Answer:
Lifting force, F = 21240 N
Explanation:
It is given that,
Mass of the helicopter, m = 1800 kg
It rises with an upward acceleration of 2 m/s². We need to find the lifting force supplied by its rotating blades. It is given by :
F = mg + ma
Where
mg is its weight
and "ma" is an additional acceleration when it is moving upwards.
So, 
F = 21240 N
So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.
Answer:
Explanation:
Gravitational force between two objects having mass m₁ and m₂ at a distance R
F = G m₁ m₂ / R²
Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²
= 1.01 x 10⁻⁶ N
b )
Force between baby and Jupiter
F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²
= 1.31 x 10⁻⁶ N
c )
Ratio = 1.01 / 1.31
= .77
Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
10.67m/s²
32N
Explanation:
Given parameters:
Mass of the body = 3kg
velocity of the mass = 4m/s
radius of circle = 0.75m
Unknown:
centripetal acceleration = ?
centripetal force = ?
Solution:
The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:
Centripetal acceleration = 
v is the velocity of the body
r is the radius of the circle
putting in the parameters:
Centripetal acceleration = 
Centripetal acceleration = 10.67m/s²
Centripetal force = m
m is the mass
Centripetal force = mass x centripetal acceleration
= 3 x 10.67
= 32N
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Answer:
The Magnifying power of a telescope is 
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective
= 
⇒
= 
⇒
= 295 cm
Focal length of eyepiece
= 2.7 cm
Magnifying power of a telescope is given by,



therefore the Magnifying power of a telescope is 