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kogti [31]
2 years ago
8

When an electric stove element is hot enough, it gives off a dull red glow. When it cools to the point that it no longer glows,

it will.
Physics
1 answer:
DochEvi [55]2 years ago
4 0

Answer:

It will have a longer wavelength

Explanation:

When an electric stove is hot and gives dull red glow. a part of the energy dissipated is emitted as visible light and part as infrared radiation in the form of heat. When the stove cools down, and no longer glows all the energy is now in the form of infrared radiation.In the electromagnetic spectrum infrared rays have a higher wavelength than visible light. Hence for the reason the radiation will have a higher wavelength since visible light is cut off.

You might be interested in
A tank contains 100 gal of water and 50 oz of salt.water containing a salt concentration of 1 4 (1 1 2 sin t) oz/gal flows into
Alchen [17]

Answer:

Explanation:

Heres the possible full question and solution:

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

a. Find the amount of salt in the tank at any time.

b. Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.

c. The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

solution

a)

Consider the tank contains 100gal of water and 50 oz of salt

Assume that the amount of salt in the tank at time t is Q(t).

Then, the rate of change of salt in the tank is given by \frac{dQ}{dt}.

Here, \frac{dQ}{dt}=rate of liquid flowing in the tank - rate of liquid flowing out.

Therefore,

Rate_{in} =2gal/min \times \frac{1}{4} (1+ \frac{1}{2}sin t)oz/gal\\\\\\ \frac{1}{2} (1+ \frac{1}{2}sin t)oz/min\\\\\\Rate_{out}=2gal/min \times\frac{Q}{100}oz/gal\\\\\frac{Q}{50}oz/min

Therefore,

\frac{dQ}{dt} can be evaluated as shown below:

\frac{dQ}{dt}=\frac{1}{2}(1+\frac{1}{2}\sin t)-\frac{Q}{50}\\\\\\\frac{dQ}{dt}+\frac{1}{50}Q=\frac{1}{2}+\frac{1}{4}\sin t

The above differential equation is in standard form:

\frac{dy}{dt}+Py=G

Here, P=\frac{1}{50},G=\frac{1}{2}+\frac{1}{4}\sin t

The integrating factor is as follows:

\mu(t)=e^{\int {P}dt}\\\mu(t)=e^{\int {\frac{1}{50}}dt}\\\mu(t)=e^{\frac{t}{50}}

Thus, the integrating factor is  \mu(t)=e^{\frac{t}{50}}

Therefore, the general solution is as follows:

y\mu(t)=\int {\mu (t)G}dt\\\\Qe^{\frac{t}{50}}=\int {e^{\frac{t}{50}}(\frac{1}{2}+\frac{1}{4}\sin t) dt}\\\\Qe^{\frac{t}{50}}=\frac{1}{2}\int {e^{\frac{t}{50}}dt + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}\\\\\Qe^{\frac{t}{50}}=25 {e^{\frac{t}{50}} + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}+C...(1)

Here, C is arbitrary constant of integration.

Solve \int {\sin te^{\frac{t}{50}}} dt}

Here u = e^{\frac{t}{50}} and v =\sin t.

Substitute u , v in the below formula:

\int{u,v}dt=u\int{v}dt-\int\frac{du}{dt}\int{v}dt\dot dt\\\\\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{1}{50}\int{e^{\frac{t}{50}}\cos t}dt...(2)

Now, take u = e^{\frac{t}{50}}, v =\sin t

Therefore, \int{e^{\frac{t}{50}}\cos t} dt=\int {e^{\frac{t}{50}}\sin t}dt - \frac{1}{50}\int{e^{\frac{t}{50}}\sin t}dt...(3)

Use (3) in equation(2)

\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{e^{\frac{t}{50}}}{50}\sin t - \frac{1}{2500}\int{e^{\frac{t}{50}}\sin t}dt\\\\\frac{2501}{2500}\int{e^{\frac{t}{50}}\sin t}dt={e^{\frac{t}{50}}\cos t}+\frac{e^{\frac{t}{50}}}{50}\sin t\\\\\int{e^{\frac{t}{50}}\sin t}dt=\frac{2500}{2501}{e^{\frac{t}{50}}\cos t}+\frac{50}{2501}e^{\frac{t}{50}}\sin t...(4)

Use (4) in equation(l) .

Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+C

Apply the initial conditions t =0, Q = 50.

50=25-\frac{625}{2501}+c\\\\c=\frac{63150}{2501}

So, Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}

Therefore, the amount of salt in the tank at any time is as follows:

Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}e^{\frac{-t}{50}}

b)

sketch the solution curve as shown in attachment as graph 1:

CHECK COMMENT FOR C

3 0
2 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
2 years ago
Assuming both graduated cylinders are holding water at room temperature, which cylinder has more thermal energy?
natta225 [31]

Answer:

The correct option is;

The graduate cylinder with more water has more thermal energy because it is holding more water molecules

Explanation:

Given that the thermal energy of the system is the energy possessed by the system by virtue of the increased motion of the particles by virtue of a transfer  of heat, when the content of the system is heated

The thermal energy, Q is given by the following equation;

Q = Mass, m × The specific heat capacity, C × The change in temperature, ΔT

Given that the graduated cylinder with more water has more mass and therefore, more water molecules, than the cylinder with less water, the cylinder with more water has more thermal energy.

3 0
2 years ago
A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du
pentagon [3]
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
8 0
2 years ago
Read 2 more answers
A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
maksim [4K]

When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0
2 years ago
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