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2 years ago

PLEASE HELP I AM SO STUMPED ABOUT THIS QUESTION PLEASE HELP

Physics

2 answers:

kiruha [24]2 years ago

7 0

This is my alt account just grinding points

Flauer [41]2 years ago

3 0

NO DONT CLICK THE LINK OR COPY IT INTO YOUR BROWSER ITS GONNA PUT A VIRUS ON UR DEVICE

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Which officeholders are elected by the electoral college

NikAS [45]

Answer:Electoral Vote

Explanation:i did it before

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2 years ago

Calculate the force exerted by a mental ball having a mass of 70kg moving with speed of 20m/s>2

lord [1]

Answer:

F = 1400 N

Explanation:

It is given that,

Mass of the ball, m = 70 kg

It is moving with an acceleration of 20 m/s². We need to find the force exerted by the ball.

Force is given by the product of mass and acceleration. So,

F = ma

$F=70\ kg\times \ 20m/s^2\\\\F=1400\ N$

So, the force of 1400 N is exerted by a metal ball.

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2 years ago

What is the resultant of the vectors shown?

ad-work [718]

Answer:

Option B

Explanation:

Option A is the wrong answer because the horizontal vector is in the opposite direction.

Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]

Option D is the wrong answer as the horizontal vector is in the opposite direction.

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1 year 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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2 years ago

A standing wave is the result of: refraction diffraction reflection resonance

natita [175]

A standing wave is the result of a reflection.

3 0

3 years ago

Read 2 more answers