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finlep [7]
3 years ago
10

A spherical hot air balloon of diameter 30.0 meters just floats (hovers) when the hot air inside has been heated to a density of

1.10 kg/m3 What is the weight of the balloon and cargo (not including the air inside) if the surrounding air has a density of 1.20 kg/m3? (VSphere = (4/3)nr3)
Physics
1 answer:
Rom4ik [11]3 years ago
5 0

Answer:

W = 166422.729 N

Explanation:

given,

diameter of the balloon = 30 m

density of the air = 1.10 Kg/m³

weight of the balloon and cargo = ?

density of the surrounding air = 1.20 kg/m³

we know,

Density = mass/volume

m = density x volume

m = \rho\times \dfrac{4}{3}\pi r^3

m =1.20 \times \dfrac{4}{3}\pi\times 15^3

m = 16964.6 Kg

Weight of the balloon

 W = m g

 W = 16964.6 x 9.81

W = 166422.729 N

Weight of the balloon and the cargo is equal to  W = 166422.729 N

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15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the
goldfiish [28.3K]

The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 800 kg is the mass of the sport car

u_1 = 13.0 m/s is the initial velocity of the car (taking its direction as positive  direction)

m_2 = 1200 kg is the mass of the truck

u_2 = 25.0 m/s is the initial velocity of the truck

v is the final combined velocity of the car and the truck, after the collision

Re-arranging the equation and substituting the values, we find the velocity after the collision:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(800)(13)+(1200)(25)}{800+1200}=20.2 m/s

And the positive sign indicates their final direction is the same as the initial direction of the two vehicles.

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5 0
3 years ago
A mother and her young child want to play on a seesaw at a playground. The child sits on the end of one side of the seesaw. Wher
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An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

5 0
3 years ago
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