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Lady_Fox [76]
3 years ago
15

A regulation basketball has a 32 cm diameter

Physics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

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Julli [10]

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6 0
3 years ago
A boy whirls a ball on a string in a horizontal circle of radius 1 m. How many revolutions per minute must the ball make if its
spayn [35]

Answer:

Nearest, the revolutions per minute will be 29.

Explanation:

Given that,

Radius of circle = 1 m

Acceleration a =g

We know that,

Angular frequency is defined as,

\omega=2\pi n

Where, n = number of revolutions in one second

We need to calculate the revolutions in one second

Using formula of centripetal acceleration

a=\omega^2r

Put the value of a and ω

g=(2\pi n)^2r

n=\sqrt{\dfrac{g}{r}}\times\dfrac{1}{2\pi}

Put the value into the formula

n=\sqrt{\dfrac{9.8}{1}}\times\dfrac{1}{2\pi}

n=0.49

We need to calculate the revolutions per minute

Using value for the revolutions per minute

n=0.49\times60

n=29.4

Hence, Nearest, the revolutions per minute will be 29.

7 0
3 years ago
Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
nignag [31]

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

           t=0 s

           t= 2 s

The ball position

          x = 40 2

          x = 80 feet

           

          y = 39 2

          y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

             x_total = 150 feet

             y _total = 80 feet

so we can see that the ball came into the field

6 0
3 years ago
You throw a ball into the air. Which two forces cause the ball to gradually stop moving upward and then fall back to Earth?
Virty [35]

Answer:

I'm pretty sure the answer is D

Explanation:

Honestly it's just a guess so let me know if it's right :3

3 0
3 years ago
Read 2 more answers
EXTREME HELP!!!!
mrs_skeptik [129]

the anwser is c is always the best anwser


8 0
3 years ago
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