I'm guessing they rounded because it is technically 2.998 X 108 miles/hour
so I would go with true!
Answer:
In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.[1] Two main classes of leptons exist: charged leptons (also known as the electron-like leptons or muons), and neutral leptons (better known as neutrinos). Charged leptons can combine with other particles to form various composite particles such as atoms and positronium, while neutrinos rarely interact with anything, and are consequently rarely observed. The best known of all leptons is the electron.
Answer:
90,000 J
Explanation:
Kinetic energy can be found using the following formula.
where <em>m </em>is the mass in kilograms and <em>v</em> is the velocity in m/s.
We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.
<em>m</em>=50
<em>v</em>=60
Substitute these values into the formula.
First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.
60^2=60*60=3,600
Multiply 50 and 3,600
Multiply 1/2 and 3,600, or divide 3,600 by 2.
Add appropriate units. Kinetic energy uses Joules, or J.
The kinetic energy of the object is 90,000 Joules
All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.
So we solve for v:
m*g*h=(1/2)*m*v², masses cancel out,
g*h=(1/2)*v², we multiply by 2,
2*g*h=v² and take the square root to get v
√(2*g*h)=v, we plug in the numbers and get:
v=9.9 m/s.
So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
<span>The energy of a single photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength.
Plugging the values in gives E = 6.63E-34 x 3.00E8 / 700E-9 = 2.84E-19 Joules
Now one mole of substance is equivalent to 6.02E23 particles, so one mole of these photons will be:
2.84E-19 x 6.02E23 = 1.71E5 Joules</span>
