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2 years ago

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An audience of 2250 fills a concert hall of volume 32000 m^3. If there were no ventilation, by how much would the temperature of

the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?

1 answer:

Nonamiya [84]2 years ago

7 0

246 and 64 minutes later

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Which of the fundamental forces explains the structure of atoms and molecules?

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The electromagnetic force<span> holds atoms and molecules together.
like a magnet's pull on steel.</span>

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A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

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2 years ago

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OverLord2011 [107]

Answer:

no habl

Explanation:

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2 years ago

A draft is made of a plastic block with a density of 650kg/m^3, and its dimensions are 2.0mx3.0mx5.0m ,what is the raft's appare

mixas84 [53]

Weight in water = mass of block - mass of volume of displaced water

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2 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

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2 years ago