Answer:
I found this don't know if its any use or not
Answer:
B = 62.9 N
Explanation:
This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid
B = ρ g V
write the equilibrium equation
T + B -W = 0
B = W- T (1)
use the density to write the weight
ρ = m / V
m = ρ V
W = ρ g V
substitute in 1
B = m g -T
B =
g V - T
To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³
calculate
B = 7850 9.8 1.20 10⁻³ - 29.4
B = 92.3 - 29.4
B = 62.9 N
Answer:
-.457 m/s^2
Explanation:
Actual weight = 60 .3 (9.81) = 591.54 N
Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift
60.3 ( 9.81 - L ) = 564
solve for L = .457 m/s^2 DOWNWARD
so L = - .457 m/s^2
Answer : Height, h = 20.4 m
Explanation :
It is given that,
Mass of an object, m = 500 g = 0.5 kg
Gravitational potential energy, PE = 100 J
The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :
PE = m g h
where,
h is the height of the cliff.

h = 20.40 m
So, the height of the cliff is 20.4 m.
The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.