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2 years ago

An airplane increases its velocity from 60m/s to 80m/s in a time of 10s.

Physics

1 answer:

Elenna [48]2 years ago

4 0

$\huge\fbox{Answer ☘}$

$acceleration \: = \frac{v - u}{t} \\ = > \frac{80 - 60}{10} \\ \\ = > \frac{20}{10} \\ \\ = > 2m \: s {}^{ - 2}$

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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

3 years ago

A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block attached to it by 0.5 m. The block is released from

morpeh [17]

Answer:

0.7000cm

Explanation:

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6 0

2 years ago

If one light bulb burns out in a string of lights, and the rest stay lit, it is reasonable to assume that the lights are wired i

Blizzard [7]

The lights are wired in PARALLEL.

In fact, when the lights are connected in parallel, they are connected on separate branches to the source of voltage, so if one light bulb burns out, the other lights continue to work because the current continues to flow in the other branches of the circuit.

Vice-versa, if the light bulbs are connected in series, they are on the same branch This means that if one of them burns out, the circuit is open in that point, so the current cannot flow anymore and the other light bulbs turn off as well.

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2 years ago

The wing of an airplane experiences the forces as depicted in the vector diagram to the right. Using both one and two dimensiona

Vedmedyk [2.9K]

Answer:

A.) 3605.6 N

B.) 33.7 degree

Explanation:

To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components

Resolving into x component :

Sum of forces = 3500 - 500 = 3000N

Resolving into y component:

Sum of forces = 2000N

Resultant force Fr = sqrt ( Fx^2 + Fy^2)

Fr = sqrt ( 3000^2 + 2000^2 )

Fr = sqrt ( 9000000 + 4000000 )

Fr = sqrt ( 13000000)

Fr = 3605.6 N

Therefore, resultant force acting on the wing is 3605.6 N

The direction of the vector will be:

Tan Ø = Fy / Fx

Substitute Fx and Fy into the formula

Tan Ø = 2000 / 3000

Tan Ø = 0.66666

Ø = tan^-1(0. 66666)

Ø = 33.7 degree.

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3 years ago

Use the concepts of kinetic energy and potential energy to describe the motion of a child on a swing. Why does the child need a

andrew-mc [135]

When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.

The child may need a push from time to time because friction with the air causes loss of energy.

4 0

2 years ago

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