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Naya [18.7K]
3 years ago
12

The number of molecules in a container is tripled and the kelvin temperature doubled. the volume remains unchanged. the new pres

sure will be how many times greater than the original pressure?
Physics
1 answer:
Nonamiya [84]3 years ago
3 0
N2 = 3*n1
T2 = 2*T1
V1 = V2

(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1

Since P2 is 6P1, it is 6 times greater than original pressure
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Which type of friction force acts when there is no motion?
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Its called static friction.
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3 years ago
An airplane of mass 39,043.01 flies horizontally at an altitude of 9.2 km with a constant speed of 335 m/s relative to Earth. Wh
Likurg_2 [28]

Answer:

1.2 x 10¹¹ kgm²/s

Explanation:

m = mass of the airplane = 39043.01

r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m

v = speed of airplane = 335 m/s

L = Angular momentum of airplane

Angular momentum of airplane is given as

L = m v r

Inserting the values

L = (39043.01 ) (335) (9200)

L =  (39043.01 ) (3082000)

L = 1.2 x 10¹¹ kgm²/s

8 0
3 years ago
Based on the map, which area has a scarcity of fresh water?
Westkost [7]

Answer:

North Africa

Explanation:

Hope this helps

6 0
2 years ago
Read 2 more answers
Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
garri49 [273]

Answer:

5.2\ \text{m/s}

70^{\circ} south of east

Explanation:

v_a = 3 m/s

\theta_a = 20^{\circ} north of east

v_b = 6 m/s

\theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

x and y component of v_a

v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

6 0
3 years ago
I need help, please!
qwelly [4]

Answer:

I don't know what you mean but I think I am going

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