Fulcrum need to be positioned balanced with weight on both the sides following law of lever.
What is the physical law of the lever?
- It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
- The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
- The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
- Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.
Learn more about the Fulcrum with the help of the given link:
brainly.com/question/16422662
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Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
If it is a headwind it means it's travelling against the motion of the plane. This means it's velocity is simply v=720-16=704 km/h due east.
Answer:
v = 7.4 m/s
Explanation:
Given that,
Mass if a volleyball, m = 5 kg
The ball reaches a height of 2.8 m
We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :
So, the required speed is 7.4 m/s. Hence, the correct option is (b).
