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3 years ago

What average force is necessary to bring a 50.2-kg sled from rest to a speed of 2.8 m/s in a period of 20.1 s?

1 answer:

7 0

To solve this problem we will use the concepts related to the Impulse-Momentum Theorem for which it is specified as the product between force and change in time

$\Delta p = F\Delta t$

And

\Delta p = m\Delta v

Where,

$F = Force$

$\Delta t = \text{Change in Time}$

$\Delta v = \text{Change in velocity}$

$m = mass$

Rearranging to find the Force we have that

$F = \frac{\Delta p}{\Delta t}$

Using the expression between mass and velocity

$F = \frac{m(v_f-v_i)}{\Delta t}$

Our values are given as,

$m = 50.2kg\\v_i = 0m/s \\v_f = 2.8m/s \\\Delta t = 20.1s$

Then replacing we have that

$F = 6.99N$

Therefore the average force is 6.99N

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a=128.6/32.5

a=3.95

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3 years ago

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3 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

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3 years ago