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3 years ago

What is another name for Kepler's third law

Physics

1 answer:

Vsevolod [243]3 years ago

6 0

Keplers third law is also called the Law of Orbits

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Solvents are the substances that dissolve solutes; therefore, solvents are always liquids.

Nina [5.8K]

False, although they are usually a liquid. Solvents can be a solid or gas as well. Also, solutes can be in any state as well.

4 0

3 years ago

A spring stretches by 0.014 m when a 3.7-kg object is suspended from its end. How much mass should be attached to this spring so

lidiya [134]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of th

Travka [436]

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

4 0

4 years ago

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$

Substitute the given values and solve the speed;

$v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s$

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

7 0

3 years ago

A space vehicle is traveling at 2980 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

Strike441 [17]

Answer:

3054.4 km/h

Explanation:

Using the conservation of momentum

momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor

initial momentum = 14900 M km/h

let v be the new speed of the motor so that the

new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )

total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M

initial momentum = final momentum

14900 M km/h = 5 Mv + 93M

14900 km/h = 5v + 93

14900 - 93 = 5v

v = 2961.4 km/h

the speed of the module = 2961.4 + 93 = 3054.4 km/h

8 0

3 years ago