Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, 
= 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
The body will take 20 seconds to cover a distance of 1000 m i.e. 1 km
u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2
Hope it helps
Answer:
W = 9.6 N
Explanation:
Given that,
Area on 1 foot, A = 0.6 m²
Pressure, P = 16 Pa
The pressure is given by force acting per unit area. So,
So, the required weight is 9.6 N.
