The 500-kg cylindrical drum is supported by a metal cable (A-B) and a rigid plate (B-C). If the contact between the drum and all
surfaces is frictionless and its radius is 500 mm, compute the elongation of the cable (A-B). The elastic modulus of A-B is 200 GPa. A, B, and C are pinned joints. Note that the cross section of Cable A-B is 5 mm2 .
1 answer:
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<u>Answer:</u>
Maximum height reached = 35.15 meter.
<u>Explanation:</u>
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g
So range of projectile,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So
Now we have
So maximum height reached = 35.15 meter.
Heat engines were developed during industrial revolution.
Generally a heat engine contains three parts i.e source, sink and working substance.
The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.
Let us consider are the temperature of source and sink.
As the source is at higher temperature as compared to sink, heat will flow from source to sink.
are the heat provided by source and heat rejected to sink.
Hence, the work done by the working substance will be -
The efficiency of a heat engine is defined as the ratio of output to the input energy.
Here output = workdone [W]
Hence, the efficiency of a heat engine is calculated as -
This is the expression for the efficiency of heat engine.
Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.
Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 × m
separation L = 0.5 mm = 0.5 × m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 × m
λ = 2042 nm