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Nezavi [6.7K]
3 years ago
14

Now slowly begin to raise the temperature. At approximately what temperature would a heated material (metal, wood, etc.) begin t

o give off visible light at a deep red color?Note: This will be the temperature where your spectrum first begins to come off of the wavelength axis in the visible region, and so is giving off a small amount of red light.
A. 500 K (440 Fahrenheit)
B. 1050 K (1430 Fahrenheit)
C. 1800 K (2780 Fahrenheit)
D. 2500 K (4040 Fahrenheit)
Physics
1 answer:
Sav [38]3 years ago
3 0

Answer:

Explanation:

We shall apply Wien's displacement law which is as follows .

λ T = b where λ is wavelength of light that is coming out of hot body to maximum extent .

Putting the value of temperature given and b

λ x 500 = 2898 μmK

λ = 5.796 μm = 5796 nm

For temp 1050 K

λ = 2760 nm

For temp 1800 K

λ = 1610  nm

For temp 2500 K

λ = 1159.2  nm

The visible range starts from 740 nm .

Hence we can expect that some amount of visible light may emerge at the temperature of 2500K because the wavelength that we have calculated above gives the value of peak wavelength of a spectrum of light coming out of hot body .  

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Setler [38]

Answer:

The minimum force that must be exerted by the mechanic at the end of a 30 cm-long wrench to loosen the nut is 110N

Explanation:

Torque = Fd sin θ

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Torque = 33 N.m

F = Force applied = ?

d = 30 cm = 0.3 m

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8 0
4 years ago
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A large truck has more inertia than small car because it has more__
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The large truck can rest at a stable equilibrium as compare to the small,from the definition of inertia, <span>Inertia is the tendency of an object to remain at rest or in motion.</span>
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3 years ago
s A horizontal insulating rod of length 11.0-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge
guajiro [1.7K]

Answer:

F = 2.26 ×  10⁻³ N

Explanation:

given,

length of rod = 11 cm

charge  = 19 nC

linear charge density = 3.9 x 10⁻⁷ C/m

electric force at 2 cm away.

E(r) = \dfrac{2K\lambda}{r}

F = E q

F= \dfrac{2K\lambda\ q}{L}\int \dfrac{dr}{r}

integrating from 0.02 to 0.02 + L

F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]

F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]

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5 0
4 years ago
The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t
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Answer:

The hydro static force on the back of the dam is 1.96\times10^{11}\ N

Explanation:

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Depth d= 200 m

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P_{avg}=\rho\times g\times d_{avg}

Put the value into the formula

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P_{avg}=980000\ Pa

We need to calculate  the hydro static force on the back of the dam

Using formula of force

F = P_{avg}\times A

Put the value into the formula

F = 980000\times1000\times200

F=1.96\times10^{11}\ N

Hence, The hydro static force on the back of the dam is 1.96\times10^{11}\ N

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3 years ago
1. Butterflies usually are identified by the colors and patterns on their wings. Butterflies have four wings. The two wings near
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Answer:

Can't answer without a picture

Explanation:

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