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HACTEHA [7]
3 years ago
6

A football player attempts a 30‐yd field goal. If he is able to impart a velocity u of 100 ft/sec to the ball, compute the minim

um angle θ. for which the ball will clear the crossbar of the goal. (Hint: Let m = tan θ)
Engineering
1 answer:
ki77a [65]3 years ago
4 0

Answer:2.53°

Explanation:The ball is undergoing a projectile motion that is why it has an angle

Range = (U^2 sin2a)/g

U= velocity of the ball=100ft/sec

Range of the ball(i.e horizontal distance the ball crossed)= 30yard=90ft

a=angle the ball is projected

g= acceleration due to gravity=9.8m/s^2

90=(100^2×sin2a)/9.8

90×9.8=10000×sin2a

882/10000=sin2a

0.0882=sin2a

Sin^-1 0.0882=2a

5.06=2a

a=5.06/2

a=2.53

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Write a Python program that does the following. Create a string that is a long series of words separated by spaces. The string i
dybincka [34]

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

\\ Then I used methods append(), insert() and extend() for adding elements to the list.

\\ Finally I converted list into a string using join() and adding space in between the elements of the list.

4 0
4 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
The combustion chamber has different shapes depending on the make and model of the engine. True or false
Alisiya [41]

Answer:

svdsdfdfsdfssdf

Explanation:

fsdsdfsdffsdsfdsdf

5 0
3 years ago
1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu
lianna [129]

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

5 0
3 years ago
Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio
SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

b)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

c)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0
3 years ago
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