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Dmitrij [34]
3 years ago
10

What is the most common grip used in racket sports?

Physics
1 answer:
Firlakuza [10]3 years ago
6 0
The continental ...........
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The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)
Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is A_{x} = 0

the y-component of the force at A is A_{y}  = 400 N

the couple acting at A is; M_{A} = 146 N-m

b)

the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑M_{L} = 0

Vector force acting at A, F_{A} = A_{x}i + A_{y}j

Now, From the image, we have;

a)

∑f = 0

F_{A} - 600j + 200j = 0i + 0j

A_{x}i + A_{y}j - 600j + 200j = 0i + 0j

A_{x}i + (A_{y} - 400)j = 0i + 0j

now by equating i- coefficients'

A_{x} = 0

so, the x-component of the force at A is A_{x} = 0

also by equating j-coefficient

A_{y} - 400 = 0

A_{y}  = 400 N

hence, the y-component of the force at A is A_{y}  = 400 N

we also have;

∑M_{L} = 0

M_{A}  - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

M_{A} - 30 N-m - 228 N-m + 112 Nm = 0

M_{A} - 146 N-m = 0

M_{A} = 146 N-m

Therefore, the couple acting at A is; M_{A} = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑M_{R} = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(A_{y} ) + M_{A}

∑M_{R} = (108  N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑M_{R} = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑M_{R}  = 0

Therefore, the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

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How is the atomic mass of an atom determined?​
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Answer:

Atomic mass is defined as the number of protons and neutrons in an atom

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19. What do vibrations create?<br><br> Wavelengths<br> sound waves<br> energy<br> electricity
OverLord2011 [107]

Answer:

Energy

Explanation:

When an object vibrates, it creates kinetic energy that is transmitted by molecules in the medium. As the vibrating sound wave comes in contact with air particles passes its kinetic energy to nearby molecules. As these energized molecules begin to move, they energize other molecules that repeat the process.

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I forgot this someone help
dexar [7]

Answer:

A. water + carbon dioxide + sunlight = oxygen + <u>glucose </u><u>!</u>

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Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant
Cerrena [4.2K]

Answer:

the  force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81  + \frac{70 \times 120^2}{100}\\\\= 10766.7 N

Hence, the  force exerted by the seat on the pilot is 10766.7 N

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3 years ago
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