In an experiment if doubling the manipulated variable results in doubling of the responding variable the relationship between th
1 answer:
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Answer:
1) ngle of incidence and reflection are equal, light carries does not change
2) the angle of this line with respect to the surface is 90º
3) protractor
4) n₂ sin θ₂ = n_1 sin θ₁, light ray must have a greater angle than the incident ray ,
Explanation:
1) When light falls on a reflective surface, the angle of incidence and reflection are equal and as it travels in the same medium, the speed that the light carries does not change
2) The normal is a line perpendicular to the point of incidence of light, so the angle of this line with respect to the surface is 90º
3) Angles are measured with a protractor
4) When light passes from one medium to another, the speed of the ray changes due to the difference in the refractive index in each medium, due to this change in speed the transmitted light ray must have a greater angle than the incident ray , since the speed increases as the density of the medium decreases
n₂ sin θ₂ = n_1 sin θ₁
Answer:
10 m/s
1.87914 s
18.7914 m
Explanation:
v = Initial velocity = 20 m/s
= Angle = 60°
Horizontal component is given by
The horizontal component is 10 m/s
y direction final displacement is zero
The time the ball is in the air is 1.87914 s
Range is given is by
The range is 18.7914 m
Answer:
d₃ = 37,729 km, θ= 5.1º North of West
Explanation:
This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction
* first displacement is
d₁ₓ = 11 km
* second offset is
cos 6 = d₂ₓ / d₂
sin 6 = d_{2y} / d₂
d₂ₓ = d₂ cos 6
d_{2y} = d₂ sin 6
d₂ₓ = 6 cos 6 = 5.967 km
d_{2y} = 6 sin 6 = 0.6272 km
* third displacement is unknown
* fourth and last displacement
cos (-11) = d₄ₓ / d₄
sin (-11) = d_{4y} / d₄
d₄ₓ = d₄ cos (-11)
d_{4y} = d₄ sin (-11)
d₄ₓ = 21 cos (-11) = 20.61 km
d_{4y} = 21 sin (-11) = -4.007 km
They tell us that at the end of the tour you are back on the island, so the displacement must be zero
X axis
x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ
0 = 11 +5.967 + d₃ₓ + 20.61
d₃ₓ = -11 - 5.967 - 20.61
d₃ₓ = -37.577 km
Y axis
y = d_{1y} + d_{2y} + d_{3y} + d_{4y}
0 = 0 + 0.6272 + d_{3y} -4.007
d_{3y} = 4.007 - 0.6272
d_{3y} = 3.3798 km
This distance can be given in the form of module and angle
Let's use the Pythagorean theorem for the module
d₃ =
d₃ =
d₃ = 37,729 km
Let's use trigonometry for the angle
tan θ = d_{3y} / d₃ₓ
θ = tan⁻¹
θ = tan-1 (-3.3798 / 37.577)
θ = 5.1º
Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is
θ= 5.1º North of West