Answer:
0.625m
Explanation:
Now Velocity of a wave ,V = frequency × wavelength
Wavelength =velocity /frequency
=500/800 =0.625m
Answer:
b. AG, work function=4.74eV
Explanation:
Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:
(wavelength of lowest-energy ultraviolet light)
So, the lowest energy of ultraviolet light can be found by using the formula
where
h is the Planck constant
c is the speed of light
Substituting,
And keeping in mind that
This energy converted into electronvolts is
The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:
b. AG, work function=4.74eV
Because for all the other metals, visible light will be enough to extract photoelectrons.
'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.
For the stone dropped from the 50m tower:
H = +50 - (1/2) G T²
For the stone tossed upward from the ground:
H = +20T - (1/2) G T²
When the stones' paths cross, their <em>H</em>eights are equal.
50 - (1/2) G T² = 20T - (1/2) G T²
Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:
50 = 20T Isn't that incredible ? ! ?
Divide each side by 20 :
<u>2.5 = T</u>
The stones meet in the air 2.5 seconds after the drop/toss.
I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?
Their height is +50 - (1/2) G T² = 19.375 meters
The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
Answer:
The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will decrease with time, the surface area between the tire and the road will increase with time, and the net force the tire exerts on the road will be constant with time.
Explanation:
when a wheel has an air leak, it means that the inside of the tire has less air, which means that there will be less air pushing the walls of the tire so that the air pressure decreases.
On the other hand, the tire begins to deform due to lack of air which increases the area of contact with the floor.
As the weight of the car remains constant and the air has a negligible mass the force towards the road will be the same
