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3 years ago

13

Anyone please helps me with this question... I'm stuck..

1 answer:

Zolol [24]3 years ago

7 0

Highest fluid potential energy: answer A
Because the fluid is pushed upwards and potential energy is function of height. Since point A is the highest, there is the highest potential energy.

highest fluid pressure: answer C
This is because it is at the bottom where you have a hydrostatic pressure component

increasing fluid speed: answer B
This is because the section of the pipe is smaller and in order to have the same fluid flow rate the speed must increase

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A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

2 years ago

Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun

yanalaym [24]

The answer is convection

6 0

3 years ago

What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?

sladkih [1.3K]

formula for gravitational P.E =mgh

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

7 0

2 years ago

A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

<u>Given :</u>

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

<u>Let's Slove :</u><u> </u>

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

3 0

1 year ago

During a workout, the football players at State U ran up the stadium stairs in 61 s. The stairs are 130 m long and inclined at a

NeTakaya

Answer:

P = 1097 Watt

Explanation:

given,

length of stairs, L = 130 m

inclination with horizontal,θ = 30°

mass of the football player = 105 Kg

time = 61 s

we know,

$Power = \dfrac{work}{time}$

Work = change in Potential energy

h = L sin 30°

h = 130 x 0.5

h = 65 m

W = m g h

W = 105 x 9.8 x 65

W = 66885 J

now,

$P = \dfrac{66885}{61}$

P = 1097 Watt

hence, the power output on the way is 1097 W

5 0

2 years ago