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artcher [175]
3 years ago
8

Imagine that a tank is filled with water. The height of the liquid column is 7 meters and the area is 1.5 square meters (m2). Wh

at's the force of gravity acting on the column of water?
Physics
2 answers:
Korolek [52]3 years ago
4 0

Answer: 1.03 × 10⁵ N

Explanation:

Force of gravity due to liquid column in a tank is given by the product of pressure due to the liquid column and area on which this pressure is acting.

Force, F = P × A

Pressure, P = ρ g h

where, ρ is the density of the liquid

g is the acceleration due to gravity

h is the height of the liquid column

It is given that, h = 7 m

Area, A = 1.5 m²

g = 9.81 m/s²

density of water, ρ = 1000 kg/m³

⇒ F =  ρ g h × A = 1000 kg/m³ × 9.81 m/s² × 7 m × 1.5 m² = 103005 N

Hence, force of gravity acting on the column of water is 1.03 × 10⁵ N

Marta_Voda [28]3 years ago
3 0
The answer to your question is 102,900N
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4 0
2 years ago
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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marusya05 [52]

Answer:

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A-200 kg
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3 years ago
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Can someone help me ASAP
Phantasy [73]

Well I don't know.  Let's actually LOOK at the picture and see if that helps.

A,  B,  C,  and D all have the same TOTAL length, but  A  has the most waves crammed into that same total length.

By golly, that means the length of <u><em>each</em></u> wave in  A  must be shorter than each wave in  B,  C,  or D.

The correct choice is <em> A </em>.  Looking at the picture did the trick !

7 0
2 years ago
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