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Hunter-Best [27]
3 years ago
9

What is the approximate mass of air in a living room 4.5m×3.4m×2.9m? the density of air is 1.29 kg/m3?

Physics
1 answer:
topjm [15]3 years ago
6 0
First we have to calculate the volume of the living room:
V = L x W x H = 4.5 m * 3.4 m * 2.9 m
V = 44.37 m³
We know that Density = 1.29 kg/m²
D = m / V
m = D · V
m = 1.29 kg/m³ · 44.37 m³
m = 57.2373 kg ≈ 57.2 kg
Answer: The approximate mass of air in living room is 57.2 kg.
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Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

7 0
4 years ago
A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

Δd = 2Δr = -12.6 * 10^{-6} mm

Explanation:

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Explanation:

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100 seconds is the right thing
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Shtirlitz [24]

Answer:

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Explanation:

6 0
3 years ago
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