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dexar [7]
3 years ago
8

Design a battery to be constructed of D-cells, rated at 3Amp-hours and voltage 1.5V, that will provide a maximum operating curre

nt of 5.0A with an emf of 4.5V. One restriction is that no cell be required to supply in excess of 1.0A. What can you expect the lifetime of your battery to be while delivering maximum current
Physics
1 answer:
Free_Kalibri [48]3 years ago
6 0

Answer:

9 hours

Explanation:

For each battery:

Current = 1 A

Voltage = 1.5 V

current and EMF required = 5.0 A and 4.5 V

Since each battery has a current of 1 A, we'll need to link five of them in parallel with our load to get a current of 5 A. so that the total current equals 5 amps . Now that each battery has a 1.5 V voltage drop and we require an emf of 4.5 V, we would use three batteries in series rather than a single battery.  So, there is a need to must substitute every single battery with three batteries in series for all five single batteries linked in parallel.

So, the total no. of batteries is

= 5 × 3 = 15 batteries.

On each battery, the charge is = 3 amp hours

∴

The total charge = 15 × 3 = 45 amp hours.

Since the charge transferred within 1 hour = 5 amp

Then, the required lifetime of the battery is:

= 45 amp-hours/ 5 amp

= 9 hours

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It is called the specific latent heat of fusion. C.

This is the heat change required in changing 1kg of a solid to its liquid and it occurs at the melting point. It is not associated with temperature change.

If it is to convert 1kg of liquid to vapor at its boiling point it is called the latent heat of vaporization.<span />
6 0
3 years ago
A cheetah can run at 105 feet per second, but only for 7 seconds, at which time the animal must stop and rest. A fully rested ch
shepuryov [24]

Answer:

5 seconds

Explanation:

The straight line distance between (0, 0) and the antelope's position (x, y) at time t can be found using distance formula:

d² = x² + y²

d² = (-39 + 40t)² + (228 + 30t)²

d² = 1521 - 3120t + 1600t² + 51984 + 13680t + 900t²

d² = 53505 + 10560t + 2500t²

The cheetah can run a total distance of:

105 * 7 = 735

The time t at this distance is:

735² = 53505 + 10560t + 2500t²

540225 = 53505 + 10560t + 2500t²

0 = -486720 + 10560t + 2500t²

0 = -24336 + 528t + 125t²

t = 12, -16.224

t can't be negative, so t = 12.

Therefore, the cheetah can wait 5 seconds before it has to start running.

5 0
3 years ago
Read 2 more answers
g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coe
Rainbow [258]

The desk is in equilbrium, so Newton's second law gives

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

==>   <em>n</em> = <em>mg</em>

==>   <em>p</em> = <em>f</em> = <em>µn</em> = <em>µmg</em> = 0.400 (80.0 kg) <em>g</em> = 313.6 N

The student pushes the desk 3.00 m, so she performs

<em>W</em> = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J

of work.

6 0
3 years ago
In a previous problem, you calculated how much energy was needed to accelerate a 2,100-kg Tesla Model S from 0 to 100 km/hr, and
Naddika [18.5K]

Answer:

t=2.348s

Explanation:

First we need to calculate the Potential Energy of a Tesla to reach the 100km/h

So we know that,

W=KE_f - KE_i = \frac{1}{2}m(V_f^2-V_i^2)=\frac{1}{2}(2100)(27.78^2)W=8.1*10^5J

Here we can to calculate the time through,

t=\frac{W}{P}

t=\frac{8.1*10^5}{345*10^3}

t=2.348s

5 0
3 years ago
A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
Read 2 more answers
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