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vova2212 [387]
3 years ago
5

A battery charger can produce 3A at 12 Volt and charges a battery fer 2 hr. Calculate work in KJ.

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0

Answer: 259.2 KJ

Explanation:

The formula calculate work don in a circuit is given by :-

W=QV, where Q is charge and V is the potential difference.

The formula to calculate charge in circuit :-

Q=It, where I is current and t is time.

Given : Current : I=3A

Potential difference : V=12\ V

Time : t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}

Now, Q=3(7200)=21,600\ C

Then,  W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}

Hence, the work done = 259.2 KJ

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A book with mass 2.3 kg sits on a table. What is the normal force on the
natita [175]

Explanation:

The table is level and there are no other forces on the book, so the normal force is equal to the weight.

N = mg

N = (2.3 kg) (9.8 m/s²)

N = 22.5 N

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3 years ago
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Write down two advantages of parallel combination ​
Simora [160]

Answer:

In parallel combination each appliance gets the full voltage.

If one appliance is switched on/of others are not affected.

The parallel circuit divide the current through the appliances.

In a parallel combination it is very easy to connect or disconnect a new appliance without affecting the working of other appliances.

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Explanation:

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3 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
SOMEONE HELP ME PLSS​
lara31 [8.8K]
1. 2500/60 joules/sec
2. 2,500Nm
7 0
3 years ago
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You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s
Jobisdone [24]
Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
Since the speed of sound and light remains constant in a particular medium, we can use
      Speed = Distance/Time

For light,
   3 x 10^8 = d/t
                t = d/(3 x 10^8)   -1 

For sound,
           339 = d/(t + 30)       -2

Putting value from 1 in 2.
               d = 10^4 m(approx)
3 0
3 years ago
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