Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer:
They will sometimes crash into other plates in the process and will rub while they are moving creating earthquakes
Explanation:
Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
Answer:280.216j/kg°C
Explanation:
Mass of metal=0.0663kg
mass of water=0.395kg
Final temperature=27.4°C
Temperature of metal=241°C
Temperature of water=25°C
specific heat capacity of water=4186j/kg°C
0.0663xax(241-27.4)=0.395x4186x(27.4-25)
0.0663xax213.6=0.395x4186x2.4
14.16168a=3968.328
a=3968.328 ➗ 14.16168
a=280.216j/kg°C
