Answer:
The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.
See attachment below for full solution
Explanation:
This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.
Answer:
The diagram represents two charges, q1 and q2, separated by a distance d. Which change would produce the greatest increase in the electrical force between the two charges? *
Explanation:
doubling charge q1, only
Answer:
ask your member
Explanation:
kayo lang din makakasagot nyan
To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
Answer:
5.09 Pa
Explanation:
Area of the circle = πr²
The diameter is given, so 100/2 = 50 m is the radius.
π x 50² = 7854 m²
Pressure = force/area
Pressure = 40000/7854
Pressure = 5.09 Pa
Hope this helps!
