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3 years ago

2.

Physics

1 answer:

anzhelika [568]3 years ago

8 0

Answer:

44.72m/s

Explanation:

use th formula:vf²=vi²at

and then substitute the values

remember the units

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Find the instantaneous velocity at 1 s . can anyone help with c-h!!!

Lelu [443]

Rise over run at 1 second
It’s the same slope from 0 to 2 seconds
10/2=5mps
As a note all time points between 0and 2 will have this instantaneous velocity

Instantaneous velocity at time 2 is 0

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3 years ago

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3 years ago

A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

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3 years ago

A hot-air balloon is floating above a straight road. To estimate their height above the ground, the balloonists simultaneously m

Karolina [17]

Answer:

3.1 miles

Explanation:

To solve this question it is important to remember that the distance between two mile markers is approximately 1 mile

Once this is known, the question becomes very easy to solve. We make two triangle, which have the following three points

Triangle 1: Hot-Air-Balloon, Ground, Milepost 1 - With angle of depression 20

Triangle 2: Hot-Air-Balloon, Ground, Milepost 2 - With angle of depression 18

As a reminder, the angle of depression is simply the angle the balloonist's head makes with the horizontal plane to be able to see the milepost.

From this we can simply drive two formulas using the Tan function

Equation 1 - $Tan(90-18) = \frac{b+1}{h}$

Equation 2 - $Tan(90-20) = \frac{b}{h}$

Solving them simultaneously we get the value of height (h) to be 3.0852 miles or 3.1 miles

8 0

3 years ago

calculate the weight of an object of mass 100kg on the surface of a planet whose mass is 4.8×10^24 kg and diameter is 12000km.

notsponge [240]

Answer:

p= mv

where p is momentum

m is mass

v is velocity

so it's given p= 100kgm/sec

v= 4m/s

so putting in the formula

100= m × 4

m = 25kg

Explanation:

3 0

3 years ago