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andrew-mc [135]
3 years ago
14

The block exerts a force F on the dart that is proportional to the dart’s velocity v and in the opposite direction, that is F =

-bv, where b is a constant.
Physics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

s = mV/2b

Assumptions: 1. the darts finally velocity is zero

2. The force being exerted on the dart by the block is constant and so the dart moves through the block with constant acceleration in the opposite direction. (Newton's second law)

3. Since the acceleration of the dart through the block is constant, then the equations of constant acceleration motion apply to the motion of the dart through the block.

Explanation:

Let a = acceleration of the dart through the block

V = velocity of the dart

m = mass of the dart

Vf = finally velocity of the dart

S = distance traveled by the dart through the block.

From Newton's second law of motion which states that the acceleration of a body is in the same direction as the net force acting on the body and is equal to the force divided by the mass. That is F = ma

Also F = -bv ........(1)

Therefore -bv = ma........(2)

From the equals of constant acceleration motion, Vf² = V² + 2aS

Vf = 0

0² = V² + 2aS

-2aS = V²

a = -V² / 2S

Substituting this expression for a in

Equation (2) above

-bV = m( - V²/2S)

On rearranging,

S = mV/2b

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Answer:

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Explanation:

We are given that

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We have to find the dh/dt when the water is 4 feet deep.

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r=\frac{2}{5}h

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Substitute the values

Volume of cone , V=\frac{1}{3}\pi(\frac{2}{5}h)^2h=\frac{4}{75}\pi h^3

Differentiate w.r.t t

\frac{dV}{dt}=\frac{12}{75}\pi\times h^2\times \frac{dh}{dt}

Substitute the values

8=\frac{12}{75}\times\pi (4)^2\times \frac{dh}{dt}

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3 0
4 years ago
How high would a skater need to start on a previous incline to make it up and around a loop that is 6.1meters high?
OlgaM077 [116]

The skater need to start 7.525 meter high.

Let,

The skater need to start on a previous incline at height = h.

Mass of the skater = m

Given, height of the loop = 6.1 meter.

Radius of the loop; r = 6.1/2 meter = 3.05 meter.

Let to make it up and around a loop that is 6.1meters high, the man required minimum velocity v on the highest point of the loop.

So, centripetal force at highest point = weight of the man

⇒ mv²/r = mg

⇒ v = √(gr)

Then, potential energy of the skater at height h is = mgh.

And, minimum energy  of the skater at the highest point of the loop is = potential energy + minimum kinetic energy.

= mg(2r) + 1/2 mv²

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According to conservation of energy,

potential energy of the skater at height h = minimum energy  of the skater at the highest point of the loop

⇒ mgh = 5/2 mgr

⇒ h = 5/2 r =( 5/2 )× 3.05 meter = 7.525 meter.

Hence, required height is 7.525 meter.

Learn more about energy here:

brainly.com/question/1932868

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If water is leaking from a certain tank at a constant rate, is it leaking at a rate that is greater than 12 liters per hour?(1 l
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Answer:

Yes  it is possible because 3.33 is greater than 2.

Explanation:

Given that,

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Suppose water is leaking from the tank at a rate that is greater than 2 milliliters per second.

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7 0
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