Ask question

Ask question

All categories

3 years ago

8

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the we

ights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. Then:

1 answer:

Alex_Xolod [135]3 years ago

7 0

Answer:

w = 2w₀ the angular velocity of man doubles

Explanation:

In this exercise, releasing the weights reduces the moment of inertia

I= I₀ / 2

Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved

L₀ = L

I₀ w₀ = I w

I₀ w₀ = I₀ / 2 w

w = 2w₀

therefore the angular velocity of man doubles

You might be interested in

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

A coiled telephone cord forms a spiral with 90.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.0 cm. Determine t

Vinvika [58]

Answer:

2.36 μ H

Explanation:

Given,

Number of turns= 90

diameter = 1.3 cm = 0.013 m

unscratched length = 57 cm = 0.57 m

Area, A = π r²

= π x 0.0065² = 1.32 x 10⁻⁴ m²

we know,

$L = \dfrac{\mu_0N^2A}{l}$

$L = \dfrac{4\pi \times 10^{-7}\times 90^2\times 1.32\times 10^{-4}}{0.57}$

L = 2.36 μ H

Hence, the inductance of the unstretched cord is equal to 2.36 μ H

8 0

3 years ago

Definition of graph?

xeze [42]

<u>Answer:</u>

1. A graph is defined as <em>" A Diagram represents a system of connections or interrelations among two or more things by a number of different dots, lines etc".</em>

2. In simple words <em>"Graph is a representation of any object or a physical structure by dots, lines, etc.</em>

6 0

3 years ago

How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?

salantis [7]

Answer:

$Q = 4.63 \times 10^6 J$

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

$s = 240 J/kg C$

now we will have

$Q = ms\Delta T + mL$

$\Delta T = 961.8 - 20$

$\Delta T = 941.8 degree C$

now from above equation

$Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)$

$Q = 3.1 \times 10^6 + 1.53 \times 10^6$

$Q = 4.63 \times 10^6 J$

3 0

3 years ago

A boy is using a rope to pull a sled to the left on a horizontal surface. The rope is parallel to the surface. What are the dire

trasher [3.6K]

Answer:

Explanation:

Check attachment

3 0

4 years ago