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7nadin3 [17]
3 years ago
11

Is Kepler's constant the same for moons orbiting a planet as it is for planets orbiting the sun? Why?

Physics
2 answers:
d1i1m1o1n [39]3 years ago
7 0
Yes because we have day and night and seasons
baherus [9]3 years ago
6 0

The planets’ orbits are elliptical.

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What is the total resistance in the circuit? (include unit in answer - ohms)
solmaris [256]

Answer:60 ohms

Explanation:

R1=30 ohms

R2=15 ohms

R3=15 ohms

Let the total resistance be R

R=R1 + R2 + R3

R=30 + 15 +15

R=60

Total resistance is 60 ohms

4 0
3 years ago
What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?
Alexandra [31]
600/3 = 200
the slope is 200m/min
 
OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour 

6 0
3 years ago
The largest stars are _____ times the mass of the Sun. 10 100 1,000 10,000
Arte-miy333 [17]
<h3><u>Answer;</u></h3>

100 times

<h3><u>Explanation;</u></h3>
  • The largest stars are 100 times the mass of the Sun.
  • <u>The giant stars are about 10 to 100 times the radius of the sun</u>, which means they are 100 times brighter than the sun.
  • <em><u>The largest known star in terms of mass and brightness is known as the Pistol Star. It is believed to be 100 times as massive as our Sun, and 10,000,000 times as bright.</u></em>
5 0
3 years ago
A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
A2 kg object moves at a constant 5 m/s across a level surface. Determine its kinetic energy.
AURORKA [14]
KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J
6 0
3 years ago
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