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marishachu [46]
3 years ago
14

Which section (A or B) of the roller coaster shows positive acceleration?

Physics
2 answers:
Step2247 [10]3 years ago
6 0

It is (A) for the first one and (B) for the second one.

SVEN [57.7K]3 years ago
6 0

Answer:

A,B

Explanation:

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A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th
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Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

V = 200\,cm^{3}+800\,cm^{3}

V = 1000\,cm^{3}

The total volume of the air is 1000 cubic centimeters.

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Which of these scientists had the greatest contribution to early microscopy?
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C

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6 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
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