The source of information was biased. It was like walking along a river bank in the country and asking everybody you meet whether they like fishing. Or asking 500 people sitting in the bleachers whether they like baseball.
I'm sure the scientist would have gotten different data if she interviewed 500 teenagers at neighborhood basketball courts, or 500 teenagers at a rock concert.
Answer:
The energy of photon, 
Explanation:
It is given that,
Voltage of anode, 
We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

e is charge of electron


So, the maximum energy of the x- ray radiation is
. Hence, this is the required solution.
Answer:
<h2>The pin's final velocity is 5m/s</h2>
Explanation:
Step one:
given data
mass of ball m1=5kg
initial velocity of ball u1=10m/s
mass of pin m2=2kg
initial velocity of pin u2= 0m/s
final velocity of ball v2=8m/s
final velocity of pin v2=?
Step two:
The expression for elastic collision is given as
m1u1+m2u2=m1v1+m2v2
substituting we have
5*10+2*0=5*8+2*v2
50+0=40+2v2
50-40=2v2
10=2v2
divide both sides by 2
v2=10/2
v2=5m/s
The pin's final velocity is 5m/s
Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have

Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have

The block's speed is 31.422 cm/s
Answer:
If conditions are just right, you can see Polaris from just south of the equator. Although Polaris is also known as the North Star, it doesn't lie precisely above Earth's North Pole. If it did, Polaris would have a declination of exactly 90 degree.
Explanation: