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mash [69]
2 years ago
7

What type of light does this light bulb produce most (i.e. at what wavelength does the spectrum have maximum intensity)?

Physics
1 answer:
lesantik [10]2 years ago
4 0

Answer: The light bulb produces the continuous light. At minimum wavelength the spectrum have maximum intensity.

Explanation:

According to Wein's displacement law, the wavelength is inversely proportional to the temperature.

The intensity depends on the frequency. The frequency is inversely proportional to the wavelength.

Therefore, when the temperature of the light bulb will be maximum then the wavelength will be minimum. At minimum wavelength the spectrum have maximum intensity.

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A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th
mr Goodwill [35]

Answer:

T=4985.5^{\circ}K

Explanation:

The equation that relates heat Q with the temperature change T-T_0 of a substance of mass <em>m </em>and specific heat <em>c </em>is Q=mc(T-T_0).

We want to calculate the final temperature <em>T, </em>so we have:

T=\frac{Q}{mc}+T_0

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 273^{\circ} to our temperature in ^{\circ}C to have it in ^{\circ}K as it must be):

T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K

3 0
2 years ago
What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

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

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

3 0
2 years ago
The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o
MrMuchimi

Answer:

The total kinetic energy of both particles is 2.43\times10^{-13}

Explanation:

Given that,

Kinetic energy of nucleusK.E= 1.24\times10^{-13}\ J

Kinetic energy of proton K.E= 2.47\times10^{-13}\ J

Radius of proton r= 0.9\times10^{-15}\ m

We need to calculate the final potential energy

Using formula of final potential energy

U=\dfrac{kq^2}{r}

Put the value into the formula

U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}

U_{f}=1.28\times10^{-13}\ J

We need to calculate the initial energy of both the particles

Using formula of energy

E_{i}=(K.E_{n}+K.E_{p})+U_{i}

E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0

E_{i}=3.71\times10^{-13}\ J

We need to calculate the total kinetic energy of both particles

Using conservation of energy

E_{i}=E_{f}

E_{i}=K.E_{f}+U_{f}

3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}

K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}

K.E_{f}=2.43\times10^{-13}

Hence, The total kinetic energy of both particles is 2.43\times10^{-13}

3 0
2 years ago
Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.
elena-14-01-66 [18.8K]

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

5 0
2 years ago
A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed
Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

8 0
2 years ago
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