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astra-53 [7]
4 years ago
15

Is time real If everything is relative to gravity

Physics
1 answer:
kozerog [31]4 years ago
4 0

Answer: yes time is real because though everything is relative to gravity they mean material things while time isn't a material thing making it different from gravity and leaving it as its own entity.

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Which of the options below has the correct number of each element for the compound?
slamgirl [31]

Answer:

there are no options buddy

7 0
3 years ago
Read 2 more answers
consider a charge of -15.0 mCmoving to the right a 2.00x10^6 m/s in a mganetic field of .0300 T pointing upwards. What is the ma
77julia77 [94]

Answer:

a = 1.13×10^-8 m/s²

Explanation:

f = ma = qvb \\ a =  \frac{qvb}{m}

q = 15×10^-3 C

v = 2×10^6 m/s

B = 0.03 T

m = 8.00 x10^10 kg

a = 900/8 x10^10

a = 1.13×10^-8 m/s²

5 0
3 years ago
Which of the following is the same in all frames and reference?
Otrada [13]

B.Speed of light  

The speed of light is used as a reference and is 3x10^8 m/s no matter what.

Hope this helps :)

7 0
3 years ago
A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces,
son4ous [18]

Answer:

It is True

Explanation:.

A  commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague,  existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone  reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

7 0
3 years ago
A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long. The other end of the rod is attach
tigry1 [53]

Answer:

Option B is the correct answer.

Explanation:

Refer the figure we have centripetal force at bottom of circle

       F_c=F_t-F_w\\\\\frac{mv^2}{r}=F_t-mg\\\\F_t=m\left ( \frac{v^2}{r}+g\right )

We have mass, m = 2 kg

Radius, r = 1.2 m

For circular motion to occur we have tension at top = 0

That is

        \frac{mv^2}{r}=mg\\\\v=\sqrt{rg}

Now let us find tension at bottom point

       F_t=2\times \left ( \frac{rg}{r}+g\right )=4g=40N

Option B is the correct answer.  

         

8 0
4 years ago
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