3+3+##+=484hcfngccnfjkr,hca,nkvn,zhjcf v

Pls help! True or False?

jasenka [17]

Answer:

True

Explanation:

i hope it helps

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7 0

2 years ago

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6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord

kiruha [24]

Answer:

Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)

Explanation:

4 0

3 years ago

The bodies in this universe attract one another name the scientist who propounded this statement

Colt1911 [192]

<h2><em>Answer: According to wikipedia Sir Isaac Newton has told this statement.</em></h2><h2><em>hope its helps you.</em></h2><h2><em>have a great day.</em></h2><h2><em> keep smiling be happy stay safe. </em></h2><h2><em /></h2>

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7 0

2 years ago

Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift

sammy [17]

Answer:

The doppler effect equation is:

$f' = \frac{v +v0}{v - vs}*f$

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0

2 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago