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A gardener rested his hose against a shed. The hose made a 50° angle with the ground
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gardener waters a lawn. the hose ejects 5*10^-4 m^3 of water per second. The area of the mouth of the hose is 5 cm^2. Find the
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To find the backward force exerted on the gardener, we need to calculate the total amount of water
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A gardener waters a lawn. the hose ejects 5*10^-4 m^3 of water per second. The area of the mouth of the hose is 5 cm^2. Find the
Top answer:
To find the backward force exerted on the gardener, we need to determine the speed at which the
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A fire hose has a circular cross-section which has a diameter of 20 cm. The hose is full of water and the water in the hose is
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Area of a cross-section is π(10)^2 = 100π cm^2 5 m = 500 cm So a volume of 100π(500) or 50000π
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A gardener holds a hose 0,75m above the ground such that the waterfront out horizontally and hits the ground at a point 2 meters
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123.56
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If you think of electrical current as a flow of water coming from a hose, which of these is voltage most like?
A the water
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A. the water pressure of the hose Voltage is most like the water pressure of the hose because it is
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A gardener holds a hose 0.75m above the ground such that the water shoot out horizontally and hits the ground at a point 20m
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time in air: hf=hi-1/2 g t^2 0=.75-4.9 t^2 solve for t then, in the horizontal 20=vi*t solve for vi
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A gardener holds a hose 0.75m above the ground such that the water shoots out horizontally and hits the ground at appoint 2.0 m
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so, how long does it take to fall 0.75 meters? 4.9t^2 = 0.75 Using that time value, the constant
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a gardener holds a hose 0.75 m above the ground such that the water shoots out horizontally and hits the ground at a point 6.0
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how long does it take to fall 0.75m? 4.9 t^2 = 0.75 find t, and then distance = speed * time
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A gardener holds a hose 0.75 m above the ground such that the water shoots out horizontally and hits the ground at the point
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how long does it take to fall 0.75m? 4.9t^2 = 0.75 t = 0.39 so the speed of the water is 2.0/0.39 =
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A gardener holds a hose 0.75 m above the ground such that water shoots out horizontally and hits the ground at a point 2 m away.
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given dy= -0.75m dx= 2m g= 9.8 m/s² sol vx formula dx=vxt dx/t=vxt/t vx=dx/t solve for t form;
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