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3 years ago

PLS HELPPP!! ITS DUE TODAY!

Physics

1 answer:

Andrews [41]3 years ago

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A is an example of physical change

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if an object is being pulled by two forces, one 4n to the left, and the other 2 n to the right, what is the net force acting on

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2 N to the left

Explanation:

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3 years ago

What is a method to predict an earthquake?

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Measurements of microtremors, slope of the land, and motion of tectonic plates predict earthquakes. But we can't predict WHEN accurately enough yet to make the prediction useful to the general public.

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3 years ago

A 65 Kg roller-blade is accelerating at 5 m/s/s across the side walk. What force would be necessary for this acceleration to occ

Neporo4naja [7]

Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
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3 years ago

Truck drivers probably cannot see your vehicle if____.

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3 years ago

The oil level in a tank is 2 m above the ground. The tank cover is air tight and the air pressure above the oil surface is 150 k

Andrew [12]

Answer:

a) 24.692 m/s

b) 19.4 m

Explanation:

To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:

$\frac{P1}{pg}+\frac{V1^2}{2g}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}+Z2$

We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:

$\frac{P1}{pg}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}$

a) The velocity (V2) is:

$\frac{P1}{pg}+Z1=\frac{V2^2}{2g}$

$(\frac{P1}{pg}+Z1)(2g)=V2^2$

$V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}$

Substituting the known values we can get the velocity at the out:

Atmospheric pressure= 101000 Pa

Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3

$V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}$

$V2=24.692 m/s$

b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:

$\frac{P2}{pg}+\frac{V2^2}{2g}+Z2=\frac{P3}{pg}+\frac{V3^2}{2g}+Z3$

We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:

$\frac{V2^2}{2g}=\frac{P3}{pg}+Z3$

$Z3=\frac{V2^2}{2g}-\frac{P3}{pg}$

Substituting the known values, the height (Z3) is:

$Z3=\frac{(24.692 m/s)^2}{2(9.81 m/s2)}-\frac{101000 Pa}{(9.81 m/s)(880 kg/m3)}$

Z3=Maximum Height=19.376=19.4 m

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4 years ago