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3 years ago

I need the answer asap plz!!

Physics

1 answer:

nikdorinn [45]3 years ago

8 0

Answer:

266.67 K

Explanation:

Recall the formula for ideal gasses:

P * V = n * R * T

from which a proportion can be obtained if the volume of the gas and its number of moles remain constant. This is:

T / P = constant and therefore T1 / P1 = T2/ P2

Then for our case:

300 / 36 = T2 / 32

solving for T2:

T2 = 300 * 32 / 36 = 266.67 K

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A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

2 years ago

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A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train

masya89 [10]

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

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2 years ago

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A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must

nika2105 [10]

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The unbalanced moment = 10.6 - 5 = 5.6 kg.m

Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)

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3 years ago

Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m

ziro4ka [17]

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

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