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stepan [7]
3 years ago
11

PLEASE HELP ME ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!! What coefficient is needed in front of NH3 to balance the

equation below?
N2 + 3H2 → ?NH3

A. 1

B.2

C.3

D.4
Physics
1 answer:
Oksanka [162]3 years ago
6 0

Answer:

c

Explanation:

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What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s
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Answer:

f=272.15Hz

Explanation:

Given data

Length of tube L=0.632 m

Speed of sound v=344 m/s

To find

Fundamental frequency f

Solution

The fundamental frequency of the tube can be given as:

f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz

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Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.
ASHA 777 [7]

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

  • Tension is also a force having Newton as S.I unit.
  • The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using  Lami's theorem.

The sum of two given angles  = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}

Where

\alpha  = \beta = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

\frac{T}{sin\alpha } =  \frac{W}{sinY}

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

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Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

3 0
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