Answer:
118 minutes( 2 hours approximately )
Explanation:
Here, we are interested in calculating the orbital period of the satellite
Please check attachment for complete solution
Answer:
1a) 857143 m
1b) 414 m
2a)
2b)
3) the medium of air has a wavelength of 0.334 m, the medium of water has a wavelength of 1.493 m, and the medium of 5.130 m.
Explanation:
Question 1a)
Given the velocity/speed, and frequency of the wave, the length can be calculated using these two quantites.
[ λ = v / f ] wavelength = <u>v</u>elocity of the wave / <u>f</u>requency of the wave in Hz.
Since 3 × 10^8 × ms^-1 is the velocity, and 350Hz is the frequency.
Anything to the negative power is reciprocated. i.e ms^-1 = m/s.
The wavelength is 300000000m/350Hz = 857142.8571428..... m ≈ 857143 m
Question 1b) Given that the frequency of the second wave in water is 1% of the first wave, and the speed of the second wave is 1450ms^-1
Therefore the second wave has a frequency of 1% of 3.5 = 350/100 Hz = 3.5 Hz
The wavelength is found using the same
formula: wavelength = 1450m/3.5Hz = 414.2857142857.... m ≈ 414 m
Question 2a)
Question 2b)
Question 3) Remember, the speed of sound of the medium = frequency of the medium × wavelength of the medium.
Therefore the wavelength of the medium = speed of sound of the medium / frequency of the medium. This has a similar correlation to the wavelength formula. We are given that all these mediums have a frequency of 1KHz = 1000Hz, where So the wavelength of each medium =
Question 4)
This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.
If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.
The energy dissipated by the circuit is the energy delivered by
the battery. We'd know what that is if we knew I₁ . Everything that
flows in this circuit has to go through R₁ , so let's find I₁ first.
-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
-- I₁ is 10volts/10Ω = 1 Ampere.
-- R1: 1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .
-- The battery is 10 volts.
7 of the 10 appear across R₁ .
So the other 3 volts appear across all the business at the bottom.
-- R₂: 3 volts across it = V₂.
Current through it is I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.
-- R3 + R4: 6Ω in the series combination
3 volts across it
Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere
-- Remember that the current is the same at every point in
a series circuit. I₃ and I₄ must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.
-- R₃: 1/2 Ampere through it = I₃ .
1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt
-- R₄: 1/2 Ampere through it = I₄
1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts
Notice that I₂ is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches I₁ coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.
Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up. But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.
The power supplied by the battery = (voltage) · (current)
= (10 volts) · (1 Amp) = 10 watts .
"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.
(30 minutes) · (60 sec/minute) = 1,800 seconds
(10 joules/second) · (1,800 seconds) = 18,000 joules in 30 min
The power (joules per second) dissipated by each individual resistor is
P = V² / R
or
P = I² · R ,
whichever one you prefer. They're both true.
If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total.
They're not asking for that. But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.
Answer:
Work Done = W
force = F
Distance = d
W = Fd
or W = F*d
W (in joules) = 3.5*4 = 14 Nm (or J)
1Nm = 1J
so newton meters and joules are the same
Power = Work (in joules) /time (in seconds)
i don’t know the time so i can’t solve it
Answer:
It travels 1414 feets.
Explanation:
Let's take the length the bullet travels <em>l </em>as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height <em>h</em> it reached, we can calculate <em>l</em> using the <em>sin(α)</em> function:
Replacing:
Solving and roundin to the nearest foot:
