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2 years ago

What type of tool seems to be the mist accurate for measuring liquids?

Physics

2 answers:

aliina [53]2 years ago

8 0

mesuring cup , beaker, graduated cylinder, burets, pipette or micropipette. <3

Anit [1.1K]2 years ago

6 0

Volumetric flasks are most accurate

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Good evening everyone, State the law of conservation of energy .Thank u

disa [49]

Energy can neither be created nor be destroyed but can be converted from one form to another .

3 0

2 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago

Which of the following examples could this free body diagram describe? (Check all that apply)

boyakko [2]

Answer:

A car accelerating to the right

Explanation:

The free-body diagram shows all the forces acting on an object. The length of each arrow is proportional to the magnitude of the force represented by that arrow.

In this free-body diagram, we see that there are 4 forces acting on the object, in 4 different directions. We also see that the two vertical forces are equal so they are balanced, while the force to the rigth is larger than the force to the left: this means that there is a net force to the right, so the object is accelerating to the right.

Therefore, the correct answer is:

A car accelerating to the right

5 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

8 0

3 years ago

You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w

solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

= 0.025×< -2, 0, -7 >

= < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

= 0.025×< 4, 4, -3 >

= < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact = < -0.05, 0, 0.175 > + < 0.1, 0.1, -0.075>

= < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

Let the final velocity = v_f

Total mass = 0.025 + 0.025 = 0.05 kg

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

= < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0

2 years ago