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zhenek [66]
3 years ago
11

A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is

rated at 51.0 A · h and each headlight requires 27.0 W of power, how long will it take the battery to completely discharge?
Physics
1 answer:
avanturin [10]3 years ago
3 0

Answer: 22.6 hours

Explanation:

The power is the measure of the rate of energy.

In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as

P=IV

where

I is the current

V = 12.0 V is the voltage of the battery

So the energy delivered by the battery can be written as

E=Pt=VIt

Where

It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s

So the energy delivered is

E=(12.0)(183,600)=2.2\cdot 10^6 J

At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:

t=\frac{2.2\cdot 10^6 J}{27.0 W}=81481 s = 22.6 h

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A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
A street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar
____ [38]

Answer:

60

so you take 120÷2 wires

4 0
3 years ago
Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T
alex41 [277]

Answer:

30643 J

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

t = Time taken = 1 ns

c = Speed of light = 3\times 10^8\ m/s

E_0 = Maximum electric field strength = 1.52\times 10^{11}\ V/m

A = Area = 1\ mm^2

Magnitude of magnetic field is given by

B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T

Intensity is given by

I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2

Power, intensity and time have the relation

E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J

The energy it delivers is 30643 J

4 0
3 years ago
which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload
Anuta_ua [19.1K]
Overload <<<<<<<<<<<<<<<<<<<<<<<<<<<<<,,,,,,,,
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3 years ago
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration
umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

F=qvB     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

F=qvB=ma       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0
3 years ago
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