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tensa zangetsu [6.8K]
3 years ago
5

A positive statement is:________. a. reflects oneâs opinions. b. can be shown to be correct or incorrect. c. a value judgment. d

. based upon an optimistic judgment.
Physics
1 answer:
kenny6666 [7]3 years ago
8 0

Answer:

b

Explanation:

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The radius of Mars is 3397 km; and its mass is 6.42x10²³ kg. What is the acceleration due to gravity at an altitude of 485 km ab
Gekata [30.6K]

Answer:nfeergnjknrtgrth

Explanation:

4 0
3 years ago
Read 2 more answers
a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin
Daniel [21]

This question involves the concepts of Wein's displacement law and characteristic wavelength.

The blackbody temperature will be "3.22 x 10⁵ k".

<h3>WEIN'S DISPLACEMENT LAW</h3>

According to Wein's displacement law,

\lambda_{max} T = c\\\\T=\frac{c}{\lambda_{max}}

where,

  • \lambda_{max} = characteristic wavelength = 9 μm = 9 x 10⁻⁹ m
  • T = temperature = ?
  • c = Wein's displacment constant = 2.897 x 10⁻³ m.k

Therefore,

T=\frac{2.897\ x\ 10^{-3}\ m.k}{9\ x\ 10^{-9}\ m}

T = 3.22 x 10⁵ k

Learn more about characteristic wavelength here:

brainly.com/question/14650107

7 0
3 years ago
6) Repeat experiment 5) a) but now use the stored energy meter. Why do you think the change takes place. charge remains constant
PilotLPTM [1.2K]

Answer:

the energy comes from the increase in the electric field

Explanation:

The capacitance is

      C = ε₀ A / d

The electric charge on the condenser plates

    Q = C ΔV

     

The stored electrical energy is

       U = ½ C ΔV²

       ΔV = E d

     U = ½ (ε₀ A / d) (E d)²

    U = ½ ε₀ A d E²

We see that the stored energy is proportional to the square of the electric field, so the capacitor can increase its energy with increasing voltage

In short, the energy comes from the increase in the electric field

5 0
3 years ago
A rock thrown with speed 7.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 18.0 m bef
Iteru [2.4K]

Answer:

height from where rock was thrown is 27.916 m

Explanation:

speed = 7.50 m/s

θ = 30°

g= 9.8 m/s²

horizontal distance = 18 m

time require for vertical displacement

time = \frac{distance}{velocity} \\t = \frac{18}{7.5\ cos30^0}

t = 2.8 sec

now for calculation of height

s = ut + 0.5 a t²

-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)

-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)

-h = -27.916 m

h= 27.916 m

height from where rock was thrown is 27.916 m

5 0
3 years ago
The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h f
zmey [24]

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where  [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

8 0
3 years ago
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