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3 years ago

I WILL MARK BRAINLIEST TO WHOEVER GETS THIS CORRECT

Physics

2 answers:

Veseljchak [2.6K]3 years ago

6 0

Answer:

If south-east Texas is where H then it is C, but if not the answer is A.

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

b or c pretty sure its b b tho

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State the laws of vibration of a stringed Instrument

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If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.

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3 years ago

A 15kg ball accelerates at a rate of 3m/s/s. What force was required?

hoa [83]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15 × 3

We have the final answer as

Hope this helps you

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3 years ago

Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.

BARSIC [14]

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Gases that are very good at absorbing long wave photons of infrared light

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Green house gases examples are Carbon monoxide, nitric oxide etc.
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They melts the glaciers and increase the sea level .

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2 years ago

Is adequate health care are available to all

tankabanditka [31]

Explanation:

I think for anyone to answer this we need more info on what you want answered. The Sentence Itself doesn't Make Since To Me

6 0

2 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