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Citrus2011 [14]
3 years ago
11

The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of

the sun. (Take e = 1, and σ = 5.67 × 10−8W/m2⋅K4).
a. 3.95 × 1026 W
b. 5.17 × 1027 W
c. 9.62 × 1028 W
d. 6.96 × 1030 W
Physics
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

a. 3.95\times10^{26}W

Explanation:

T = temperature of the surface of sun = 5800 K

r = Radius of the Sun = 7 x 10⁸ m

A = Surface area of the Sun

Surface area of the sun is given as

A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}

e = Emissivity = 1

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the sun is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W

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Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

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\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

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\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
3 years ago
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