If the water measures -5 feet at low tide and 3ft at high tide what is the tidal range
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Answer:
Decreases, Increases
Explanation:
Resistance is parallel can be calculated using
1/Req = 1/R1 + 1/R2 + 1/R3 +....
Then, as more resistor is added in parallel the equivalent resistance is reduced.
Let use a simple sample
Let all the resistor have equal resistances
Let say R = R1 = R2 = R3 =...Rn
Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....
1/Req = 1/R + 1/R + 1/R +.... 1/Rn
Req = R/n
Check attachment on how I got that.
This implies that, the equivalent resistance will always be less than the original resistance, since n>1
So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.
B. Now, to know if the current reduces or increases
Using Ohms law
V = iR
Then, I = V/R
So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.
So current increase as we add more resistor in parallel to a circuit
To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.
From the definition we know that the wavelength is described under the equation,
Where,
c = Speed of light (vacuum)
f = frequency
Our values are,
Replacing we have,
<em>Therefore the wavelength of this wave is </em>
The impulse of a force is due to the change in the motion of an object
A. The persons speed after impact is approximately 59.38 mi/h
B. The expected speed is <u>29.89 mi/h</u> which is less than the findings
Reason:
Known parameters are;
The speed of the bus, v = 30 mi/h
The force with which the person was hit, F = 58,000 lbs
Mass of the bus, M = 40,000 lbs
Mass of the person, m = 150 lbs
Duration of the impact, Δt = 0.007 seconds
A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;
The impulse of the force = F × Δt = m × Δv
For the person, we get;
58,000 lbf ≈ 1866094.816 lb·ft./s²
58,000 lbf × 0.007 s = 150 lbs × Δv
1,866,094.816 lb·ft./s²
Δv = v₂ - v₁
The initial speed of the person at the instant, can be as v₁ = 0
The final speed, v₂ = Δv - v₁
∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s
≈ <u>87.084 ft./s</u>
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The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>
B. Where the momentum is conserved, we have;
m₁·v₁ + m₂v₂ = (m₁ + m₂)·v
The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>
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