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3 years ago

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A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position

and given an initial velocity of 1.0 m/s back towards equilibrium. a) What is the total mechanical energy of the motion

1 answer:

Svet_ta [14]3 years ago

7 0

Answer:

Explanation:i think this would help u

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A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?

Olenka [21]

car starts from rest

$v_i = 0$

final speed attained by the car is

$v_f = 5 m/s$

acceleration of the car will be

$a = 3.6 m/s^2$

now the time to reach this final speed will be

$t = \frac{v_f - v_i}{a}$

$t = \frac{5 - 0}{3.6}$

$t = 1.39 s$

so it required 1.39 s to reach this final speed

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3 years ago

In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin

Genrish500 [490]

Energy can not be created or destroyed but can change from one form to another.

example: as a roller coaster cart loses height the more speed it gains, the potential energy is transferred into kenetic energy

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3 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

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

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

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3 years ago

A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi

sleet_krkn [62]

Answer:

Equilibrium temperature will be $T=52.2684^{\circ}C$

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So $2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)$

$0.334T-3.67=1688-32.031T$

$T=52.2684^{\circ}C$

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3 years ago

Determine the magnitude of the electrostatic force on a 0.06000 C charged object when it is placed in an electric field of magni

joja [24]

Answer:

Explanation:

Use the following equation:

$E=\frac{q}{F}$ and solve for F:

$F=\frac{q}{E}$ and filling in:

$F=\frac{.0600}{1500}$

F = 4.0 × $10^{-4$ N

3 0

3 years ago