Answer:
Intensity, ![I=1.101\ W/m^2](https://tex.z-dn.net/?f=I%3D1.101%5C%20W%2Fm%5E2)
Explanation:
Power of the light bulb, P = 40 W
Distance from screen, r = 1.7 m
Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :
![I=\dfrac{P}{A}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BP%7D%7BA%7D)
![I=\dfrac{P}{4\pi r^2}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BP%7D%7B4%5Cpi%20r%5E2%7D)
![I=\dfrac{40\ W}{4\pi (1.7\ m)^2}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B40%5C%20W%7D%7B4%5Cpi%20%281.7%5C%20m%29%5E2%7D)
![I=1.101\ W/m^2](https://tex.z-dn.net/?f=I%3D1.101%5C%20W%2Fm%5E2)
So, the intensity of light is
.
Answer:
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. Since the force required to lift it is equal to its weight, it follows that the gravitational potential energy is equal to its weight times the height to which it is lifted.
Answer:
3 km/h
Explanation:
Let's call the rowing speed in still water x, in km/h.
Rowing speed in upstream is: x - 2 km/h
Rowing speed in downstream is: x + 2 km/h
It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:
8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)
Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)
8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)
Dividing by 8
(x+2) + (x-2) = (6/5)*(x² - 4)
2*x = (6/5)*x² - 24/5
0 = (6/5)*x² - 2*x - 24/5
Using quadratic formula
![x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B%28-2%29%5E2%20-%204%286%2F5%29%28-24%2F5%29%7D%7D%7B2%286%2F5%29%7D)
![x = \frac{2 \pm 5.2}{2.4}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2%20%5Cpm%205.2%7D%7B2.4%7D)
![x_1 = \frac{2 + 5.2}{2.4}](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B2%20%2B%205.2%7D%7B2.4%7D)
![x_1 = 3](https://tex.z-dn.net/?f=x_1%20%3D%203)
![x_2 = \frac{2 - 5.2}{2.4}](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B2%20-%205.2%7D%7B2.4%7D)
![x_2 = -1\; 1/3](https://tex.z-dn.net/?f=x_2%20%3D%20-1%5C%3B%201%2F3)
A negative result has no sense, therefore the rowing speed in still water was 3 km/h
A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
![\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B32%20s%7D%3D0.20%20rad%2Fs)
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
![\theta= \omega t](https://tex.z-dn.net/?f=%5Ctheta%3D%20%5Comega%20t)
and substituting t = 75 seconds, we find
![\theta= (0.20 rad/s)(75 s)=15 rad](https://tex.z-dn.net/?f=%5Ctheta%3D%20%280.20%20rad%2Fs%29%2875%20s%29%3D15%20rad)
In degrees, it is
![15 rad: x = 2\pi rad : 360^{\circ}\\ x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}](https://tex.z-dn.net/?f=15%20rad%3A%20x%20%3D%202%5Cpi%20rad%20%3A%20360%5E%7B%5Ccirc%7D%5C%5C%0A%3C%2Fp%3E%3Cp%3Ex%3D%5Cfrac%7B%2815%20rad%29%28360%5E%7B%5Ccirc%7D%29%7D%7B2%5Cpi%7D%3D860%5E%7B%5Ccirc%7D%20%3D%20140%5E%7B%5Ccirc%7D)
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where we have
![\omega=0.20 rad/s](https://tex.z-dn.net/?f=%5Comega%3D0.20%20rad%2Fs)
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
![v=(0.20 rad/s)(13.5 m)=2.7 m/s](https://tex.z-dn.net/?f=v%3D%280.20%20rad%2Fs%29%2813.5%20m%29%3D2.7%20m%2Fs)