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emmasim [6.3K]
3 years ago
12

When a downward force is applied at a point 0.60 m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10–7 k

g at a point 0.40 m to the right of the fulcrum. What is the magnitude of the downward force
Physics
1 answer:
Sliva [168]3 years ago
4 0

Answer:

force = 6.53×  10^{-7} N

Explanation:

Given data

downward force = 0.60 m

mass m  = 10^{-7} kg

distance h = 0.40 m

to find out

magnitude of the downward force

solution

we know here mg is apply 0.4 m away from support and

so applied force is d =  0.6 m from support

so

by balancing torque we can calculate force

that is

force = mass  × g ×  h / d

put here all these value

force =  mass  × g ×  h / d

force =  ( 10^{-7}  × 9.81 ×  0.4 )  / 0.6

force = 6.53×  10^{-7} N

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3 years ago
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3 years ago
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

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Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

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3 years ago
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