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2 years ago

Diana raises a 1000. N piano a distance of 5.00 m using a set of pulleys. She pulls

Chemistry

1 answer:

Alina [70]2 years ago

5 0

Answer:

a) 250 N

b) 50 N

c) 5000 J

d) 4

e) 3.33

Explanation:

Given that weight of piano ( $F_r$ ) = 1000 N, distance moved by pulley ( $d_r$ ) = 5 m and the rope used ( $d_e$ ) =20 m

a) The effort applied ( $F_e$ ) if it was an ideal machine is:

$F_e=\frac{F_rd_r}{d_e} = \frac{1000*5}{20}=250N$

b) If the actual effort = 300 N

The force to overcome friction = actual effort - $F_e$ = 300 - 250 = 50 N

c) Work output = $F_rd_r=1000*5=5000J$

d) ideal mechanical advantage (IMA) = $\frac{d_e}{d_r} =\frac{20}{5} = 4$

e) Input force = 300 N, Therefore:

actual mechanical advantage = $F_r$ / input force = 1000 / 300 = 3.33

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PLEASE ASSIST MEEEEEEE!!!!!!!!! 40 POINTS

lana66690 [7]

Answer:

Wavelength of this beam of light: $\rm 4.39\times 10^{-7}\; m$ .

Explanation:

The speed of light in vacuum is approximately $\rm 2.998\times 10^{8}\;m \cdot s^{-1}$ .

Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.

On the other hand, the frequency of a wave is the number of periods in unit time. $1\rm \; Hz$ means one oscillation per second. The frequency of this particular wave is $\rm 6.83\times 10^{14}\; Hz$ . In other words, there are $6.83\times 10^{14}$ oscillations in each second.

The period of oscillation will be equal to

$\displaystyle t = \frac{1}{f} = \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}}$ .

In that period of time, a beam of light in vacuum would have traveled

$\displaystyle \rm 2.998\times 10^{8}\; m\cdot s^{-1} \times \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}} = 4.39\times 10^{-7}\; m$ .

In other words, if this beam of light of frequency $\rm 6.83\times 10^{14}\; Hz$ is in vacuum, its wavelength will be equal to $\rm 4.39\times 10^{-7}\; m$ .

8 0

2 years ago

The equilibrium constant, Kc, for the reaction H2 (g) + I2 (g) ⇄ 2HI (g) at 425°C is 54.8. A reaction vessel contains 0.0890 M H

Mars2501 [29]

Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the reaction quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

The expression for $Q$ is written as:

$Q=\frac{[HI]^2}{[H_2]^1[I_2]^1}$

$Q=\frac{[0.0890]^2}{[0.215]^1[0.498]^1}$

$Q=0.074$

Given : $K_{eq}$ = 54.8

Thus as $Q$ , the reaction will shift towards the right i.e. towards the product side.

4 0

2 years ago

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =

IgorC [24]

According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2

8 0

2 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

6 0

2 years ago

If honey has a density of 1.36 g/ml what is the mass of 1.25 qt reported in kilograms

yawa3891 [41]

Answer;
1.6 kg.

Solution;
The density is 1.36 g/ml;

The volume is 1.25 qt
However; 1 qt = 946.35 ml

Mass is given by; density × volume;
= 1.25 qts × 946.25 ml/qt × 1.36 g/ml =1609 g
but; 1 kg = 1000 g
Hence the mass = 1609/1000 = 1.609 Kg or 1.61 (sig figs)

7 0

2 years ago