If one bulb goes out then all the others won't light up because electricity will be cut off. It's a disadvantage because in a parallel circuit if one bulb burns out all the others will still be on because they won't be affected. I hope I've helped you ☺
For each load, Work = (mass) x (gravity) x (distance .
Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .
Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.
The work required is equal in both cases.
The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .
Answer:
2700
Explanation:
because calculate the minute1=60×45=2700
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
Answer:
There are several options that the teacher can use to incorporate the concept into students' understanding.
Explanation:
1. The students can draw all the plants that they know.
2. Children can be asked to bring the flowers to school so that they can identify the plants themselves.
3. The children can plat the flowers in makeshift pots and then take the best plants and transplant them in the garden or elsewhere.
4. The children can take occasional trips and observe and record any changes to the plants.
4. The teacher can ask the students to draw the flowers and emphasize on the productive parts like the stamens, leaves, pistils, stems.
