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Lena [83]
3 years ago
7

A 4.9 A current is set up in a circuit for 5.4 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energ

y of the battery reduced?
Physics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

E =9525.6 J        

Explanation:

Given that

Current ,I = 4.9 A

time ,t = 5.4 min

Voltage difference ,ΔV = 6 V

energy delivered by battery in time t is given as

E=V I t

where V is voltage, I is the current and t is the time.

now putting the values in above equation

E= 6 x 4.9 x 5.4 x 60  J   ( 1 min = 60 s , 5.4 min  = 5.4 x 60 s)

E =9525.6 J

Therefore the energy reduced will be 9525.6 J

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Answer:

m = 28.7[kg]

Explanation:

To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.

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Ek = kinetic energy = 1800 [J]

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v = 11.2 [m/s]

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Answer:

L = Pt/M

Explanation:

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While participating in a blood drive at school, Keona learns that blood has a density of 1.06 g/mL. She donates one pint of bloo
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Answer:

m≈501.57 g

Explanation:

The density formula is:

d=m/v

Let’s rearrange the formula for m. m is being divided by v. The inverse of division is multiplication, so multiply both aides by v.

d*v= m/v*v

d*v=m

The mass can be found by multiply the density and the volume.

m=d*v

The density is 1.06 grams per milliliter and the volume is 473.176 milliliters.

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Substitute the values into the formula.

m= 1.06 g/mL * 473.176 mL

Multiply. When multiplying, the mL will cancel out.

m= 501.56656 g

Let’s round to the nearest hundredth. The 6 in the thousandth place tells us to round the 6 to a 7 in the hundredth place.

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The mass is about 501.57 grams.

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