Answer: 10.29 sec.
Explanation:
Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.
If the car starts from rest, this means the following:
ΔK = 1/2 m*vf ²
As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:
100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec
Now, we calculate the change in energy:
ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J
<h2>If P= ΔK/Δt, </h2><h2>Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.</h2>
Answer:
The minimum value of wall thickness t=3.63 mm.
Explanation:
Given:
D=200 mm
P=4 MPa
t= Wall thickness
maximum shear stress=27.5 MPa
We know that
hoop stress ![\sigma _{h}=\frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma%20_%7Bh%7D%3D%5Cfrac%7BPd%7D%7B2t%7D)
Longitudinal stress
So maximum shear tress in plane![\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}](https://tex.z-dn.net/?f=%5Ctau%20_%7Bmax%7D%3D%5Cdfrac%7B%5Csigma%20_h-%5Csigma%20_l%7D%7B2%7D)
![\tau _{max}=\dfrac{Pd}{8t}](https://tex.z-dn.net/?f=%5Ctau%20_%7Bmax%7D%3D%5Cdfrac%7BPd%7D%7B8t%7D)
Now by putting the value
![27.5=\dfrac{4\times 200}{8t}](https://tex.z-dn.net/?f=27.5%3D%5Cdfrac%7B4%5Ctimes%20200%7D%7B8t%7D)
So t=3.36 mm
The minimum value of wall thickness t=3.63 mm.
Answer:
Explanation:
Attached is the solution to the question
Answer:
b)1.08 N
Explanation:
Given that
velocity of air V= 45 m/s
Diameter of pipe = 2 cm
Force exerted by fluid F
![F=\rho AV^2](https://tex.z-dn.net/?f=F%3D%5Crho%20AV%5E2)
So force exerted in x-direction
![F_x=\rho AV^2](https://tex.z-dn.net/?f=F_x%3D%5Crho%20AV%5E2)
![F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2](https://tex.z-dn.net/?f=F_x%3D1.2%5Ctimes%20%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%200.02%5E2%5Ctimes%2045%5E2)
F=0.763 N
So force exerted in y-direction
![F_y=\rho AV^2](https://tex.z-dn.net/?f=F_y%3D%5Crho%20AV%5E2)
![F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2](https://tex.z-dn.net/?f=F_y%3D1.2%5Ctimes%20%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%200.02%5E2%5Ctimes%2045%5E2)
F=0.763 N
So the resultant force R
![R=\sqrt{F_x^2+F_y^2}](https://tex.z-dn.net/?f=R%3D%5Csqrt%7BF_x%5E2%2BF_y%5E2%7D)
![R=\sqrt{0.763^2+0.763^2}](https://tex.z-dn.net/?f=R%3D%5Csqrt%7B0.763%5E2%2B0.763%5E2%7D)
R=1.079
So the force required to hold the pipe is 1.08 N.
Can I get a picture? Of the question so I can understand it more?