Explanation:
Given that,
Size of object, h = 0.066 m
Object distance from the lens, u = 0.210 m (negative)
Focal length of the converging lens, f = 0.14 m
If v is the image distance from the lens, we can find it using lens formula as follows :
(a) Magnification,
(b) Magnification, 
h' is image height
Hence, this is the required solution.
The only force opposing the block's sliding as it slows down is friction with magnitude <em>f </em>. By Newton's second law, the net force in this direction is
∑ <em>F</em> = -<em>f</em> = <em>ma</em> = (4.00 kg) <em>a</em>
Assuming constant acceleration <em>a </em>, the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2<em>a</em> (4.00 m)
Solve for the acceleration :
<em>a</em> = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-<em>f</em> = (4.00 kg) (-14.8 m/s²)
<em>f</em> ≈ 59.4 N
and was directed opposite the block's motion.
Answer:
a. Value of MA is always lesser than VR because mechanical advantage decreases due to the friction and weight of moving parts of the machine.
b.Efficiency of lever is never 100 or more because it will always lose some tiny (sometimes immeasurable) percentage of power due to the lever bending.
c.Because it helps to apply more torque.
d.Because the upright snip has the blades rotated 90° from the handles.
e.because some of the input work is used to overcome friction.
f.because the load arm is always longer than effort arm.
g.because the effort arm is always longer than load arm of these Lever.
h. because if the effort distance is greater than load distance there will be magnification of force and the metals could easily cut.
i. because the wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull.
The working equation for this problem is a derived equation from the rectilinear motion at constant acceleration. The equation is written below:
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
Of course, the plane starts from rest, so v₀ = 0. The final velocity is the liftoff speed which is v = 120 km/h. Substituting the values:
2a(270 m) = (120 km/h)² - 0²
Solving for a,
a = 26.67 km/h²
To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.
Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.
Part A) Conservation of the energy
Part B) Kinematic equation of linear motion,
Here
v= 0 Because at 1.5m reaches highest point, so v=0
Therefore the velocity after the collision with the floor is 3.7m/s
PART C) Total change of impulse is given as,
