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3 years ago

11.) An object that is NOT accelerating could be sitting still or:

Physics

2 answers:

mart [117]3 years ago

7 0

Moving slowly is the correct answer

Aleonysh [2.5K]3 years ago

4 0

Answer:

moving slowly is the answer

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

6 0

4 years ago

What is the value of the initial horizontal velocity?

Dima020 [189]

Initial Velocity is the velocity at time interval t = 0 and it is represented by u. It is the velocity at which the motion starts.

<h2>I hope this helps...</h2><h2>Good day :)</h2>

4 0

3 years ago

A gold bar 20.0kg at 35.0°c is placed in a large insulated 0.8kg glass container at 15°c and 2.0kg of water at 25°c.. calculate

Oksanka [162]

Answer:

The final equilibrium temperature is approximately 26.69 °C

Explanation:

The heat transferred, ΔQ, from a hot body to a cold one is given by the following formula;

ΔQ = m·c·ΔT

Where;

m = The mass of the body

c = The specific heat capacity of the body

ΔT = The temperature change of the body

The given mass of the gold bar, m₁ = 20.0 kg

The initial temperature of the gold bar, T₁ = 35.0 °C

The specific heat capacity of gold, c₁ = 0.13 kJ/(kg·K)

The mass of the glass container, m₂ = 0.8 kg

The initial temperature of the glass container, T₂ = 15°C

The specific heat capacity of glass, c₂ = 0.792 kJ/(kg·K)

The mass of the added water, m₃ = 2.0 kg

The initial temperature of the added water, T₃ = 25°C

The specific heat capacity of water, c₃ = 4.2 kJ/(kg·K)

The heat lost by the gold = The heat gained by the glass and the water

Let 'T' represent the temperature at the final equilibrium, we have;

m₁·c₁·ΔT₁ = m₂·c₂·ΔT₂ + m₃·c₃·ΔT₃

Where;

ΔT₁ = T₁ - T

ΔT₂ = T - T₂

ΔT₃ = T - T₃

∴ 20.0 × 0.13 × (35 - T) = 0.8 × 0.792 × (T - 15) + 2.0 × 4.2 × (T - 25)

Expanding and collecting like terms (using a graphing calculator) gives;

91 - 2.6·T = 9.0336·T - 219.504

9.0336·T + 2.6·T = 219.504 + 91 = 310.504

11.6336·T = 310.504

T = 310.504/11.6336 ≈ 26.69

The final equilibrium temperature, T ≈ 26.69 °C.

4 0

3 years ago

A book with a mass of 2 kg sits on top of a book shelf at a height of 3 m. How much potential energy does the book have

Arada [10]

Answer:

Answer is 98 joules

Explanation:

P. E=2*9.8*3=98

S.i unit of potential energy ⚡ is joules

So answer is 98 joules

5 0

3 years ago

A hammer taps on the end of a 4.0-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pu

Mashcka [7]

Answer: c*m on now

Explanation:

u dumb b***h

6 0

4 years ago