Answer:
Explanation:
Behind the second car , no of cars = 13
total mass of 13 cars = 13 x 37000 Kg
acceleration = .46 m /s²
If T be the tension
This tension is pulling 13 cars
force = mass x acceleration.
T = 13 x 37000 x .46
= 221260 N .
Answer:
A = 150 cm at 120 degc
Where A + B = C find B
B = C - A add -A to both sides
Ax = 150 cos 60 = -75
Ay = 150 sin 60 = 129.9
Likewise
Cx = 114.7
Cy = 80.3
Bx = Cx - Ax = 114.7 + 75 = 189.7
By = Cy - Ay = 80.3 -129.9 = -49.6
B = (189.7^2 + 49.6^2)^1/2 = 196.7 length of B vector
tan B = By / Bx = -49.6 / 189.7 B = -14.65 deg
Also
sin B = By / B = -49.6 / 196.1 = -14.65 deg
So B is 196.1 cm at -14.65 deg
Explanation:
using the first eqn for motion
Answer:
The number of photons per second that strike the given area is 2.668 x 10⁸ photons/second
Explanation:
Given;
intensity of the sunlight, I = 2.00 kJ·s−1·m^−2
area of incident, A = 5.2 cm² = 5.2 x 10⁻⁴ m²
Energy of incident photons per second on the given area;
E = IA
E = (2000)( 5.2 x 10⁻⁴)
E = 1.04 J/s
Energy of a photon is given is by;
![E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{(510*10^{-9})}\\\\E = 3.898*10^{-19} \ J/photon](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%20%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B%286.626%2A10%5E%7B-34%7D%29%283%2A10%5E8%29%7D%7B%28510%2A10%5E%7B-9%7D%29%7D%5C%5C%5C%5CE%20%3D%203.898%2A10%5E%7B-19%7D%20%5C%20J%2Fphoton)
The number of photons per second that strike the given area is;
![n = \frac{1.04 \ J/s}{3.898*10^{-19} \ J/photon} \\\\n = 2.668*10^{18} \ photons/second](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B1.04%20%5C%20J%2Fs%7D%7B3.898%2A10%5E%7B-19%7D%20%5C%20J%2Fphoton%7D%20%5C%5C%5C%5Cn%20%3D%202.668%2A10%5E%7B18%7D%20%5C%20photons%2Fsecond)
Therefore, the number of photons per second that strike the given area is 2.668 x 10⁸ photons/second