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natta225 [31]
3 years ago
6

Two forces, one of which double the other has resultant of 280N. if the direction of the large force is reversed and the remaini

ng unaltered, the resultant reduces to 160N. determine the magnitude of the force and the angle between the forces.
A body weighing 210N is resting on a horizontal table.A pull of 70 applied at an angle of 21 with the horizontal just causes the body to slide over the table. calculate the normal reaction and the coefficient of friction.​
Engineering
1 answer:
White raven [17]3 years ago
4 0

Answer:

point is 16 N.If their resultant is normal to the smaller force and has a magnitude of 8 N. Then two forces are.

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The denominator of a fraction is 4 more than the numenator. If 4 is added to the numenator and 7 is added to the denominator, th
kati45 [8]

Answer:

\frac{3}{7}

Explanation:

Lets take the numerator of the fraction to be = x

So the denominator of the fraction is 4 more than the numerator = x+4

The fraction is ;\frac{x}{4+x}

Now add 4 to the numerator and add 7 to the denominator as;

\frac{x+4}{4+x+7} =\frac{x+4}{x+11}

This new fraction is equal to 1 half =1/2

write the equation as;

\frac{x+4}{x+11} =\frac{1}{2}

perform cross-product

2(x+4 )=1( x+11 )

2x+8 = x + 11

2x-x = 11-8

x=3

The original fraction is;  

\frac{x}{4+x} =\frac{3}{3+4} =\frac{3}{7}

3 0
3 years ago
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water,
Scorpion4ik [409]

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water  so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

4 0
3 years ago
The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false
vovangra [49]

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

6 0
3 years ago
This is it dont anwser this is for my other account
Nezavi [6.7K]

Answer:

thanks for the poiunts

Explanation:

6 0
3 years ago
Ethane (component A - C2H6) and hydrogen (component B) are fed to a differential reactor where they react on the catalyst to for
Fofino [41]
HELP ILL GIVE MOST BRAINLY AND 50 POINTS
HURRY PLEASE component c it is a compound so it will break
4 0
3 years ago
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