Answer:
The brightness of bulb 1 dies because it is switched off.
<span>The answer is: ultraviolet
The energy (E) of a photon is directly proportional to its frequency f, by Planck's
formula: E = hf, where h is Planck's constant (6.625 * 10**-34 joule-second).
The frequency is inversely proportional to the wavelength w by: f = c/w, where
c is the speed of light, 3.0 * 10**8 meters per second.
Combine these formulas and we see that the energy is inversely proportional to
the wavelength by: E = hc/w
If the energy is inversely proportional to the wavelength, a photon with twice the
energy has half the wavelength of our 442-nm. photon in this example.
So its wavelength is 221 nm. which is in the ultraviolet range.</span>
Answer:
We want to describe how to graph a linear equation.
Explanation:
The given equation is:
y = -4x - 1
a) To graph it, we need to find two points that belong to the line, then we graph the points, and then we connect them with a line.
To get the points, we just need to evaluate the function in two different values of x.
for x = 0
y = -4*0 - 1 = -1
So we have the point (0, -1)
for x = 1
y = -4*1 - 1 = -5
So we have the point (1, - 5)
Now we just need to find these two points and connect them with a line, the graph can be seen below.
b) To check if the graph is correct we can see two things:
in y = -4*x - 1
The y-intercept is -1, this means that the graph should intersect the y-axis at y = -1
The slope is -4, this means that for each unit increase on x, we should see that the y-value decreases by 4.
Checking those two things we can see if our graph is correct or not.
pls give brainliest!
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
