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MakcuM [25]
3 years ago
6

Keplers laws follow which law discovered by sir Isaac Newton

Physics
2 answers:
Zolol [24]3 years ago
8 0

“Universal Gravitational Law” discovered by Sir Issac Newton followed Kepler’s law.

<u>Explanation:</u>

The acceleration of the moon and objects on the earth are compared by Newton. Gravitational force directly depends on both the object's masses and inversely proportional to square of distance between the center of the objects.

Kepler’s Laws of Motion has been reviewed and Newton developed the proof that elliptical form of planetary orbits resulting the centripetal force which is inversely proportional to square of radius vector.

V125BC [204]3 years ago
7 0

The Keplers Laws about the motion of planets leds to Newton to discovered the three laws of motion. Newton based on Kepler came to the conclusion that all motion whereas it was follow the same principle.

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1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
3 years ago
Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted
kkurt [141]

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

#PLATOLIVESMATTER

7 0
3 years ago
A 55.4 g sample of water at 99.61 °C is placed in a constant pressure calorimeter. Then, 23.4 g of zinc metal at 21.6 °C is adde
Zolol [24]

Answer:

The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C

Explanation:

From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture

Heat loss by water = Heat gain by zinc metal

m1C1(T1 - T3) = m2C2(T3 - T2)

m1 is mass of water = 55.4 g

C1 is specific heat capacity of water = 4.2 J/g.°C

m2 is mass of zinc metal = 23.4 g

C2 is specific heat capacity of zinc metal

T1 is the initial temperature of water = 99.61 °C

T2 is the initial temperature of zinc metal = 21.6 °C

T3 is the final temperature of the mixture = 96.4 °C

55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)

746.9028 = 1750.32C2

C2 = 746.9028/1750.32 = 0.427 J/g.°C

3 0
3 years ago
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri
lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point.  If the board is just about to tip, the normal force from the left support will be going to zero.  So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0  ⇒     τ₁ + τ₂ = 0  

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒  τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒  τ₂ = Wman*x = 522 N*x   (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒  x = 0.6034 m

7 0
3 years ago
Wha is the frequency of a wave having a period equal to 18 seconds?
Rainbow [258]

Answer:

5.5 × 10-2 hertz

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

= 0.055 per second          (1 cycle per second = 1 Hertz)

Thus, we can conclude that the frequency of the wave is 5.5 X 10^{-2} hertz.

Hopes this helps, love <3

5 0
3 years ago
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