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3 years ago

6

Work in the amount of 280 J is done in lifting an object a distance of 4.0 m in a time. How much force did it take to lift the o

bject?

1 answer:

ki77a [65]3 years ago

7 0

W=fd
W/d=F
280/4=F
70N=F

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What is the speed of an object that travels 60 metres in 4 seconds

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Answer:

$s = \frac{d}{t} = \frac{60}{4} \\ \boxed{speed = 15m. {sec}^{ - 1} }$

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3 years ago

Read 2 more answers

The voltage across the terminals of a 250nF capacitor is푣푣=�50푉푉, 푡푡≤0(푚푚1푒푒−4000푡푡+푚푚2푡푡푒푒−4000푡푡)푉푉, 푡푡 ≥0The initial current

olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

Answer:

A) initial energy = 0.3125 mJ

B) A1 = 50 and A2 = 1,800,000

C) Capacitor Current is given by the expression;

I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

U = (1/2)Cv²

Where C is capacitance and v is voltage.

So initial kinetic energy;

U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ

B) from the question, we know that;

A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

v(0) = 50

Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

Current (I) = C(dv/dt)

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

Thus, Initial current (Io) is;

Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]

We know that Io = 400mA from the question or 0.4 A

Thus;

0.4 = (250 x 10^(-9))[-4000A1 + A2]

0.4 = 0.001A1 - (250 x 10^(-9)A2)

Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

Thus, making A2 the subject, we obtain;

(0.4 + 0.05)/(250 x 10^(-9))= A2

A2 = 1,800,000

C) We have derived that ;

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

So putting values of A1 = 50 and A2 = 1,800,000 we obtain;

I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]

I = e^(-4000t)[0.05 + 0.45 - 1800t]

I = e^(-4000t)[0.95 - 1800t]

5 0

3 years ago

Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational

vampirchik [111]

Answer:

I) $c=1385.667\frac{J}{kg K}$

II)The difference from the value obtained on part I is: 2000-1385.67 =614.33 $\frac{J}{Kg K}$

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

Explanation:

From the problem we have the molar mass given $M=18\frac{gr}{mol}$ of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.

Part I

Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water is 18.0 g/mol=0.018kg/mol.

Let $C_v (\frac{J}{Kg K})$ the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.

Let $C (\frac{J}{Kg K})$ the specific heat capcity this value represent the heat capacity aboserbed by mass.

For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by $C_v =\frac{R}{2}$ so the total for the 6 degrees of freedom would be:

$C_v =6*\frac{R}{2}=3R=3x8.314\frac{J}{mol K}=24.942\frac{J}{mol K}$

And by definition we know that the specific heat capacity is defined:

$c=\frac{C_V}{M}$

If we replace all the values we have:

$c=\frac{24.942\frac{J}{mol K}}{0.018\frac{kg}{mol}}=1385.667\frac{J}{kg K}$

So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is $c=1385.667\frac{J}{kg K}$

Part II

The actual specific heat of water vapor at low pressures is about 2000 J/(kg * K). Compare this with your calculation.

The difference from the value obtained on part I is: 2000-1385.67 =614.33 $\frac{J}{Kg K}$

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

4 0

3 years ago

If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre

creativ13 [48]

Answer:

$-112.876J$

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

$16KNO_3(s) + 24C(s) + S_8(s) \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)$

Let us define $P - V$ work as;

$w_{pv} = - P_{external} \triangle Volume$

where $\triangle (Volume) = (V_{final} - V_{initial})$

External pressure is given as $1.00$ $atm$ , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence, $V_i = 0.$

To find the volume of the products, we need to first find the amount of moles of the product made from $2.40_gKNO_3,$ using the molar mass of $KNO_3$ which is 101.1032 g/mol

$2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3$

Now let us convert moles of $KNO_3$ into moles of $CO_2$ and $N_2$ using the stoichiometric ratios from our balanced equation of the reaction.

$0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2$

$0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2$

$K_2S$ is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of $CO_2$ and $N_2$ into grams using their respective molar masses.

$0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2$

$0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2$

We will now proceed to convert grams into volume using the density values provided.

$1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2$

$0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2$

Summing up the two volumes, we get the final volume

$0.856L + 0.258L = 1.114L = V_f$

Plugging everything into the $w_{pv}$ equation, we get:

$w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm$

Finally, let us convert $L.atm$ into joules using the conversion rate of;

$1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J$

7 0

3 years ago