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scZoUnD [109]
3 years ago
15

A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet w

ill the ball have traveled when it hits the ground for the fourth time
Physics
1 answer:
Crank3 years ago
4 0

The total distance travelled by the ball after the fourth impact is 275 feet.

<u>Explanation:</u>

Given-

Height, h = 100 feet

Rebounds half the distance

Distance in feet for the fourth time, x = ?

For the first time, the distance travelled by the ball is, x = 100 feet

For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)

So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet

For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)

So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet

For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)

So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet

Therefore, total distance travelled by the ball after the fourth impact is 275 feet.

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Which statement correctly describes the relationship between current, voltage, and resistance? If we
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Answer: Option (b) is the correct answer.

Explanation:

According to ohm's law, the relationship between voltage, resistance, and current is that current passing through a conductor is directly proportional to the voltage over resistance.

Mathematically,           I = \frac{V}{R}

From this relationship we can see that when we decrease the voltage, and do not change the resistance, the current will also decrease. As current is directly proportional to voltage and inversely proportional to resistance.


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So, the ship will have to travel 156.2 km to the northwest direction.

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If a proton is released at the equator and falls toward earth under the influence of gravity, the magnetic force on the proton w
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a race car driver achieved a speed of 53m/s in 14 seconds after taking off from rest from the starting line. what was the averag
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. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0
3 years ago
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