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3 years ago

The law of conservation of energy stated that ?

Physics

1 answer:

givi [52]3 years ago

5 0

Answer:

The total energy of an isolated remains constant. It is said to be conserved over time.

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A planet has a period of revolution about the Sun equal to and a mean distance from the Sun equal to R .T^2 varies directly as _

shutvik [7]

T² caries directly as R³ .

This is Kepler's 3rd law of planetary motion .

4 0

3 years ago

A 100-meter sprint is a race using only the straight side of a racetrack. A 400-meter sprint is a race that makes one complete l

Shalnov [3]

Speed uses distance and velocity uses displacement in its calculation.

For 100 m race, distance = displacement. Hence speed = velocity

For 400m race, distance ≠ displacement. distance = 400m whereas displacement = 0m. Hence speed ≠ velocity

3 0

3 years ago

Read 2 more answers

A 1000 kg race car rounds a curve with a radius of 50 m.

Kruka [31]

Answer:

5644556677888777766554433

5 0

2 years ago

Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4

arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

initial velocity of the car, u = 76 m/s

acceleration of the car, a = - 9 m/s²

time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

height above the ground, h = 500 m

velocity of spaceship, u = 150 m/s

time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

$h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) - (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m$

Learn more here: brainly.com/question/24527971

5 0

2 years ago

The planet Mercury takes 0.24 sidereal years to go around the sun. What is the distance from the center of Mercury to the center

lyudmila [28]

Answer:

5.77×10¹⁰ m

Explanation:

From the question,

Applying

Kepler's third law

P² = d³...................... Equation 1

Where P = Planet's period, d = distance between the center of the planet and the sun.

make d the subject of the formula n equation 1

d = $\sqrt[3]{P^{2} }$ .................. Equation 2

Given: P = 0.24 sidereal.

Substitute the value of P into equation 2

d = $\sqrt[3]{0.24^{2} }$

d = 0.386 Au

d = 0.386×1.496×10¹¹

d = 5.77×10¹⁰ m

6 0

3 years ago