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tiny-mole [99]
3 years ago
12

A hot air balloon has a volume of 65,00 cubic feet of hot air. The balloon is at sea level on a standard day. If the temperature

of the hot air in the balloon is 1100F, determine the lifting force of the balloon.
Physics
1 answer:
seraphim [82]3 years ago
6 0
If you heat that air by 100 degrees F, it weighs about 7 grams less. Therefore, each cubic foot of air contained in a hot air balloon can lift about 7 grams. That's not much, and this is why hot air balloons are so huge -- to lift 1,000 pounds, you need about 65,000 cubic feet of hot air.
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what will be the final temperature is 0.05kg water at 0 degree centigrade is added to 0.25kg pf water at 90 degree centigrade sp
scoray [572]

Q in = Q out

0.25 x 4200 x (90-t)=0.05 x 4200 x (t-0)

1050 (90-t)=210t

94500-1050t=210t

94500=1260t

t=75°

7 0
2 years ago
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tatuchka [14]

Answer:

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Explanation:

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The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.
Doss [256]

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

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So it's initial momentum would be:

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It's final momentum is given by:

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Suppose a particle is accelerated through space (no gravity) by a 10 N force. Suddenly the particle encounters a second force of
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3 0
3 years ago
I need help with this question how to solve it for Brass and Cooper
Ksenya-84 [330]

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

3 0
1 year ago
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