided by 2, we have;
F_g = (G × m1/2 × m2)/(d/3)²
Thus gives;
F_g = ½(G•m1•m2)/((1/9)d²)
Simplifying this gives;
F_g = (9/2)G•m1•m2/d²
From earlier, we saw that;
G•m1•m2/d² = 340
Thus;
F_g = (9/2) × 340
F_g = 1530 N
Explanation:
Given that,
Diameter of the wheel, d = 70 cm = 0.7 m
Initial angular speed, 
Final angular speed, 
Time, t = 4 s
(a) Angular acceleration,

(b) Tangential acceleration is :
Angular speed of the wheel after 2 seconds is :

Radial acceleration will be :

Hence, this is the required solution.