1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
diamong [38]
3 years ago
10

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]3 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
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find
____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s)  u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

 t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t  = time

           θ = 1 rev.

Since 1 rev = 2π (rad)

           t = 1.37 s

 Average angular velocity = 2π/t

π = 3.143

 Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

   Average angular velocity ≈ 4.59 rad/s

8 0
3 years ago
If you could be the first person to go somewhere where would it be and why??? Example plz
vivado [14]
New York because of the lady of all the people and excitement
3 0
3 years ago
Read 2 more answers
A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mea
Vlada [557]

The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p =  \sqrt{3RTm}

According to De broglie hypothesis, P = h ÷ λ

Therefore,    λ  = h ÷ \sqrt{3RTm}

                     = 6.6× 10^{-34} ÷  \sqrt{3 \times 1.38 \times  10^{-23}  \times 300 \times 1.6 \times  10^{-27} }

                     = 0.15 × 10^{-9}

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

To learn more about De broglie wavelength, refer: <u>https://brainly.in/question/6131028</u>

#SPJ4

6 0
1 year ago
Other questions:
  • A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/
    8·1 answer
  • How is Mercury different from all of the other metals in the periodic table <br><br>​
    6·2 answers
  • High values of Ksp greater than 1.0 infer: The solid phase is relatively insoluble. The solid phase is relatively soluble. The c
    13·2 answers
  • What's the speed of a sound wave through water at 25 Celsius
    9·1 answer
  • The acceleration of gravity on the surface of Earth is 9.8 m/s / s . On the Moon the acceleration of gravity is 1/6 of this valu
    7·1 answer
  • What is a ig?<br><br><br> my friend wont stop saying "what i your ig"
    5·1 answer
  • un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
    7·1 answer
  • Light bending at a boundary is caused by differences in the speed of light in different materials. The greater the difference in
    9·1 answer
  • M=3000km v=25m/s what’s the momentum
    8·1 answer
  • An object that is dropped from a height H falls with a constant acceleration of g. The final
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!