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jeyben [28]
4 years ago
11

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov

ements. If the linear velocity of the ball relative to the elbow joint is 17.1 m/s at a distance of 0.470 m from the joint and the moment of inertia of the forearm is 0.500 kg·m^2. What is the rotational kinetic energy of the forearm?
Physics
1 answer:
Brums [2.3K]4 years ago
6 0

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J

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Answer:

Magnetic field will be ZERO at the given position

Explanation:

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Whats the role of the suns gravity in the solar system
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Answer:

v₂ = 5131.42 m/s

Explanation:

given,

radius of the planet = r₁ = 9.00×10⁶ m

mass of the satellite = m₁ = 68 Kg

orbital radius = r₁ = 8 x 10⁷ m

orbital speed = v₁ = 4800 m/s

mass of second satellite = m₂ =  84.0 kg

orbital radius = r₂ = 7.00×10⁷ m

orbital speed of second satellite = v₂ = ?

using orbital speed of satellite

            v = \sqrt{\dfrac{GM}{r}}

     so,

            v \alpha \dfrac{1}{\sqrt{r}}

now,

            \dfrac{v_1}{v_2} =\sqrt{\dfrac{r_2}{r_1}}

            \dfrac{v_2}{4800} =\sqrt{\dfrac{8\times 10^7}{7 \times 10^7}}

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3 years ago
6. How much voltage would be necessary to generate 100.5 amps of current in a circuit that
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2 years ago
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho
cupoosta [38]

Answer:

xf = 5.68 × 10³ m  

yf = 8.57 × 10³ m  

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

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simplfy it we get

xf = 0 + vicosθ × t + 0

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xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

yf = 8.57 × 10³ m  

5 0
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