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Mademuasel [1]
3 years ago
11

An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th

at the battery delivers a steady current for its lifetime of 4.00 h, how much charge passed through the toy? (c) How much energy was delivered to the toy?
Physics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c)  The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

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We have given that initial velocity of the object u = 160 ft/sec'

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At maximum height we know that velocity is zero

So -32t+160=0

t = 5 sec

So it will take 5 sec to reach maximum height

Now maximum height will occur at t = 5 sec

So maximum height = 16\times 5^2+160\times 5=1200feet

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3 years ago
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