Answer:
Uniform rectilinear movement (m. r. u.)
Explanation:
It is a continuous movement without acceleration, that is, it moves at a constant speed. The speed does not change over time, for this reason, there is no change in acceleration.
Vf = final velocity = 50/30 [km/s] = 1.67 [km/s]
Vo = initial velocity = 50/30 [km/s] = 1.67 [km/s]
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
D = 2.38 m
Explanation:
This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit
a sin θ = m λ
Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle
θ = λ / a
Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant
θ = 1.22 λ / D
Where D is the circular tightness
Let's apply this equation to our case
D = 1.22 λ / θ
To calculate the angles let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (4.30 10⁻² / 140 10³)
θ = tan⁻¹ (3.07 10⁻⁷)
θ = 3.07 10⁻⁷ rad
Let's calculate
D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷
D = 2.38 m
Answer:
6370 J
Explanation:
By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m
where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference
