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worty [1.4K]
3 years ago
9

As a freely falling object speeds up, what is happening to its acceleration - does it increase, decrease, or stay the same? (a)

Ignore air resistance. (b) Consider air resistance.
Essay ANSWER
Physics
1 answer:
Doss [256]3 years ago
5 0

Answer:

(a)When ignoring air resistance its accelerating increases steadily .

(b)When considering air resistance then its acceleration decreases this could either be uniformly or unevenly.

Hope this helped.

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Unmmm to eat so we don't die
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The shape of a planet's orbit is influenced by gravity.
stiks02 [169]

Answer:

True

Explanation:

Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit.

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A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su
Sedaia [141]

Answer:

\boxed {\boxed {\sf 18 \ m/s}}

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

F_c= \frac{mv^2}r}

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

  • F_c= 85 kg*m/s²
  • m= 0.5 kg
  • r= 1.9 m

Substitute the values into the formula.

85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}

Isolate the variable v. First, multiply both sides by 1.9 meters.

(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m

(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}

161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2

Divide both sides by 0.5 kilograms.

\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}

\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2

323 \ m^2/s^2 = v^2

Take the square root of both sides of the equation.

\sqrt {323 \ m^2/s^2} =\sqrt{ v^2

\sqrt {323 \ m^2/s^2} =v

17.9722007556 \ m/s =v

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

18 \ m/s =v

The maximum speed is approximately <u>18 meters per second.</u>

8 0
3 years ago
A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca
Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, f_o=30\ cm

The focal length of the eyepiece lens of an astronomical telescope, f_e=5\ mm=0.5\ cm

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

m=\dfrac{-f_o}{f_e}

m=\dfrac{-30}{0.5}

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

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The image below models how air can move by convection at the Earth's equator.
Xelga [282]

Answer: C

Explanation: i hoped that helped!

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