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3 years ago

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What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the m

otor?

1 answer:

QveST [7]3 years ago

8 0

Answer:

From ohms law,

V=IR

R=V/I =12.0/150 =0.08 ohm.

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Radar uses radio waves of a wavelength of 2.4 \({\rm m}\) . The time interval for one radiation pulse is 100 times larger than t

blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x $10^{8}$ m/s)

f= 3 x $10^{8}$ / 2.4

f=1.25 x $10^{8}$ hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x $10^{8}$

T= 8 x $10^{-9}$ s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x $10^{-9}$ => 800 x $10^{-9}$ s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

$x_{min}$ = τc/2 => (800 x $10^{-9}$ x 3 x $10^{8}$ )/2

$x_{min}$ =120m

8 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

You are aboard the Amtrak Southwest Chief, travelling from Los Angeles to the Grand Canyon. Getting hungry, you decide to walk t

masya89 [10]

Answer:

An apple, potato, and onion all taste the same if you eat them with your nose plugged

Explanation:

4 0

3 years ago

Which of the following surfaces reflects the most light

Gre4nikov [31]

Answer:

D

Explanation:

it could be the yellow cloth but we see are shadow on the sidewalk most of the time and shadows are made from reflecting light so your answer should be D

3 0

3 years ago

Read 2 more answers

A spherical shell has inner radius 1.5 m, outer radius 2.5 m, and mass 850 kg, distributed uniformly throughout the shell. What

Nimfa-mama [501]

Answer:

The magnitude of the gravitational force is 4.53 * 10 ^-7 N

Explanation:

Given that the magnitude of the gravitational force is F = GMm/r²

mass M = 850 kg

mass m = 2.0 kg

distance d = 1.0 m , r = 0.5 m

F = GMm/r²

Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.

F = (6.67 × 10^-11 * 850 * 2)/0.5²

F = 0.00000045356 N

F = 4.53 * 10 ^-7 N

3 0

3 years ago