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2 years ago

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A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 7

50 kg car from rest to 25.0 m/s, make it climb a 2.00×102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00×102 N force for an hour.

1 answer:

Yuliya22 [10]2 years ago

3 0

Answer:

3894531 coulombs

Explanation:

1 hour = 3600 seconds

Let g = 10m/s2

The distance that the car travel at 25 m/s over an hour is

s = 25 * 3600 = 90000 m

The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:

$\sum E = E_k + E_p + E_W$

$\sum E = mv^2/2 + mgh + Fs$

$\sum E = 750*25^2/2 + 750*10*200 + 500*90000 = 46734375 J$

This energy is drawn from the battery over an hour (3600 seconds), so its power must be

$P = E / t = 46734375/3600 = 12982 W$

The system is 12V so its current is

$I = P/U = 12982 / 12 = 1081.8 A$ or 1081.8 Coulombs/s

The the total charge it needs for 1 hour (3600 s) is

C = 1081.8 * 3600 = 3894531 coulombs

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However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

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A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

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the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

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Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

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(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

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$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

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$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

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