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MAXImum [283]
3 years ago
5

What is the degree of precision of the graduated cylinder 0.01 mL 0.1 mL 1 mL 10 mL

Physics
2 answers:
Elodia [21]3 years ago
5 0

Answer:

0.01 ml

Explanation:

Graduated cylinder is used to measurement of volume of liquid.Generally it gives precise result up to 2 digit or up to 3 digit.

Graduated cylinder are of different types .10 mL graduated cylinder measure up to 2 digit while 100 mL graduated cylinder measure up to 1 digit .

So here it is not size of cylinder is not specified so for most precise result the degree of precision of graduated cylinder will be 0.01 mL (up to 2 didgit).

enot [183]3 years ago
4 0
I believe it is 0.1mL or letter B on Edgenuity.
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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

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Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

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For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

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We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

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