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3 years ago

7

An object travels in a straight line for 1200 m at a velocity of 50 m/s, what is its time of travel?

1 answer:

Crank3 years ago

5 0

Given that,

distance = d = 1200m

velocity/speed = s = 50m/s

time = t

Also,

s = d/t

And,

t = d/s

t = 1200/50

t = 120/5

t = 24

Therefore, it takes 24s to travel.

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A van is traveling with an initial velocity of 12 m/s. The driver takes a time of 45 seconds to speed up to a velocity of 20 m/s

Rufina [12.5K]

Initial velocity=u=12m/s
Final velocity=v=20m/s
Time=t=45s

$\\ \rm\hookrightarrow Acceleration=\dfrac{v-u}{t}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{20-12}{45}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{8}{45}$

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Now

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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 $v_{1}$

as

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

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

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

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

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

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

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

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3 years ago

Anyone know these questions?

salantis [7]

400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent

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