Answer:
letter C. The mass of the object.
Explanation:
i hope it helps.
Answer:
u" + 40u' + 49u = 2 sin(t/6)
upp + 40up + 49u = 2 sin(t/6)
Explanation:
Step 1: Data given
mass = 5 kg
L = 20 cm = 0.2 m
F = 10 sin(t/6)N
Fd(t) = - 6 N
u(0) = 0.03 m/s
u(0) = 0
u'(0) = 3 cm/s
Step 2:
ω =kL
k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²
Since Fd(t) = -γu'(t) we know:
γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m
The initial value problem which describes the motion of the mass is given by
5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03
This is equivalent to:
u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03
upp + 40up + 49u = 2 sin(t/6)
With u in m and t in s
Answer:
Archimedes Principle states that "any body completely or partially submerged in water is acted upon by an upthrust force which is equal to the magnitude of Weight of the body."
Answer:
The moment arm is 0.6 m
Explanation:
Given that,
First force 
Second force 
Distance r = 0.2 m
We need to calculate the moment arm
Using formula of torque
So, Here,
We know that,
The torque is the product of the force and distance.
Put the value of torque in the equation
Where, 
=First force
=First force
=Second force
= distance
Put the value into the formula
Hence, The moment arm is 0.6 m
solution:
We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v
so,
we use the equation:
v = v0 + at
v = 0 + 9.8*4.0
v = 39.2 m/s
Now we just need to solve for d, so we use the equation:
d = v0t + 1/2*a*t^2
d = 0*4.0 + 1/2*9.8*4.0^2
d = 78.4 m
