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saveliy_v [14]
3 years ago
8

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

y space? The Stefan-Boltzmann constant is 5.67 × 10-8 W/(m2 ∙ K4).
A.) 9.9 mW

B.) 3.7 W

C.) 7.1 W

D.) 1.4 kW
Physics
1 answer:
Yuri [45]3 years ago
6 0
The total power emitted by an object via radiation is:
P=A\epsilon \sigma T^4
where:
A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW
So, the correct answer is D.
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A horizontal force F~ is applied to a block of mass m = 1 kg placed on an inclined
vova2212 [387]

Hi there!

To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.

Recall:

\large\boxed{F_D = \mu N}}

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

\large\boxed{N = Mgcos\theta + Fsin\theta}}

We can plug in the known values to solve for one part of the normal force:

N = (1)(9.8)(cos30)  + F(.5) = 8.49  + .5F

Now, we can plug this into the equation for the dynamic friction force:

Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F

For a block to move with constant speed, the summation of forces must be equivalent to 0 N.

If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:

Fcosθ = 1.697 + .1F

Solve for F:

Fcos(30) - .1F = 1.697

F(cos(30) - .1) = 1.697

F = 2.216 N

6 0
3 years ago
One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t
Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2\frac{m}{s^{2} }

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=FΔx=(0.400m)\\

W_{ext}=20.4 Nm

U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2}  \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}

To determine, the acceleration we solve the following equation for a:

F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }

8 0
4 years ago
Which of the following changes when an unbalanced force acts on an object?
KengaRu [80]
What are the choices? then i will answer
3 0
4 years ago
A sports car moving at constant speed travels 164 m in 13.77 s. If it then brakes and comes to a stop in 3.6 s. What is its acce
Sergio039 [100]

Answer:

a = - 0.3376 g's

Explanation:

The sports car has a constant speed when travelling. Covered 164 m in 13.77 s. Thus, speed = 164/13.77 m/s

It brakes and now comes to a stop in 3.6 s.

Thus final velocity = 0 m/s

Formula for acceleration is;

a = (v - u)/t

a = (0 - (164/13.77))/3.6

a = -3.308 m/s²

In terms of g's", where 1.00 g = 9.80 m/s², we have;

a = -3.308/9.8 g's

a = - 0.3376 g's

4 0
3 years ago
*please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a dis
pishuonlain [190]

(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

(d) The potential difference between A and B is 56.7 V.

<h3>Electric force on the sphere</h3>

The electric force on the sphere is calculated as follows;

F = Eq

where;

  • E is electric field
  • q is the charge

F = 300 x (44 x 10⁻⁹)

F = 1.32 x 10⁻⁵ N

The direction of the force is towards the right.

<h3>Work done on the sphere</h3>

W = Fd

W = 1.32 x 10⁻⁵ N  x 0.189 m

W = 2.5 x 10⁻⁶ J

<h3>Change of the electric potential energy </h3>

The change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.

<h3> Potential difference between A and B</h3>

VB − VA =  Ed

VB − VA =  300 N/C  x   0.189 m

VB − VA =  56.7 V

Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

The potential difference between A and B is 56.7 V.

Learn more about potential difference here: brainly.com/question/9060304

#SPJ1

6 0
2 years ago
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