Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²
Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m
(a) Angular frequency w= 2πf
w= 2π × 760 = 4775.2Hz
(b) Maximum speed v is given as the product of angular frequency and maximum displacement
V=wx
V= 4775.2 × 0.00085
V= 4.06m/s
(c) The maximum acceleration a
= w²x
= (4775.2)² × (0.00085)
a= 19382.15m/s².
Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
If we are to place dots to teh places that have been struck by an earthquake these past 100 years, the dots would be concentrated in the east and southeast Asia region. This is because of the presence of the Pacific ring of fire. This is a major area in the Pacific Ocean where most of the earthquakes are likely to occur.