Answer:
a = the lowest critical speed of the shaft 882.81 rad/s
b = new diameter 0.05m or 50mm
c = critical speed 1765.62rad/s
Explanation:
see the attached file
The air temperature over a lake decreases linearly with height after sunset, since air cools faster than water. The mean molar m
2 answers:
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Answer:
Expression of work done is
Work done to move the sled is given as 1.94 J
Explanation:
As we know that the formula of work done is given as
here we know that
F = 6 N
d = 0.4 m
so we will have
Answer:
the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south
Explanation:
given information:
Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus
A= 27
Ax = 27 sin 60 = - 23.4
Ay = 27 cos 60 = 13.5
Jane walks 16.0 m in a direction 30.0 ∘ south of west, so
B = 16
Bx = 16 cos 30 = -13.9
By = 16 sin 30 = -8
the direction that should be walked by Ricardo to go directly to Jane
R = √A²+B² - (2ABcos60)
= √27²+16² - (2(27)(16)(cos 60))
= 23.52 m
now we can use the sines law to find the angle
tan θ =
= By - Ay/Bx -Ax
= (-8 - 13.5)/(-13.9 - (-23.4))
θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))
= 24° east of south
Answer:
a = 5.05 x 10¹⁴ m/s²
Explanation:
Consider the motion along the horizontal direction
= velocity along the horizontal direction = 3.0 x 10⁶ m/s
t = time of travel
X = horizontal distance traveled = 11 cm = 0.11 m
Time of travel can be given as
inserting the values
t = 0.11/(3.0 x 10⁶)
t = 3.67 x 10⁻⁸ sec
Consider the motion along the vertical direction
Y = vertical distance traveled = 34 cm = 0.34 m
a = acceleration = ?
t = time of travel = 3.67 x 10⁻⁸ sec
= initial velocity along the vertical direction = 0 m/s
Using the kinematics equation
Y = t + (0.5) a t²
0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²
a = 5.05 x 10¹⁴ m/s²