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victus00 [196]
3 years ago
9

What instrument is used to measure volume by displacement

Physics
1 answer:
kicyunya [14]3 years ago
8 0
Volumes of liquids such as water can be readily measured in a graduated cylinder.
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Westkost [7]

Answer:

I think D it could maybe B

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2 years ago
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A paragraph that can serve the specific purpose of creating a special effect within the essay is called?
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Its b.functional paragraph because writers use this for interest presents and special effects
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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?
poizon [28]

Answer:

\boxed {\boxed {\sf 7.84 \ Joules}}

Explanation:

The formula for potential energy is:

PE=m*g*h

where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m

Substitute the values into the formula.

PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)

Multiply the first two numbers.

  • 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
  • If we convert the units now, the problem will be much easier later on.
  • 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

PE=(3.92 \ N )(2 \ m)

Multiply.

  • 3.92 N* 2 m=7.84 N*m
  • 1 Newton meter is equal to 1 Joule (this is why we converted the units).
  • Our answer is equal to<u> 7.84 Joules.</u>

PE=7.84  \ J

6 0
2 years ago
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A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie
ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, A=7.4\ cm^2=7.4\times 10^{-4}\ m^2

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, B_1=0.5\ T

Final magnetic field, B_2=3\ T

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_2-B_1}{t}

\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}

\epsilon=1.65\times 10^{-3}\ V

Induced current in the loop of wire is given by :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.65\times 10^{-3}}{2.4}

I=6.87\times 10^{-4}\ A

So, the induced current in the loop of wire over this time is 6.87\times 10^{-4}\ A. Hence, this is the required solution.

7 0
3 years ago
HEre now can somebody help!! <br> Pic is what i need help with
AleksAgata [21]

Answer:

which of the cars are speeding up: c

which of the cars or slowing down: a

which of the cars are maintaning a constant speed: b

Explanation:

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