Given :
Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.
A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².
To Find :
The magnitude of F.
Solution :
Torque on hoop is given by :
( Moment of Inertia of hoop is MR² )
Putting value of M, R and α in above equation, we get :
Therefore, the magnitude of force F is 25 N.
Hence, this is the required solution.
Answer: It will be take 2.6 hours
Explanation: Please see the attachments below
Answer:
Explanation:
Let L be the length of the wire.
velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s
mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m
m = 7.25 x 10⁻⁷ kg / m
Tension in the wire = Mg , M is mass hanged from lower end.
= .4 x 9.8
= 3.92 N
expression for velocity of wave in the wire
, T is tension in the wire , m is mass per unit length of wire .
40.48 L =
1638.63 L² = 3.92 / (7.25 x 10⁻⁷)
L² = 3.92 x 10⁷ / (7.25 x 1638.63 )
L² = 3299.64
L = 57.44 m /s
Answer:
455165.278 m
Explanation:
P = Power = 3.7 W
v = Velocity = 10.7 m/s
Amount of fat = 4 g
1 gram of fat provides about 9.40 (food) Calories
Energy given by 4 g of fat
Time required to burn the fat
Distance traveled by the bird
The bird will fly 455165.278 m