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vesna_86 [32]
3 years ago
12

A television remote control uses infrared light with a wavelength of 940 nm. What is the frequency of the light?

Physics
1 answer:
Ronch [10]3 years ago
5 0

Answer:

Frequency = 3.19 * 10^14 Hz or 1/s

Explanation:

Relationship b/w frequency and wavelength can be expressed as:

C = wavelength * frequency, where c is speed of light in vacuum which is 3.0*10^8 m/s.

Now simply input value (but before that convert wavelength into meters to match the units, you do this by multiply it by 10^-9 so it will be 940*10^-9)

3.0 * 10^8 = Frequency * 940 x 10^-9

Frequency = 3.19 * 10^14 Hz or 1/s

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IrinaK [193]

Answer : The specific heat capacity of the alloy 1.422J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of alloy = ?

C_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of alloy = 21.6 g

m_2 = mass of water = 50.0 g

T_f = final temperature of system = 31.10^oC

T_1 = initial temperature of alloy = 93.00^oC

T_2 = initial temperature of water = 22.00^oC

Now put all the given values in the above formula, we get

21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC

c_1=1.422J/g^oC

Therefore, the specific heat capacity of the alloy 1.422J/g^oC

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3 years ago
Which of the following is located in the temperature climate zone?
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What is the domain theory of ferromagnetism?
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A region within a magnetic material in which magnetization is in a uniform direction this means the individual magnetic moments of the atoms are aligned with one another and they point the same direction. when cooled bwlow a temperature called the curie temperature the magnetization of a piece of ferromagnetic material.<span />
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3 years ago
An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-f
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Resistance = (voltage) / (current)

Resistance = (6.0 v) / (2.0 A)

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7 0
3 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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