All categories

3 years ago

What is the photoelectric effect?

Physics

2 answers:

Veronika [31]3 years ago

5 0

Answer:

emissions of electrons from a metal

Explanation:

I did grad point

n200080 [17]3 years ago

4 0

the emission of electrons or other free carriers when light shines on a material.

You might be interested in

You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

We know that the electric field by the linear charge is given as:

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

Now the force on the given charge can be given as:

$F=E.q$

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

3 0

3 years ago

What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?

Rzqust [24]

49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma â‡’ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.

5 0

2 years ago

Read 2 more answers

As an electron moves, what does it make or cause ?

jolli1 [7]

It causes or makes a magnetic field.

3 0

2 years ago

Help i am having a mental breakdown help! need this done...

guajiro [1.7K]

1. he traveled a total of 24 miles
2. peter is not moving between 10 and 30 minutes

4 0

2 years ago

Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

$E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J$

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

$\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg$

HOPE THIS HELPS!!

3 0

3 years ago

Read 2 more answers