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Lorico [155]
3 years ago
6

Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as

the initial height of the block. Will A make it completely around the loop without losing contact with the track?

Engineering
1 answer:
Zolol [24]3 years ago
4 0

Answer:No

Explanation:

Given

Block A is at height h and released from rest

Initial Energy possessed by block A is equal to Potential energy of Block which is given by

E_i=mgh

where m=mass of block

After releasing the block, block first reaches to the bottom of the circle and then uses this energy to reach a top point of the loop.

But as soon as block reaches the top point of the loop it acquires the energy which is equal to Initial energy i.e. all the energy is stored in the form of potential energy and there is no kinetic energy so the block will not able to move further and fall from the top point.

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A large retirement village has a total retail employment of 120. All 1600 of the households in this village consist of two nonwo
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See explaination

Explanation:

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Roads in rural areas are _______.
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Read 2 more answers
What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Dominik [7]

Answer:

0

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)=\frac{s}{(s+3)^2}

U(s)=\frac{1}{s} for unit step function

output =H(s)×U(s)

=\frac{s}{(s+3)^2}×\frac{1}{s}

=\frac{1}{(s+3)^2}

taking inverse laplace of output

output=t×e^{-3t}

at t=0 putting the value of t=0 in output

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3 0
3 years ago
A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p
Natali [406]

Answer:

Mechanical Efficiency =  83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = V=8ft^3/s\\

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

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Efficiency = 0.8351 = 83.51%

5 0
3 years ago
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