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nikdorinn [45]
3 years ago
12

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Physics
1 answer:
Rama09 [41]3 years ago
7 0

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, \theta=21^{\circ}

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

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Answer:

ans:

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At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper
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Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
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Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
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We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.


PV = n*R*T

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Total Volume = 3.41 + 4.00 = 4.41


V1 * P1 = V2 * P2

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P2 = 101 kPa

V1 = 7.41 L

V2 = ????


<span> <span> 7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L </span> </span>


<span>You had 4 L now you need 31.94 more.</span> 
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