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BigorU [14]
3 years ago
9

A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the

balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.5 m. The total mass of the empty balloon and basket is mb = 123 kg and the total volume is Vb = 0.068 m3. Assume the average person that gets into the balloon has a mass mp = 74 kg and volume Vp = 0.077 m3.What is the volume of helium in the balloon when fully inflated?
Physics
1 answer:
max2010maxim [7]3 years ago
3 0

Answer:

V = 381.70 m³

Explanation:

ρ air = 1.28 kg / m³

ρ helium = 0.18 kg / m³

R = 4.5 m

Vb = 0.068 m³

mb = 123 kg

To determine the volume of helium in the balloon when fully inflated

V = 4 / 3 π * R ³

V = 4 * π / 3 ( 4.5 m )³

V = 381.70 m³

To determine the mass total

m = ρ helium * V

m = 0.18 kg / m³  * 381.70  m³

m = 68.70 kg

mt = ( 68.70 + 123 )kg

mt = 191.70 kg

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Find the kinetic energy of a 0.1-kilogram toy truck moving at the speed of 1.1 meters per second.
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Answer:

0.061 J

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

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v is its speed

For the toy truck in the problem, we have

m = 0.1 kg is its mass

v = 1.1 m/s is its speed

Putting the numbers into the equation, we find

K=\frac{1}{2}(0.1 kg)(1.1 m/s)^2=0.061 J

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Explain the method to measure the external diameter of a sphere
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the sphere is the distance r is the radius of the ball and the circle is the center of the mathematical ball, as the center and the radius of the sphere is to respectively.

The ball and sphere have not to be maintained mathematical references as solid references. A sphere of any radius is centered at the number of zero.

Explanation: Hope this helps and good luck :)

3 0
3 years ago
A pendulum consists of a 2.0-kg block hanging on a 1.5-m length string. A 10-g bullet moving with a horizontal velocity of 900 m
pav-90 [236]

Answer:

The maximum height above its initial position is:

h_{max}=1.53\: m

Explanation:

Using momentum conservation:

m_{b}v_{ib}=m_{B}v_{fB}+m_{b}v_{fb} (1)

Where:

  • m(b) is the mass of the bullet
  • m(B) is the mass of the block
  • v(ib) is the initial velocity of the bullet
  • v(fb) is the final velocity of the bullet
  • v(fB) is the final velocity of the block

Let's find v(fb) using equation (1)

m_{b}(v_{ib}-v_{fb})=m_{B}v_{fB}

v_{fB}=\frac{m_{b}(v_{ib}-v_{fb})}{m_{B}}

v_{fB}=\frac{0.1(900-300)}{2}

v_{fB}=30\: m/s

We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.

\frac{1}{2}m_{B}v_{fB}=m_{B}gh_{max}

h_{max}=\frac{1}{2g}v_{fB}

h_{max}=\frac{1}{2(9.81)}30

h_{max}=1.53\: m

I hope it helps you!

8 0
3 years ago
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