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2 years ago

Calculate the power provided by a 12 volt car battery with a rating of 350 amperes.

Physics

1 answer:

mamaluj [8]2 years ago

8 0

Answer:

Option D

4200 W

Explanation:

Power, P is also given as the product of voltage and current, expressed as P=VI

Here, P is power, V is voltage in the xircuit and I is current theough voltage.

Taking 12 V for voltage across and 350A for current across circuit then power will be

P=350*12=4200 W

Therefore, option D is correct.

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Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie

Bond [772]

Answer:

Explanation:

For parallel inductors ,

$\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}$

$\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}$

$L_R=16.67 mH.$

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω $L_{total}$ where ω is angular frequency

= 2πf $L_{total}$

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0

2 years ago

A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?

Eduardwww [97]

relation between linear velocity and angular velocity is given as

$v = R\omega$

here

v = linear speed

R = radius

$\omega$ = angular speed

now plug in all data in the equation

$25.0 = 0.410 \omega$

$\omega = \frac{25}{0.410}$

$\omega = 60.9 rad/s$

so rotating speed is 60.9 rad/s

6 0

2 years ago

Which of the following objects cannot make a shadow? Broken piece of glass from a window/wooden pane of the same window? Explain

Lorico [155]

Answer:

broken pieces of glass from a window can't form a shadow.

Explanation:

the reason is that shadow is formed only when light rays hits an opaque object, which doesn't let light to pass through it, but glass is a transparent object, hence light rays passes through it forming no shadow..

i hope it helped....

4 0

2 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )