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Studentka2010 [4]

3 years ago

10

3. Light travels from the Sun to Earth in 8.3 min. Given that the speed of light is 3.00108 m/s, what is the distance in meters

between the Sun and Earth?

1 answer:

nasty-shy [4]3 years ago

4 0

Answer:

13

Explanation:

13.0120481928 it is the distance

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Time period of a simple pendulum if it makes 40 oscillations in 20 seconds.

Hunter-Best [27]

(20 sec)/(40 swings) = 0.5 sec/swing

5 0

3 years ago

Find the perimeter of the rectangle if each block represents one foot(show your calculation)

iren2701 [21]

I'm pretty sure that the "block" of which you speak is one in a pattern
of them that covers the drawing you have of the rectangle, and now
I need to explain something to you:

The REASON for printing that drawing next to the question that you
partially copied is that the drawing has information that's needed to
answer the question with, and rather than repeat all that information
in the question, it just says "LOOK AT THE DRAWING !"

In fact, the whole point of the question may not be just to remind you of
what "perimeter" means. It's more likely that the purpose of this problem
is to make you pick the information you need off of a drawing.

Either way, if you'll kind of "read between the lines" of the part of the
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going nowhere in the direction of a solution without SEEing the drawing.

So my bottom-line conclusion regarding a solution for this problem is:
Not possible with the given information.

4 0

3 years ago

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

7 0

3 years ago

What would the weight of an astronaut be on Neptune if his mass is 68 kg and acceleration of gravity of Neptune is 11.15 m/s^2?

Dima020 [189]

Weight = Mass of object * Gravity of experimented place
Here, m = 68 Kg
g' = 11.15 m/s²

Substitute their values,
w = 68 * 11.15
w = 758.2 N

In short, Your Answer would be 758.2 Newtons

Hope this helps!

6 0

3 years ago

Explain the limitations of electrical energy production created exclusively by renewable energy sources.

Diano4ka-milaya [45]

Answer:

The main limitation while the production of electrical energy is that the source from which electrical energy products cannot be available all the time

Explanation:

The main limitation while the production of electrical energy is that the source from which electrical energy products cannot be available all the time. There can be many factors that can hinder the availability of the source. for example, the Availability of solar energy can be hindered due to the cloud.

one the other important thing that needs to be taken into consideration is that although renewable energy is considered to be environment-friendly but to rely on this point cannot be true always

5 0

3 years ago