Answer:
15000 m/s
Explanation:
You just need to multiply the wavelength with the frequency.
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.
Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms
Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A
Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.
Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A
Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
If the scale reads 650N, then the mass of whoever it is standing on the scale is
(weight) / (gravity) = (650N) / (9.8 m/s²) = 66.3 kilograms .
It's not MY mass, even if I'm the one standing on the scale.
If I stand on a scale and it reads 650 N, the scale is broken.
Force, F = ma
Where m = mass in kg, a = acceleration in m/s², Force, F is in N.
F = ma
2000 = m*2.2
2.2m = 2000
m = 2000/2.2
m ≈ 909.09
Mass is ≈ 909.09 kg.
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>
<h3>B )
The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>
C ) the direction of motion can be changed without changing the direction of the applied force .</h3>
D ) the system would continue to move without any further, after and initial force has set in motion.
