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svp [43]
3 years ago
8

What is relative velocity?​

Physics
2 answers:
hjlf3 years ago
6 0
The relative velocity is the velocity of an object
joja [24]3 years ago
5 0
Search it up bruhhhhhhhhhhhhh
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A humanoid skeleton is found buried in the ashes of a volcano that erupted between 10,000 and 12,000 years ago. when scientists
Inga [223]
From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
                                           A(t) = A(0)(1/2)^t
where t is the number of half-lives, in this case 2. Thus, only about 1/4 of the original amount will be left. 
5 0
3 years ago
Read 2 more answers
The mass of a car is 2,400 kg and its momentum is 22,240 kg<br> m/s. What is its velocity?
Crazy boy [7]

Answer:

9.27

Explanation:

6 0
3 years ago
The relative highness or lowness of a sound is called ______. Multiple Choice pitch timbre dynamics octave
Assoli18 [71]
Pitch is the answer…….
8 0
2 years ago
As a projectile falls, what happens to the components of velocity?
netineya [11]

Answer:

Option (c).

Explanation:

An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,

v_x=v\cos\theta\ \text{and}\ v_y=v\sin\theta

\theta is the angle of projection

The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.

As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same .

7 0
3 years ago
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

5 0
3 years ago
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