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diamong [38]
2 years ago
13

If someone looks far enough into space, they should be able to see the beginning of the universe true of false

Physics
1 answer:
kodGreya [7K]2 years ago
8 0
The answer is no, it would be impossible to see the beginning of the universe
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What 2 things need to be known about an object in order to determine its kinetic energy?
NikAS [45]
I believe the answer is the mass of the object and the speed at which it is moving. 
7 0
3 years ago
A private aviation helicopter's main rotor blades rotate at approximately
Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity \omega a helicopter's main rotor blades:

\omega=450 rpm=450 \frac{rev}{min}

However, we are asked to express this \omega in the International Systrm (SI) units. In this sense, the SI unit for time is second (s):

\omega=450 \frac{rev}{min} \frac{1 min}{60 s}

\omega=7.5 \frac{rev}{s}

4 0
3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a
Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

d = v*t \rightarrow v = \frac{d}{t}    

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

t = \sqrt{\frac{2d}{g}}

<u>Where:</u>

g: is gravity = 9.8 m/s²

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s

Now, the horizontal velocity of the rock is:

v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s

So, the distance is:

d = v*t = 6.67 m/s*1.43 s = 9.54 m    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!        

7 0
3 years ago
If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s
Sladkaya [172]
 the orbital period is 5170 s
6 0
3 years ago
A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners
galina1969 [7]

Answer:

Explanation:

As the source is situated on x - axis , it must be situated in between the two listeners .

So the x coordinate of source is

(-7 + 3 )/2

= - 2 m

The equation of the wave- front will be that o a circle having centre at (-2,0)

and radius = distance between -2 and 3 , that is 5 m

equation of circle

=( x+2 )² + y² = 25

It cuts y axis when x = 0

Putting x = 0

4 + y² = 25

y² = 21

y = + √21 , or - √21

7 0
3 years ago
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