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ss7ja [257]
3 years ago
7

A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 3 m/s just before s

triking the floor, and a speed of 6.5 m/s just after bouncing, find the average force of the floor on the ball. Assume that the ball is in contact with the floor for 0.32 s, and that the mass of the ball is 0.42 kg
Physics
1 answer:
Ganezh [65]3 years ago
6 0

Answer:

F=12.5N

Explanation:

Net force = rate of change of momentum

F = m*a

so find the change of momentum P

Pdown

P=m*v_1=0.42kg*3m/s

Pup

P=m*v_1=0.42kg*6.5m/s

dP = change in P

dP= 0.42kg (3- -.6.5)m/s =3.99 kg m/s

dT = 0.32 s

so

F = \frac{dP}{dt}=\frac{3.99Kg*m/s}{0.32s} =12.468 N

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Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?
Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion s=ut+\frac{1}{2}at^2

40=0\times 10+\frac{1}{2}a10^2

a=0.8944m/sec^2

Now from first equation of motion v=u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So v=u+at=0+0.8944\times 10=8.944m/sec

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3 years ago
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radiation

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2 years ago
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Explanation:

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2 years ago
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3 years ago
A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is in
Gnoma [55]

Answer:

v₂ = 97.4 m / s

Explanation:

Let's write the Bernoulli equation

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Index 1 is for tank and index 2 for exit

We can calculate the pressure in the tank with the equation

        P = F / A

Where the area of ​​a circle is

       A = π r²

E radius is half the diameter

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      A = π d² / 4

We replace

    P = F 4 / π d²2

    P₁ = 397 4 /π  0.058²

    P₁ = 1.50 10⁵ Pa

The water velocity in the tank is zero because it is at rest (v1 = 0)

The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

Since the pipe is horizontal y₁ = y₂

We replace on the first occasion

   P₁ = P₂ + ½ ρ v₂²

  v₂ = √ (P1-P2) 2 / ρ

  v₂ = √ [(1.50-1.013) 10⁵ 2/1000]

  v₂ = 97.4 m / s

6 0
3 years ago
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