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3 years ago

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The mass of an object is determined by: weighing the object with a scale multiplying the height, width, and length measuring the

object with an equal-arm balance using the water displacement method

1 answer:

rosijanka [135]3 years ago

7 0

The true scientific way is the last: using the water displacement method

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Please help me I’ll mark brainless .

Ivan

The mass is 10.811 hope this helps

7 0

2 years ago

A substance with a pH of 9 has

Lady bird [3.3K]

A) no H30+ ions or OH- ions.

3 0

3 years ago

A platinum ball weighing 100 g is removed from a furnace and dropped into 400 g of water at 0 degree C. If the equilibrium tempe

kakasveta [241]

Answer:

T = 1010 degree Celsius

Explanation:

mass of ball (Mb) = 100 g

mass of water (Mw) = 400 g

temp of water = 0 degree

specific heat of platinum (C) = 0.04 cal/g degree celsius

we can calculate the temperature of the furnace from the equation before

Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)

100 x 0.04 x ( T - 10) = 400 x (10 - 0)

4 (T - 10) = 4000

T - 10 = 1000

T = 1010 degree Celsius

3 0

3 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

I have a bottle of gas, the bottle can expand and contract. Initially the gas is at 1 kpa of pressure and a volume of 1 Liter, a

drek231 [11]

Answer:

P₂ = 1.22 kPa

Explanation:

This problem can be solved using the equation of state:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

where,

P₁ = initial pressure = 1 KPa

P₂ = final pressure = ?

V₁ = initial Volume = 1 liter

V₂ = final volume = 1.1 liter

T₁ = initial temperature = 290 k

T₂ = final temperature = 390 k

Therefore,

$\frac{(1\ kPa)(1\ liter)}{290\ k} =\frac{(P_2)(1.1\ liter)}{390\ k}\\\\P_2= \frac{(1\ kPa)(1\ liter)(390\ k)}{(290\ k)(1.1\ liter)}$

<u>P₂ = 1.22 kPa</u>

7 0

3 years ago