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: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec
Answer:
<em>B) 1.0 × 10^5 V</em>
Explanation:
<u>Electric Potential Due To Point Charges
</u>
The electric potential produced from a point charge Q at a distance r from the charge is
![\displaystyle V=k\frac{Q}{r}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3Dk%5Cfrac%7BQ%7D%7Br%7D)
The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.
We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is
![d=\sqrt2 a](https://tex.z-dn.net/?f=d%3D%5Csqrt2%20a)
where a is the length of the side.
The distance from any corner to the center is half the diagonal, thus
![\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7Ba%7D%7B%5Csqrt%7B2%7D%7D)
![\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%3D0.707%5C%20m)
The total potential is
![V_t=V_1+V_2+V_3+V_4](https://tex.z-dn.net/?f=V_t%3DV_1%2BV_2%2BV_3%2BV_4)
Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of
. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.
![\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V_1%3DV_2%3Dk%5Cfrac%7BQ%7D%7Br%7D%3D9%5Ctimes%2010%5E9%20%5Cfrac%7B4%5Ctimes%2010%5E%7B-6%7D%7D%7B0.707%7D)
![V_1=V_2=50912\ V](https://tex.z-dn.net/?f=V_1%3DV_2%3D50912%5C%20V)
The total potential is
![V_t=50912\ V+50912\ V=1\times 10^5\ V](https://tex.z-dn.net/?f=V_t%3D50912%5C%20V%2B50912%5C%20V%3D1%5Ctimes%2010%5E5%5C%20V)
![\boxed{V_t=1\times 10^5\ V}](https://tex.z-dn.net/?f=%5Cboxed%7BV_t%3D1%5Ctimes%2010%5E5%5C%20V%7D)
Answer:
The induced current direction as viewed is clockwise
Explanation:
Lenz's Law states that the induced e. m. f. causes current to be driven in the loop of wire in such a way as to generate magnetic field that are oppose the magnetic flux change which is the source of the induced current
Therefore, as the magnet approaches the coil with the south pole, the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise
Therefore;
The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise
Answer:before throwing and after catching the ball
Explanation:
When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.
After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.