Hey there!
Here is your answer:
<u><em>The proper answer to this question is option C "</em></u><span><u><em>0.00349".</em></u>
Reason:
</span><span><u><em>1 L = 100 cL. Or 1 cL = 0.01 L</em></u>
</span><span><u><em>34.9 cL = 34.9 / 100 L = 0.349 L</em></u>
</span><span><u><em> 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL</em></u>
<em>Therefore the answer is option C!</em>
If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit</span>
Answer:
A) 12.57 m
B) 5 RPM
C) 3.142 m/s
Explanation:
A) Distance covered in 1 Revolution:
The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:
s = rθ
where,
s = distance covered = ?
r = radius of circle = 2 m
θ = Angle = 2π radians (For 1 complete Revolution)
Therefore,
s = (2 m)(2π radians)
<u>s = 12.57 m</u>
B) Angular Speed:
The formula for angular speed is given as:
ω = θ/t
where,
ω = angular speed = ?
θ = angular distance covered = 15 revolutions
t = time taken = 3 min
Therefore,
ω = 15 rev/3 min
<u>ω = 5 RPM</u>
C) Linear Speed:
The formula that gives the the linear speed of an object moving in a circular path is given as:
v = rω
where,
v = linear speed = ?
r = radius = 2 m
ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s
Therefore,
v = (2 m)(1.571 rad/s)
<u>v = 3.142 m/s</u>
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Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
