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3 years ago

A machine has a mechanical advantage of 0.6. What force should be applied to the machine to make it apply 600 N to an object?

Chemistry

1 answer:

Vinil7 [7]3 years ago

8 0

Answer:

1000N is needed to be applied.

Explanation:

Machines make doing work easier. They allow us use small effort to carry out work on huge amount of load.

The mechanical advantage of a machine;

(M.A) =load/effort

M.A = 0.6

Load =600N

effort =?

0.6 = 600/effort

effort = 600/0.6

effort = 1000N

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4. A 14.5-L balloon is in the air where it is 20.0 C at 0.980-atm. A wind passes over and the balloon's pressure becomes 740.0-m

jeka94

The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

<h3>What is Combined gas law?</h3>

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

Initial volume V₁ = 14.5L

Initial pressure P₁ = 0.980atm

Initial temperature T₁ = 20.0°C = 293.15K

Final volume V₂ = 14.3L

Final pressure P₂ = 740.mmHg = 0.973684atm

Final temperature T₂ = ?

We substitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂

14.21Latm / 293.15K = 13.92368Latm / T₂

14.21Latm × T₂ = 13.92368Latm × 293.15K

14.21Latm × T₂ = 4081.72679LatmK

T₂ = 4081.72679LatmK / 14.21Latm

T₂ = 287.24K

T₂ = 14.09°C

Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

Learn more about the combined gas law here: brainly.com/question/25944795

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4 0

2 years ago

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. ____CO +____ O2 → ____CO2

Black_prince [1.1K]

<h3><u>Answer;</u></h3>

2, Blank, 2 ;

<h3><u>Explanation;</u></h3>

The balanced chemical equation would be;

2 CO + O2 → 2 CO2

Balancing a chemical equation ensures that the number of atoms of each element are equal on both the reactants side and the products side. This ensures that the law of conservation of mass is obeyed in chemical reactions.

4 0

3 years ago

Read 2 more answers

Why can liquid boil at a low pressure?

Paladinen [302]

Answer: I found this online. Hope it helps you.

Explanation:

This pressure is transmitted throughout the liquid and makes it more difficult for bubbles to form and for boiling to take place. If the pressure is reduced, the liquid requires less energy to change to a gaseous phase, and boiling occurs at a lower temperature.

7 0

3 years ago

Explain why most metals are malleable and ductile but ionic crystals are not

Dima020 [189]

Throughout the metallic structure allowing the atoms to slide past each other. This sliding is why metals are ductile and malleable. Ioniccompound must break bonds to slide past one another, which causes the ionic material to split and crack.

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3 years ago

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an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago