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3 years ago

To ancient peoples, why were planets special?

Physics

1 answer:

coldgirl [10]3 years ago

7 0

Answer:

Planets were like gods.

Explanation:

To the people of many ancient civilizations, the planets were thought to be deities. Our names for the planets are the Roman names for these deities. For example, Mars was the god of war and Venus the goddess of love.

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I need to submit my homework in an hour, this is only question i cant solve. someone pls help!!

Sauron [17]

Hope this helps.......☺

8 0

3 years ago

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

alukav5142 [94]

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

$1\ light\ year=3.72\times 10^{17}\ inches$

So, $4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches$

$4.218\ light\ year=1.56\times 10^{18}\ inches$

We need to convert 4.218 light-years barley corns.

Since, 1 barleycorn = 1/3 inch

$1\ inch=3\ barleycorn$

$1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn$

So, the nearest star to the Earth is at a distance of $4.68\times 10^{18}\ barleycorn$ . Hence, this is the required solution.

3 0

3 years ago

You throw a ball into the air. Which two forces cause the ball to gradually stop moving upward and then fall back to Earth?

Virty [35]

Answer:

I'm pretty sure the answer is D

Explanation:

Honestly it's just a guess so let me know if it's right :3

3 0

3 years ago

Read 2 more answers

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

Which instrument is used to measure depth of ocean ?

skelet666 [1.2K]

Echo sounding is a type of SONAR used to determine the depth of water by transmitting sound pulses into water. The time interval between emission and return of a pulse is recorded, which is used to determine the depth of water along with the speed of sound in water at the time.

5 0

2 years ago