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3 years ago

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5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal whil

e you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.

1 answer:

Lelu [443]3 years ago

7 0

Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.

By Newton's second law, the net vertical force is

• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0

where a is the acceleration of the wagon.

Solve for n (the magnitude of the normal force) :

n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N

Then

f = 0.500 (58.0 N) = 29.0 N

Meanwhile, the horizontal component of the applied force has magnitude

(80.0 N) cos(30.0°) ≈ 69.3 N

Now calculate the work done by either force.

• friction: -(29.0 N) (10.0 m) = -290. J

• pull: (69.3 N) (10.0 m) = 693 J

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Answer:

Part a)

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Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

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$T_L cos65 = T_R cos80$

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now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

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$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

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3 years ago

A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi

Vika [28.1K]

Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

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When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

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Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

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AlladinOne [14]

Answer:

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Explanation:

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3 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

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Answer:

289282

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d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

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Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

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3 years ago

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WARRIOR [948]

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Explanation:

Given:

$\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}$

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The cross product $\textbf{A}×\textbf{B}$ is given by

$\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|$

$= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}$

$= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

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3 years ago