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Yuki888 [10]
3 years ago
14

Suppose that a student is doing the photoelectric effect experiment by shining light onto a metallic surface and measuring the s

peed of the ejected electrons. if the intensity of the incident light is doubled without changing the type of light, the kemax of the electrons will be ______________ original.
Physics
1 answer:
makkiz [27]3 years ago
6 0
The maximum kinetic energy of the electrons will be the same as the original.

In fact, in the photoelectric effect one photon hits the surface of the metal and gives its energy to one photoelectron, according to the equation:
hf = \phi + K
where hf is the photon energy, with h being the Planck constant and f the frequency of the light, \phi is the work function (the energy needed to extract the photoelectron from the metal) and K is the kinetic energy of the photoelectron.

So the kinetic energy of the photoelectron is
K= hf - \phi
and we can see that this value doesn't change when we change the intensity of the light: in fact, if the intensity of the light is double, then the energy of the single photon hf remains the same (we have only doubled the number of photons, but not the energy of the single photon), so the kinetic energy of the electrons doesn't change.
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A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi
zimovet [89]

Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

6 0
4 years ago
You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie
Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance.  i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by =  $\frac{1}{0.06^2}$

                       = $3.6 \times 10^{-3}$ times

3 0
3 years ago
A cart of mass 4.0 kg is being pulled with a force of 24 N. The cart accelerates at 3.0 m s-2. What is the net force on the cart
Gre4nikov [31]

Answer:

waitt in comments... .................

5 0
2 years ago
A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit
Amanda [17]

A) +1.67 rad/s^2

The angular acceleration of the wheel is given by

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

\omega_f = 4.00 rad/s is the final angular velocity (anticlockwise, so with a positive sign)

\Delta t= 6.00 s - 0=6.00 s is the time interval

Substituting into the equation, we find the angular acceleration:

\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant t_1 where the angular velocity becomes positive (so, \omega_i=0) and the time corresponding to the final instant t_2 = 6.0 s, where \omega_f = +6.00 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant t_1=0 when \omega_i=-4.00 rad/s) and the time corresponding to the instant in which the velovity becomes positive t_2, when \omega_f = 0 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s

D) 5.6 rad

The angular displacement of the wheel is given by the equation

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

where we have

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel

\omega_f = 4.00 rad/s is the final angular velocity

\alpha=+1.67 rad/s^2 is the angular acceleration

Solving for \theta,

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad

3 0
3 years ago
Strontium chloride, SrCl2<br> Is ionic or covalent
Cloud [144]

Answer:

it is a ionic bond because electrons are transferred from chlorine to strontium

5 0
3 years ago
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