Answer:
V at C is 3.6 m/s
Explanation:
At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J
By conservation of energy.
KE+PE=Constant
At C PE=0.6 J
the KE=2.943-0.6=2.343 J
KE=0.5*m*v^2
v=√[KE/(0.5*m)]=3.06 m/s
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)
Answer:

Explanation:
For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:
(1)
With
the initial kinetic energy,
the initial potential energy,
the final kinetic energy and
the final potential energy. Note that initialy the masses are at rest so
, when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so
. So, equation (1) becomes:
(2)
At initial moment all the potential energy is gravitational because the spring is not stretched so
and at final moment we have potential gravitational energy and potential elastic energy so
, using this on (2)
(3)
Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached),
and we have by (3) :
(4)
Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

dividing both sides by d


, with k the constant of the spring and g the gravitational acceleration.
Answer: 1. The movie one is virtual and the bathroom mirror is real
2. The image is distorted in a way
3. Behind the spoon
Explanation: