Answer:
Approximately
(assuming that the projectile was launched at angle of
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing
is the same as the altitude
at which this projectile was launched:
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is
(upwards,) the vertical velocity right before landing would be
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be
. In other words,
.
Hence, the time it takes to achieve a (vertical) velocity change of
would be:
.
Hence, this projectile would be in the air for approximately
.
Answer:
Tension on tendon = 1669800N
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
A telescope's angular resolution.
Explanation:
Diffraction limit is a minimum angular separation of two sources and it can be distinguished by the telescope. This angle is known as the diffraction limit. It is proportional to the wavelength of light and it has an inverse relation with the diameter of the telescope. Mathematically it is defined as
θ = 1.22λ/d
where θ is the angle, λ wavelength and d is the diameter of the objective mirror (lenz).
Light travels as transverse waves and faster than sound. It can be reflected, refracted and dispersed. Ray diagrams show what happens to light in mirrors and lenses. Eyes and cameras detect light.
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:

where the force is F=qE, d=0.556 and

. Using the value of q and E given by the problem, we find