Answer:
Work done by the machine (W) = 500 × 1.5 = 750 J
Work supplied to the machine (W) = 100 × 10 = 1000 J
Here, work supplied to the machine is input work = 1000 J
Answer:
3.066×10^21 photons/(s.m^2)
Explanation:
The power per area is:
Power/A = (# of photons /t /A)×(energy / photon)
E/photons = h×c/(λ)
photons /t /A = (Power/A)×λ /(h×c)
photons /t /A = (P/A)×λ/(hc)
photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)
= 3.066×10^21
Therefore, the number of photons per second per square meter 3.066×10^21 photons/(s.m^2).
Answer:
formation of gas bubbles at electrodes
deposition of metals at electrodes
changes in solution colour
electroplating
electrolysis
Answer:
Hey there
The ozone layer or ozone shield is a region of Earth's stratosphere that absorbs most of the Sun's ultraviolet radiation
Can u have brainly
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
