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Murrr4er [49]
4 years ago
7

The law of harmonies is referred to as _____.

Physics
1 answer:
AnnZ [28]4 years ago
3 0

Answer:

Kepler's third law

Explanation:

As per Kepler's third law all planets revolve around the Sun in such a way that their square of time period of revolution around the Sun is proportional to the cube of the mean radius.

It is given as

T^2 = cR^3

here

c = proportionality constant

R = mean radius

T = time period around the sun

So here all celestial body and all planets revolve as per the Kepler's third law which is also known as law of harmonies

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A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0
FromTheMoon [43]

Answer:

v_0 =1.71

Explanation:

the parabolic movment is described by the following equation:

y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2

where y is the height of the ball, a is the angle of launch, v_0 the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2

Finally, solving for v_0, we get:

v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}

v_0 =1.71

4 0
4 years ago
A squirrel is trying to locate some nuts he buried for the winter. He moves 4.3 m to the right of a stone and dogs unsuccessfull
krek1111 [17]

Answer:

The total displacement from the starting point is 1.5 m.

Explanation:

You need to sum and substract, depending on the movement (to the right, sum; to the left, substract).

First, it moves 4.3 m right and return 1.1 m. So the new distance from the starting point is 3.2 m.

Second, it moves 6.3 m right, so the new distance is 9.5 m.

Finally it moves 8 m to the left, so 9.5 m - 8 m= 1.5 m.

Summarizing, at the end the squirrel is 1.5 m from its starting point.

8 0
4 years ago
Describe a ball's motion as it rolls up a slanted
emmasim [6.3K]

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

  • initial velocity of the ball, u = 1.25 m/s
  • time of motion of the ball, t = 4.22 s

As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

a = \frac{v_f -v_0}{t} \\\\

<em>at the highest point on the incline plane, the final velocity </em>v_f<em> is zero</em>

a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

Learn more here:brainly.com/question/23860763

4 0
3 years ago
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

                                        T^{2} \alpha R^{3}

                                         ⇒ T^{2} = KR^{3} where is k is the proportionality  constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence T_{1} =365.5 days

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence R_{1} =1 AU

The distance of venus from sun is 0.723 AU i.eR_{2} =0.723

From keplers law we know that-\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }

                            ⇒T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }

Putting the values mentioned above we get-

                                      T_{2} ^{2} =50,350.132851075

                                         ⇒ T_{2} =\sqrt{50,350.132851075}

                                        ⇒T_{2} = 224.388352752710 days.

Hence the time period of venus is 224.388352752710 days

                                         

                     






                           

7 0
4 years ago
Read 2 more answers
This for my previous question.. im so sorry
Marta_Voda [28]

Answer:

the answer for the question is C

3 0
3 years ago
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