Here,
height at failure, h1 = 525 m,
upward acceleration, a = 2.25 m/s^2,
velocity = v m/s,
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2,
<span>
SO, </span>
<span>
heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence,
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters
b)
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec
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The magnitude of the unknown height of the projectile is determined as 16.1 m.
<h3>
Magnitude of the height</h3>
The magnitude of the height of the projectile is calculated as follows;
H = u²sin²θ/2g
H = (36.6² x (sin 29)²)/(2 x 9.8)
H = 16.1 m
Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.
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Answer:
100 cc
Explanation:
Heat released in cooling human body by t degree
= mass of the body x specific heat of the body x t
Substituting the data given
Heat released by the body
= 70 x 3480 x 1
= 243600 J
Mass of water to be evaporated
= 243600 / latent heat of vaporization of water
= 243600 / 2420000
= .1 kg
= 100 g
volume of water
= mass / density
= 100 / 1
100 cc
1 / 10 litres.
Answer:
52.5°C
Explanation:
The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state
mh1 + W = mh2
h2 = h1 + W/m
h1 + Wα1/V1
242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg
=287.4 kJ/kg
From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation
i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)
= 50°C + 60 - 50/295.15 - 284.79
(287.4 - 284.79)°C
= 52.5°C
Answer:
The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.
Explanation:
I hope this helps a little bit.
