Acceleration = (change in speed) / (time for the change).
Change in speed = (15 - 30) = -15 km/hr
Time for the change = 4 sec
Acceleration = (-15 km/hr) / (4 sec) = -3.75 km/hr per second .
Is that a lot ? Not much ?
Let's convert it to a unit that we can think about:
(-15 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) =
(-15 x 1,000) / (3,600) = -(4 and 1/6) m/sec .
So the acceleration of the bus is -(4 and 1/6) m/sec² .
The negative sign means that it slowed down.
(4 and 1/6) m/sec² is about 42% of the acceleration of gravity ...
the acceleration the bus would have if it drove off of a cliff.
When the car or the bus you're riding in slows down at that rate,
you feel 42% of your weight pulling you forward against your
seat belt. That's quite a drastic acceleration !
Condensing because when particles vibrate they condense.
Answer:
Binding Energy = 2.24 eV
Explanation:
First, we need to find the energy of the photon of light:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)
E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 3.1 eV
Now, from Einstein's Photoelectric Equation:
E = Binding Energy + Kinetic Energy
Binding Energy = E - Kinetic Energy
Binding Energy = 3.1 eV - 0.86 eV
<u>Binding Energy = 2.24 eV</u>
Answer:
nuclear energy is a energy that holds together of atoms
light energy is a kind of kinetic with the ability to make types of light visible to human eyes
Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
