Answer:
563.86 N
Explanation:
We know the buoyant force F = weight of air displaced by the balloon.
F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m
So, F = ρgV = ρg4πr³/3
substituting the values of the variables into the equation, we have
F = 1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3
= 1691.58 N/3
= 563.86 N
Pressure= hqg
H=depth
q=density
g=gravity
h=0.2
q=7
g=10
0.2*7*10= 14pa
FINAL ANSWER = 14pa
Gravitational potential energy = mass x acceleration due to gravity x height
GPE=mgh
4620=mx9.81x8.4
4620/(9.81x8.4)=m=56.1 kg
Answer:
d = 10.2 m
Explanation:
When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:
where,
m = mass of car
v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)
v = 10 m/s
F = force on the car in direction of inclination = W Sin θ
W = weight of car = mg
θ = Angle of inclinition = 30°
d = distance covered up the ramp = ?
Therefore,
<u>d = 10.2 m</u>
Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
