Let us say that x is the cut that we will make on the
sides to make a box, therefore the new dimensions are:
l = 15 – 2x
w = 8 – 2x
It is 2x since we cut on two sides.
We know that volume is:
V = l w x
V = (15 – 2x) (8 – 2x) x
V = 120x – 30x^2 – 16x^2 + 4x^3
V = 120x – 46x^2 + 4x^3
Taking the 1st derivative:
dV/dx = 120 – 92x + 12x^2
Set dV/dx = 0 to get maxima:
120 – 92x + 12x^2 = 0
Divide by 12:
x^2 – (92/12)x + 10 = 0
(x – (92/24))^2 = -10 + (92/24)^2
x - 92/24 = ±2.17
x = 1.66, 6
We cannot have x = 6 because that will make our w
negative, so:
x = 1.66 inches
So the largest volume is:
V = 120x – 46x^2 + 4x^3
V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3
V = 90.74 cubic inches
A frog can be many different colours. It appears green under normal 'white' light because it absorbs all the other colours in the light's spectrum apart from green. It reflects the green light back and that is picked up by your eye.
If the light is red, there is no green in the spectrum of the light, only red. So, the red light will be absorbed and there is no green to be reflected back for you to see. Therefore, the frog will not look green.
Answer:
Answer is
A. I = 6.3×10^8 A
B. Yes
C. No
Refer below.
Explanation:
Refer to the picture for brief explanation.
Answer:
The magnitude of the gravitational force is 4.53 * 10 ^-7 N
Explanation:
Given that the magnitude of the gravitational force is F = GMm/r²
mass M = 850 kg
mass m = 2.0 kg
distance d = 1.0 m , r = 0.5 m
F = GMm/r²
Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.
F = (6.67 × 10^-11 * 850 * 2)/0.5²
F = 0.00000045356 N
F = 4.53 * 10 ^-7 N
Answer:
Part a)
![f = 4.76 \times 10^{14} Hz](https://tex.z-dn.net/?f=f%20%3D%204.76%20%5Ctimes%2010%5E%7B14%7D%20Hz)
Part b)
![d = 3.48 \times 10^{-4} m](https://tex.z-dn.net/?f=d%20%3D%203.48%20%5Ctimes%2010%5E%7B-4%7D%20m)
Part c)
![\theta = 0.311 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200.311%20degree)
Explanation:
Part a)
As we know that the speed of light is given as
![c = 3 \times 10^8 m/s](https://tex.z-dn.net/?f=c%20%3D%203%20%5Ctimes%2010%5E8%20m%2Fs)
![\lambda = 630 nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20630%20nm)
now the frequency of the light is given as
![f = \frac{c}{\lambda}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D)
so we have
![f = \frac{3 \times 10^8}{630 \times 10^{-9}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B3%20%5Ctimes%2010%5E8%7D%7B630%20%5Ctimes%2010%5E%7B-9%7D%7D)
![f = 4.76 \times 10^{14} Hz](https://tex.z-dn.net/?f=f%20%3D%204.76%20%5Ctimes%2010%5E%7B14%7D%20Hz)
Part b)
Position of Nth maximum intensity on the screen is given as
![y_n = \frac{n\lambda L}{d}](https://tex.z-dn.net/?f=y_n%20%3D%20%5Cfrac%7Bn%5Clambda%20L%7D%7Bd%7D)
so here we know for 3rd order maximum intensity
![y_3 = 0.76 cm](https://tex.z-dn.net/?f=y_3%20%3D%200.76%20cm)
n = 3
L = 1.4 m
![0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}](https://tex.z-dn.net/?f=0.76%20%5Ctimes%2010%5E%7B-2%7D%20%3D%20%5Cfrac%7B3%28630%20%5Ctimes%2010%5E%7B-9%7D%29%281.4%29%7D%7Bd%7D)
![d = 3.48 \times 10^{-4} m](https://tex.z-dn.net/?f=d%20%3D%203.48%20%5Ctimes%2010%5E%7B-4%7D%20m)
Part c)
angle of third order maximum is given as
![d sin\theta = 3 \lambda](https://tex.z-dn.net/?f=d%20sin%5Ctheta%20%3D%203%20%5Clambda)
![3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})](https://tex.z-dn.net/?f=3.48%20%5Ctimes%2010%5E%7B-4%7D%20sin%5Ctheta%20%3D%203%28630%20%5Ctimes%2010%5E%7B-9%7D%29)
![\theta = 0.311 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200.311%20degree)