Answer:
<em>The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.</em>
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. <em>Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight</em>
Answer:
The angle between the mirrors is 90°.
Explanation:
Given that,
Number of image = 5
Rahul kept two plane mirrors at a certain angle in front of a burning candle.
We need to find the angle between the mirrors
Using given data
When Rahul kept two plane mirrors at a certain angle in front of a burning candle then see the 5 image of candle in the mirrors.
If rahul kept two plane mirrors at a 90° then we get the 5 image of candle.
Firstly we see the two images in both mirror and now first image of first mirror is see in second mirror and second image of second mirror is see in first mirror.
So, we can say that we see the 5 image of candle in the mirrors.
Hence, The angle between the mirrors is 90°.
Answer:
C. 10⁻³ rads
Explanation:
Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:
θ = λ/a
where,
θ = angular resolution of Cat's eye = ?
λ = wavelength = 500 nm = 5 x 10⁻⁷ m
a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m
Therefore,
θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)
Therefore,
θ = 0.001
θ = Sin⁻¹(0.001)
θ = 0.001 rad = 1 x 10⁻³ rad
Hence, the correct answer is:
<u>C. 10⁻³ rads</u>
Answer:
The correct option is;
C. 1 mile clear of clouds
Explanation:
Given that the indicated airspace location is at or below 700 feet AGL therefore, it is taken as being in the region of a class G airspace which covers the airspace regions from the base up to and equal to 1,200 feet beneath the class E airspace and the requirement for VFR flight for class G are 1 mile and clear of clouds.
Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.
