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Luden [163]
3 years ago
7

A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is

rated at 61.0 A · h and each headlight requires 31.8 W of power, how long will it take the battery to completely discharge?
Physics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

23.02 h.

Explanation:

Power is defined as the time rate at which energy is dissipated.

Mathematically,

Power = Energy/time

Energy = I * V * t

Where,

I * t = current

= 61 Ah

V = potential difference

= 12 V

Power = 31.8 W

Therefore,

31.8 = (61 * 12)/t

t = 732/31.8

= 23.02 h

Converting to seconds,

= 23.02 h * 3600 s/1 h

= 82867.9 s.

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The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u
irinina [24]

Explanation :

The forces acting on hot- air balloon are:

Weight, (W)

Force due to air resistance, (F)

Upthrust force, (U)

Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.

In this case, the hot-air balloon descends vertically at constant speed.

so, a=0

and F=ma=0

so, W = F + U ....................(1)

when it is ascending let the weight that it is releasing is R, so

(W-R) + F = U..........(2)

solving equation (1) and (2)

(W-R)+F=W-F

R=2F            

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3 years ago
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The energy of a photon was found to be
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A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 12 ft. (a) Find the force exer
7nadin3 [17]

Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

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4 0
3 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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