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SpyIntel [72]
3 years ago
9

As a planet's semimajor axis gets smaller, its speed will _______ and its period will _________ .

Physics
1 answer:
Rufina [12.5K]3 years ago
4 0
Well first of all, a planet doesn't have a semimajor axis, although it's orbit does.

In an orbit with a smaller semimajor axis, the planet moves faster, and its orbital period is shorter.

That's why the International Space Station circles the Earth in less time than the Moon does.
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the power rating of an electric lawn mower is 2200 watts. if the lawn mower was used for 30 mins, and 650 Newtons of force was u
Tasya [4]

Answer:

Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m ... A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min ... A power mower does 9.00 x 105 J of work in 0.500 h. ... p: W 2200ch: w will320,000 T/ ... How much electrical energy (in kilowatt hours) would a 60.0 W light bulb use in ..

Explanation:

4 0
2 years ago
Question 4. Calculate the kinetic energy of a Year 11 pupil (mass 55Kg)
bogdanovich [222]

Answer:

990 J

Explanation:

Kinetic energy is:

KE = ½ mv²

Given m = 55 kg and v = 6 m/s:

KE = ½ (55 kg) (6 m/s)²

KE = 990 J

4 0
3 years ago
What is meant by pressure?write its unit.
Mars2501 [29]

Answer:

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed

Unit: Pascal (Pa)

1Pa = 1N/m^2

5 0
3 years ago
I need help with questions b and d, that’s all.<br><br> Thank you.
Mkey [24]

b).  The power depends on the RATE at which work is done.  

Power = (Work or Energy) / (time)

So to calculate it, you have to know how much work is done AND how much time that takes.

In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A.  But the question doesn't tell us anywhere how much time that takes.  So there's NO WAY to calculate the power needed to do it.

The more power is used, the faster the car is lifted.  The less power is used, the slower the car creeps up the first hill.  If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill.  That would just take a long time, but it could be done if the power is small enough.

Without knowing the time, we can't calculate the power.

...

d).  Kinetic energy = (1/2) · (mass) · (speed squared)

On the way up, the car stops when it reaches point-A.  

On the way down, the car leaves point-A from "rest".

WHILE it's at point-A, it has <u><em>no speed</em></u>.  So it has no (<em>zero</em>) kinetic energy.

7 0
2 years ago
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle conn
suter [353]

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × \frac{V^{2} }{2g}

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

5 0
2 years ago
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