Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
wavelength = 24 m
Period = 10 s
f = 0.1 Hz
Amplitude = 4 m
Explanation:
Wavelength:
Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:
<u>wavelength = 24 m</u>
Period:
The period is given as:
<u>Period = 10 s</u>
<u></u>
Frequency:
The frequency is given as:
<u>f = 0.1 Hz</u>
<u></u>
Amplitude:
Amplitude will be half the distance between extreme points, that is, crest and trough:
Amplitude = 8 m/2
<u>Amplitude = 4 m</u>
Answer:
F_A = 8 F_B
Explanation:
The force exerted by the planet on each moon is given by the law of universal gravitation
F =
where M is the mass of the planet, m the mass of the moon and r the distance between its centers
let's apply this equation to our case
Moon A
the distance between the planet and the moon A is r and the mass of the moon is 2m
F_A = G \frac{2m M}{r^{2} }
Moon B
F_B = G \frac{m M}{(2r)^{2} }
F_B = G \frac{m M}{4 r^{2} }
the relationship between these forces is
F_B / F_A = = 1/8
F_A = 8 F_B
The correct answer is sulfur dioxide pollution
Hope this helped :)
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is
Clock time of computer B is
Effective CPI of computer A is
Effective CPI of computer B is
CPU time of A is
CPU time of B is
Hence Computer A is Faster by
Computer A is 1.41 times faster than the Computer B