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4 years ago

What is the total amount of energy stored in a 12-V, 75 A*h car battery when it is fully charged?

Physics

1 answer:

erma4kov [3.2K]4 years ago

8 0

Explanation:

Power is current times voltage. Power is also energy per time.

P = IV = E/t

Solving for energy:

E = V It

E = (12 V) (75 Ah) (3600 s / h)

E = 3,240,000 J

E = 3240 kJ

You might be interested in

Which of the following element has seven total valence electrons? Your answer: argon helium oxygen bromine

GalinKa [24]

Answer:

bromine

Explanation:

Any element in the halogen group will have seven valence electrons. These elements include fluorine, chlorine, bromine, iodine, and astatine

6 0

3 years ago

Aunt Matilda goes to a well and throws a penny straight down the well at 3.0 m/s. She hears a splash after 0.5 seconds. How deep

nevsk [136]

Answer : The correct option is (d) 2.73 m

Explanation :

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$

where,

s = distance or height = ?

u = initial velocity = 3.0 m/s

t = time = 0.5 s

a = acceleration due to gravity = $9.8m/s^2$

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

$s=(3.0m/s)\times (0.5s)+\frac{1}{2}\times (9.8m/s^2)\times (0.5s)^2$

$s=2.73m$

Therefore, the correct option is (d) 2.73 m

8 0

3 years ago

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

Roman55 [17]

Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

8 0

3 years ago

Read 2 more answers

What is the minimum force required to increase the energy of a car by 84 J over a distance of 38 m? Assume the force is constant

konstantin123 [22]

Answer:

2.210N

Explanation:

Workdone = Force x distance

Distance = 38m , Workdone = 84J

Hence 84J = Force x 38m

Force = 84J / 38m

Force = 2.210N =2.2N

4 0

3 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago