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2 years ago

What is one of Kepler's laws of planetary motion?

Physics

2 answers:

Murljashka [212]2 years ago

7 0

Answer a is one of keplers 3 laws

Lelechka [254]2 years ago

4 0

so whats the answer .

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A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di

Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0

2 years ago

Read 2 more answers

FAST PLEASE What is the change in internal energy if 90 J of thermal energy is added to a

vladimir1956 [14]

A is right because I took the test

5 0

1 year ago

What is the kinetic energy of a ball that has a mass of 3kg and Is<br> moving at 5m/s?

Arada [10]

Answer:

E kin = 1/2 • m • v^2

E kin: kinetic energy

m: mass

v: velocity

Explanation:

6 0

2 years ago

If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?

Sphinxa [80]

The line that should be secured first in pushing the boat away from the dock in preparation to dock is the bow line. When the bow line is secured, it is best to reverse it and turn to the dock, this will engage the line to tighten in a way that will help it swing back in the dock.

4 0

3 years ago

Read 2 more answers

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago