Answer:
Lithium
Explanation:
The equation for the photoelectric effect is

where
is the energy of the incident photon, with
h being the Planck constant
c is the speed of light
is the wavelength of the photon
is the work function of the metal (the minimum energy needed to extract the photoelectron from the metal)
is the maximum kinetic energy of the emitted photoelectrons
In this problem, we have
is the wavelength of the incident photon
is the maximum kinetic energy of the electrons
First of all we can find the energy of the incident photon

Converting into electronvolts,

So now we can re-arrange the equation of the photoelectric effect to find the work function of the metal

So the metal is most likely Lithium, which has a work function of 2.5 eV.
Answer:

Explanation:
To solve this problem we use the formula for accelerated motion:

We will take the initial position as our reference (
) and the downward direction as positive. Since the rock departs from rest we have:

Which means our acceleration would be:

Using our values:

Answer:
Pebble A has 1/3 the acceleration as pebble B.
Explanation:
F = m×a
mass of a = 3 × mass of b (m_a = 3 × m_b)
Same starting force, F
m_a = mass of a
m_b = mass of b
a_a = acceleration of a
a_b = acceleration of b
F = m_a × a_a = m_b × a_b
3 × m_b × a_a = m_b × a_b
3 × a_a = a_b
OR
a_a = a_b / 3
Answer:
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg
velocity of meteor v=40km/s \approx 40000 m/s
Kinetic Energy of Meteor
K.E.=\frac{mv^2}{2}
K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}
K.E.=2\times 10^6 J
Kinetic Energy of Car
=\frac{1}{2}\times Mu^2
=\frac{1}{2}\times 900\times u^2
\frac{1}{2}\times 900\times u^2=2\times 10^6
900\times u^2=4\times 10^6
u^2=\frac{4}{9}\times 10^4
u=\frac{2}{3}\times 10^2
u=66.67 m/s