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3 years ago

There is no gravity in outer space . explain why this statement is true or false

1 answer:

lisabon 2012 [21]3 years ago

5 0

There is no gravity in outer space but when you near a planet or star you will start to feel or experience different pulls of gravity. hope this helps :) Please mark brainlist

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Can someone please give me the answers to this? ... please ...

alexira [117]

3. 2 meters per second 4. i think object 7. i’ll try to figure it out 8. .77 9. 25km 10. 10m each second

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3 years ago

A technician services the carburetor, and then, performs a complete governor system adjustment. The governor system on the engin

madreJ [45]

Answer: Either of the two

Explanation:

Either of the two - the governed idle spring or the normal primary governor spring can be adjusted first.

The main function of the carburetor is to mix air with fuel.

5 0

3 years ago

A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu

Tpy6a [65]

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

8 0

4 years ago

Which of the following sensory receptors would lead you to squint in bright light?. A. thermoreceptors B. mechanoreceptors C. ph

kipiarov [429]

<span>If my memory serves me well, sensory receptors which would lead you to squint in bright light are called </span><span>C. photoreceptors</span>

3 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago