time=distance/speed
1.6/100 secs = 0.016secs=16millisecs
Acceleration = Change in Velocity / time
a = (v - u) / t
Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²
A proper record of the changes in velocity with the corresponding time would help find the acceleration.
Answer:
It corresponds to a distance of 100 parsecs away from Earth.
Explanation:
The angle due to the change in position of a nearby object against the background stars it is known as parallax.
It is defined in a analytic way as it follows:
Where d is the distance to the star.
(1)
Equation (1) can be rewritten in terms of d:
(2)
Equation (2) represents the distance in a unit known as parsec (pc).
The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).
For the case of (
):
Hence, it corresponds to a distance of 100 parsecs away from Earth.
<em>Summary:</em>
Notice how a small parallax angle means that the object is farther away.
Key terms:
Parsec: Parallax of arc second
When the incident light is yellow the width of the central band greater. Single-wavelength light sources are known as monochromatic lights, where mono stands for one and chroma for color. Monochromatic lights are defined as visible light that falls inside a specific range of wavelengths. It has a wavelength that falls within a constrained wavelength range.
A laser beam is the ideal illustration of monochromatic light. A monochromatic light beam produced by a single atomic transition with a particular single wavelength is what makes up a laser. A color scheme that consists solely of different shades of one color is referred to as monochromatic.
To learn more about monochromatic, click here.
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Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
