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vova2212 [387]
3 years ago
15

Two resistors, A and B, individually connect to a 9V battery. A student notices that resistor A is warmer than resistor B. Which

resistor has the greatest resistance?
Engineering
1 answer:
dybincka [34]3 years ago
4 0

Answer:

Resistor B

Explanation:

Since resistance is the opposition to the flow of current in a circuit,

first let assume the two resistors are connected in parallel to the voltage, recall that when connection is in parallel, the different amount of current pass through the resistors depending on the value with the small resistor having  a lower resistance effect hence higher current will pass through

The energy dissipated in each resistor can be calculated as

E=\frac{1}{2}IR^{2}t.

from the formula we can conclude that the energy value will be higher for the resistor with small resistance value. hence more heating effect which will cause it to be warm.

Also when connected individually the current flow from the voltage source will pass through the resistor which when we calculate the energy dissipated, the resistor with smaller value will be higher because it will draw more current which will in turn lead to a heating effect and cause the resistor to be warm. Hence we can conclude that the resistance B has greatest resistance value.

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What is the magnetic force on a moving electric charge called
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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

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A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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3 years ago
The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement
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The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

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