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Rama09 [41]
3 years ago
12

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A?h). A 50 A?h

battery can supply a current of 50 Afor 1.0 h, or 25 A for 2.0 h, and so on.A) What total energy can be supplied by a 13V , 60A?h battery if its internal resistance is negligible?Answer= ...... JB) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 90 kg/m 3.)Answer= ........ LC) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?Answer= ........ h
Physics
1 answer:
V125BC [204]3 years ago
5 0

Answer: (A) 780J

(B) 1.89×10^-11L

(C)1.67×10^-4 h

Explanation:

Energy of the battery = IVt

=13×60 = 780J

Heat combustion of

1g of gasoline relax 46000J

Therefore 780J will release 780/46000

= 0.017g

Density = mass/volume

Volume = mass/density

Volume =0.017× 10^-3 / 900

= 1.89× 10^-8 m3

= 1.89×10^-11 litres

P=IVt

t=P/IV

= 450/60×13

1.67×10^-4 hours

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Debbie plotted a graph to show how the mass of aluminium rivets affected the distance the cup moved down.
Elan Coil [88]

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What is the IMA of the following pulley system?<br><br>34567
Lynna [10]

Answer:

    IMA of given system =   \frac{F_{r} }{F_{e} }

Explanation:

  • The "Ideal Mechanical advantage" (IMA) of given pulley is \frac{F_{r} }{F_{e} } .
  • Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by F_{r}. The input force is analogous or equivalent to the effort applied i.e. F_{e} .
  • Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is  \frac{F_{r} }{F_{e} } .
  • Its mathematical reference would be:

                                                IMA =   \frac{F_{r} }{F_{e} }

6 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m
Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A  where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0
4 years ago
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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
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