Answer:
3.34×10^-6m
Explanation:
The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain
can be expressed as
shear stress/(shear strain)
= (F/A)/(Lo/ . Δx)
Stress=Force/Area
The sheear stress can be expressed below as
F Lo /(A *Δx)
Where A=area of the disk= πd^2/4
F=shearing force force= 600N
Δx= distance
S= shear modulus= 1 x 109 N/m2
Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m
If we make Δx subject of the formula we have
Δx= FLo/(SA)
If we substitute the Area A we have
Δx= FLo/[S(πd^2/4]
Δx=4FLo/(πd^2 *S)
If we input the values we have
(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2
= 3.35×10^-6m
Therefore, its shear deformation is 3.35×10^-6m
A=area of the disk= πd^2/4
= [3.142×(4×10^-2)^2]/4
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Answer:
80%
Explanation:
Efficiency = Power output / Power input × 100 %
To calculate efficiency we need to find power output of electric pump.
We can use,
Work done = Energy change
Work done per second = Energy change per second
Work done per second = Power
Therefore, Power = Energy change per second
= Change in potential energy of water per second
=mgh / t
= 200× 10×6 / 10
= 1200 W = 1.2 kW
Now use the first equation to find efficiency,
Efficiency =
× 100%
= 80 %
Answer:
x = 0.396 m
Explanation:
The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring
Data the putty has a mass m1 and velocity vo1, the block has a mass m2
. t's start using the moment to find the system speed.
Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash
p₀ = m1 v₀₁
Moment after shock
= (m1 + m2) 
p₀ =
m1 v₀₁ = (m1 + m2) 
= v₀₁ m1 / (m1 + m2)
= 4.4 600 / (600 + 500)
= 2.4 m / s
With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring
Before compressing the spring
Em₀ = K = ½ (m1 + m2)
²
After compressing the spring
= Ke = ½ k x²
As there is no rubbing the energy is conserved
Em₀ = 
½ (m1 + m2)
² = = ½ k x²
x =
√ (k / (m1 + m2))
x = 2.4 √ (11/3000)
x = 0.396 m
Under water turbans that are placed at the above to middle of the ocean they are used to capture kinetic motion