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charle [14.2K]
4 years ago
12

Would you rather be beautiful but you cant look at yourself or be ugly and no one else can see you

Physics
1 answer:
WARRIOR [948]4 years ago
6 0
Whats the point of being either of them?

If your Beautiful, you cant look at yourself...

And if your ugly, no one else can see you...

I'd rather not exist LOL XD
You might be interested in
Which of the following statements are true?
inessss [21]

Answer:

a. If an object's speed is constant, then its acceleration must be zero.

FALSE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity

b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

c. If an object's velocity is constant, then its speed must be constant.

TRUE

Since velocity is constant then it shows that its magnitude and direction both are constant so its speed is also constant.

d. If an object's acceleration is zero, its velocity must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

a = \frac{d\vec v}{dt}

Since we know that if acceleration is 0 then velocity must be constant

e. If an object's speed is constant, then its velocity must be constant.

FALSE

Speed is just the magnitude so we can not say about its direction and hence if speed is constant then velocity may or may not change

7 0
3 years ago
In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

8 0
3 years ago
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
3 years ago
Describe at least TWO hazardous conditions that exist in space AND the technological features a spacecraft must have in order to
nexus9112 [7]

gamma radiation and heat flares from the sun, they use refelective gold sheets

3 0
3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
Read 2 more answers
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