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2 years ago

Does a car have antennae?what senses does it have?

Physics

1 answer:

brilliants [131]2 years ago

4 0

Answer:

The ground of an antenna needs to be attached to the car in such a way that the voltage of the antenna can connect to the earth. This connection allows the earth to receive and store excess voltage rather than the extra voltage bouncing back through your antenna and radio.

Explanation:

pls give brainliset

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Determine the answer to the equation 30 km/h × 17 h =

Jobisdone [24]

30 km/h * 17 h = 30*17 km/h *h
= 510 km

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3 years ago

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3 years ago

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If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 wil

Lina20 [59]

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

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3 years ago

A force F~ = Fx ˆı + Fy ˆ acts on a particle that

Hatshy [7]

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

$tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\$

Then we find the angle β of the displacement vector

$tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\$

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

$W=F*d*cos (theta)$

The absolute value of F will be:

$F=\sqrt{8^{2}+1^{2} } \\F= 8.06 N$

The absolute value of d will be:

$d=\sqrt{(6 )^{2}+(2)^{2} } \\d= 6.32m\\$

Now we have:

$W=8.06*6.32*cos(25.56)\\W=46 J$

4 0

3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

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3 years ago