Answer:
Explanation:
Possible causes for overcurrent include short circuits, excessive load, incorrect design, an arc fault, or a ground fault. Fuses, circuit breakers, and current limiters are commonly used overcurrent protection (OCP) mechanisms to control the risks
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v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;
and solve for "v".
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Explanation:
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The formula is: KE = (½) * (m) * (v²) ;
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"Kinetic energy" = (½) * (mass) * (velocity , "squared")
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Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since: a "negative" multiplied by a "negative" equals a "positive".
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So, we have the formula:
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KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
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we arrange the formula ;
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(KE) = (½) * (m) * (v²) ; ↔ (½)*(m)* (v²) = (KE) ;
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→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
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→ We divide each side of the equation by: "[(½)* (m)]" ;
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→ [ (½)*(m)*(v²) ] / [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
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to get:
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→ v² = (KE) / [(½)* (m)]
→ v² = 2 KE / m
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Take the "square root" of each side of the equation ;
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→ √ (v²) = √ { 2*(KE) ] / m }
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→ v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];
and solve for "v".
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Answer:
Average speed of Olympic runner will be 23.08 km/hr
Explanation:
We have given distance d = 27.3 miles
We know that 1 mile = 1.6 km
So 27.3 mile = 
Given time = 2 hour 45 minutes and 35 sec
We know that 1 hour = 60 minutes
So 45 minutes 
We also know that 1 hour = 3600 sec
So 1 sec 
So total time t =
We know that speed 
Answer:
For C1, Q = 1.6125×10⁻³ C
For C2, Q = 6.25×10⁻⁴ C
Explanation:
Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.
From the question,
Q = CV........................ Equation 1
Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.
For the first capacitor,
Q = C1V............. Equation 2
Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V
Substitute into equation 2
Q = (6.45×10⁻⁶ )(250)
Q = 1.6125×10⁻³ C.
For the the second capacitor,
Q = C2V............. Equation 3
Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V
Q = (2.5×10⁻⁶ )(250)
Q = 6.25×10⁻⁴ C
