This assumption is correct. The emissivity of a substance is the ratio of the energy radiated from a body to the energy radiated from a black body (perfect emitter). Because these stars reflect negligible radiation compared to the amount they emit.
The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>
All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.
The deBroglie wavelength λ of a particle is given by,
Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.
Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.
The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>
Answer:
I think it's d) down the page
I assume you mean that the car's motor is not running ... the car is just
sitting there.
If that's so, then the car's mechanical energy is just like the mechanical
energy of any other object. It has potential energy if it's in a high place
from which it can roll or fall, and it has kinetic energy if it's moving.
-- If you make the car move by pushing it, then you gave it kinetic energy
that it didn't have while it was just sitting there.
-- If it's already moving slowly, and you're able to make it move faster by
pushing, then you increased its kinetic energy.
-- If you're able to push it up a hill, no matter how small the hill is but just
to any higher place, then you gave it more gravitational potential energy
than it had before you came along.
In all of these cases, if you exert a force and keep exerting it through some
distance while the car moves, then you have done "work", which is just
another name for mechanical energy, and your work adds to the mechanical
energy of the car.
But if you didn't move the car, then no matter how hard you pushed, no work
was done, and the car's mechanical energy didn't change.
Answer:
13.5
Explanation:
Mass: 5kg
Initial Velocity: -15
Final Velocity: 12
Force: 10
We can use the equation: Vf = Vi + at
We need to find acceleration, and we can use the equation, F=ma,
We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.
Now we have all the variables to find time.
Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)
Plugging them in into desmos gives 13.5 for time.
