answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
saveliy_v [14]
2 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]2 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.
You might be interested in
A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height
Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

v_x = 23.2 m/s

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

v_y(t)= v_{y0} -gt (1)

where

v_{y0}=0 is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s

Substituting t into (1) we find the final vertical velocity

v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}

6 0
2 years ago
A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0
2 years ago
Read 2 more answers
Two very large, flat plates are parallel to each other. Plate A, located at y=1.0 cm, is along the xz-plane and carries a unifor
Dmitry [639]

Answer:

 E ≈ 1.70 10⁵ N/C

Explanation:

The electric field is a vector quantity, so we can calculate the field of each plate and then add them. To calculate the field of a plate let's use Gauss's law

       Φ = ∫ E. dA = q_{int} / ε₀

To apply this law we must create a Gaussian surface that takes advantage of the symmetry of the problem. The electric field lines on the surface are perpendicular, so the Gaussian surface that will be a cylinder with the base parallel to the plate.

On this surface the normal to the base (A) is parallel to the field lines whereby the scalar product is reduced to the ordinary product. The normal on the sides of the cylinder is perpendicular to the field, therefore, the product scale is zero.

        ∫I E dA = q_{int}  /ε₀

Let's look for the load under the cylinder, let's use the concept of load density

        σ =  q_{int} / A

         q_{int} = σ A

Let's write Gauss's law for this case

       E A =  q_{int} /ε₀  

       E A = σ A / ε₀

       E = σ / ε₀

As the field is emitted for each side of the plate the value to only one side is

      E = G / 2ε₀  

This expression is the same for each plate, now let's add the electric field at the requested point

     R = (0.50, 0.00, 0.00) cm

We see that this point is on the X axis, between the plates that are at the points y = -1.0 cm and y = 1.0 cm, as the plates are very large the test point is between them

The negative plate has an incoming field and the positive plate has an outgoing field, the test load is always positive. The field due to the negative plate goes to the left, the field through the positive plate goes to the left at this point whereby two are added

     E = E_ + E +

     E = σ1 / 2ε₀  + σ2 / 2ε₀  

     E = 1 / 2o (σ1 + σ2)

Let's calculate the value

     E = 1/2 8.85 10⁻¹² (1.00 10⁻⁶ + 2.00 10⁻⁶)

     E = 3 10⁻⁶ / 17.7 10⁻¹²

     E = 1,695 10⁵ N / C

     E ≈ 1.70 10⁵ N/C

6 0
2 years ago
HEY GUYS IS THIS TRUE OR FALSE????????
slavikrds [6]
Answer:
false

explanation:
8 0
2 years ago
Read 2 more answers
Which type of wave can transmit energy through a vacuum?
m_a_m_a [10]
The wave that can transmit energy through a vacuum is a electromagnetic wave
4 0
2 years ago
Read 2 more answers
Other questions:
  • Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
    5·2 answers
  • Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is
    7·1 answer
  • In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired
    8·1 answer
  • An electron is pushed into an electric field where it acquires a 1-v electrical potential. suppose instead that two electrons ar
    5·2 answers
  • A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq
    9·2 answers
  • A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis
    9·1 answer
  • It is initiated by the pressure gradient force. b. It blows from regions of high pressure to regions of low pressure. c. The dir
    14·1 answer
  • A weightlifter lifts a 13.0-kg barbel from the ground an moves it a distance of 1.3 meters. What is the work se does on the barb
    9·1 answer
  • A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c
    6·2 answers
  • Using all three of Newton's laws of motion, carefully describe the motion of a baseball beginning when the ball is resting in th
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!