How is the kelvin scale different from the Celsius scale?

A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0

3 years ago

What do you mean by MA of a lever is 3 , VR of a lever is 4 and efficiency of a machine is 60%

Montano1993 [528]

Explanation:V.R of a lever is 3 means effort distance of the lever is 3 times more than the effort distance of the load. ... Efficiency of a lever is 60% means 40% of the energy is wasted due to friction as no machine is Friction less.

6 0

3 years ago

Help me please it's due tomorrow

maria [59]

If you can get a clearer image, I would be able to complete most, if not, all problems on the paper.

5 0

2 years ago

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Uranium was used to generate electricity in a nuclear reactor of a nuclear power station. The nuclear reactor was 29% efficient

Minchanka [31]

Answer:

$C_{lifetime} = 2,618,017,174\,USD$

Explanation:

The energy contained per second in the fuel is:

$\dot E_{fuel} = \frac{800\times 10^{6}\,W}{0.29}$

$\dot E_{fuel} = 2.758\times 10^{9}\,W$

It is know that fission of a gram of uranium liberates $5.8\times 10^{8}\,J$ . Mass flow rate is computed herein:

$\dot m_{U}= \frac{2.758\times 10^{9}\,W}{5.8\times 10^{8}\,\frac{J}{g}}$

$\dot m_{U} = 4.755\,\frac{g}{s}$

The needed quantity of fuel per second is:

$\dot m_{fuel} = \frac{4.755\,\frac{g}{s} }{0.04}$

$\dot m_{fuel} = 118.875\,\frac{g}{s}$

The annual fuel consumption is now obtained:

$\Delta m_{fuel,year} = (0.119\,\frac{kg}{s})\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{24\,h}{1\,day} )\cdot(\frac{365\,days}{1\,year} )$

$\Delta m_{fuel,year} = 3,752,784\,\frac{kg}{year}$

The cost of fuel for the lifetime of the plant is:

$C_{lifetime} = (3,752,784\,\frac{kg}{year} )\cdot (\frac{0.453\,lb}{1\,kg} )\cdot [(20\,years)\cdot(\frac{12\,USD}{1\,lb} )+(20\,years)\cdot (\frac{65\,USD}{1\,lb} )$

$C_{lifetime} = 2,618,017,174\,USD$

6 0

3 years ago

THE MASS OF VENUS IS 81.5% OF THE EARTH AND ITS RADIUS IS 94.9% THAT OF THE EARTH. A)COMPUTE THE ACCELERATION DUE TO GRAVITY ON

vazorg [7]

I'd just keep in mind the relationships for: g = 9.81 m/s²
and universal gravitational acceleration: 
ie that: 
for MASS, its directly proportional: 
a mass that is 0.815 of Earth's would have a gravitational acceleration that's: 
0.815g 
and 
for RADIUS (R) of planet, it's inversely proportional to the square of R 
a planet with radius that's 0.949 of Earth's would have a gravitational acceleration of: 
(1/0.949)² = 1.11g 

the combination of these two effects would be: 
(0.815 • 1.11)g = 0.905g = 8.88 m/s²

6 0

3 years ago

Read 2 more answers