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3 years ago

The self-referencing effect refers to ________.

1 answer:

5 0

The self-reference effect is the tendency an individual to have better memory for information that relates to oneself than information that is not personally relevant.

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gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are

AveGali [126]

Answer:

$\lambda=1.03\times 10^{-3}\ m$

Explanation:

Given that,

The energy of the microwave oven is $1.2\times 10^{-3}\ eV$ .

We need to find the wavelength of these photons.

$1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J$

The energy of a wave is given by :

$E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}$

Put all the values,

$\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m$

So, the wavelength of these photon is $1.03\times 10^{-3}\ m$ .

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3 years ago

A banked curve is designed for a roadway. The curve is designed so that a car traveling at 19.2 m/s can negotiate it without rel

egoroff_w [7]

I don’t know wait for someone else

8 0

3 years ago

As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

Hunter-Best [27]

Periodic time is the time taken for one complete oscillation by a body in circular motion. In this case the merry-go round takes 2 minutes to cover 15 complete oscillations. 2 Minutes = 120 seconds
Hence, 15 oscillations takes 120 secs
thus 1 oscillation takes 120/15 = 8 seconds
therefore the period of the merry-go-round = 8 seconds

8 0

3 years ago

Read 2 more answers

An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle

Katarina [22]

Answer:

The change on the second particle is $2.93\times 10^{-16}\ C$ .

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

$T=\dfrac{2\pi m}{Bq}$

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

$m_s=m_p$

If both particles take the same amount of time to go once around their respective circles. So,

$T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C$

So, the change on the second particle is $2.93\times 10^{-16}\ C$ .

7 0

3 years ago

When three identical bulbs of 60W,200V rating are connected in series at a 200V supply the power drawn by them will be?

Law Incorporation [45]

Answer:

P = 180 [w]

Explanation:

To solve this problem we must use ohm's law, which is defined by the following formula.

V = I*R & P = V*I

where:

V = voltage = 200[volts]

I = current [amp]

R = resistance [ohm]

P = power [watts]

Since the bulbs are connected in series, the powers should be summed

P = 60 + 60 + 60

P = 180 [watts]

Now we can calculate the current

I = 180/200

I = 0.9[amp]

Attached is an image where we see the three bulbs connected in series, in the circuit we see that the current is the same for all the elements connected to the circuit.

And the power is defined by P = V*I

we know that the voltage is equal to 200[V], therefore

P = 200*0.9

P = 180 [w]

8 0

3 years ago