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2 years ago

Is hunting good for the environment

Physics

1 answer:

Ipatiy [6.2K]2 years ago

4 0

Answer:

yes but no

Explanation:

if there weren't carnivorous animals then there would be overpopulation of animals but since there are maybe no

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An object that is slowing down is accelerating.<br> True<br> False

Art [367]

Answer:

True

Explanation

When an object slows down the Acceleration is in the other direction which “ slows it down

8 0

3 years ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

3 years ago

A coin is made of ______

lys-0071 [83]

Answer:

coins made with the combination of copper zinc and Nickel

A cloth is made of fabrics

A towel is made from cotton

an umbrella is made up of fiberglass

a sweater is made up of wool

hope it helps

please mark me as the brainliest

5 0

3 years ago

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f

Usimov [2.4K]

Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

$\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s$