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Brilliant_brown [7]
3 years ago
14

The spool has a mass of 50 kg and a radius of gyration of ko = 0.280 m. if the 20-kg block a is released from rest, determine th

e velocity of the block when it descends 0.5 m.
Physics
1 answer:
Rzqust [24]3 years ago
5 0
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis 
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is 
T = 1/2 I₀w² + 1/2mAVA²

= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation  is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
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Answer:

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Explanation:

Given the following data:

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Initial temperature, T1 = 10°C to Kelvin = 10 + 273 = 283K

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Specific heat capacity = 4182 j/kgK

To find the final temperature;

Heat capacity is given by the formula;

Q = mcdt

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m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making dt the subject of formula, we have;

dt = \frac {Q}{mc}

Substituting into the equation, we have;

dt = \frac {120000}{0.9*4182}

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But, dt = T2 - T1

T2 = dt + T1

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8 0
3 years ago
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The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

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Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

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so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

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a = \frac{1.46}{0.145}

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Part B)

As per kinematics we know that

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4 0
3 years ago
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Answer:

2Micro Farahds

Explanation:

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