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s344n2d4d5 [400]

3 years ago

5

Explain Kepler's second law in detail with figures.

1 answer:

Jet001 [13]3 years ago

5 0

Answer:

Kepler's second law of planetary motion describes the speed of a planet traveling in an elliptical orbit around the sun. It states that a line between the sun and the planet sweeps equal areas in equal times. Thus, the speed of the planet increases as it nears the sun and decreases as it recedes from the sun.

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39 liters of water at

Harman [31]

Answer:

Temperature of the mixture=8 °C

Explanation:

Let, Temperature of the mixture=t °C

Q₁ = Q₂

=> m₁SΔΘ₁ = m₂SΔΘ₂

=> 39 × (20+t) = 21 × (60-t)

∴ t = 8 °C

hope you have understood this

pls mark it as the brainliest

3 0

3 years ago

1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?

labwork [276]

Answer:

1.35 kJ

Explanation:

KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

7 0

3 years ago

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EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

jek_recluse [69]

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3 years ago

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S waves arrive at distant points before other seismic waves, true or false

Step2247 [10]

False. p waves, than s waves

3 0

3 years ago

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

Harman [31]

Answer: a) 11.76 m/s b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

4 0

3 years ago