Answer:
Angle between incident ray and reflected ray will be 104°
Explanation:
We have given angle of incidence = 52 °
From law of reflection angle of incidence will be equal to angle of reflection
So angle of reflection will be also 52°
We have to find the angle between incident ray and reflected ray
As the incidence angle and reflected angle both is from normal of the surface and opposite to each other
So angle between incident ray and reflected ray will be 52°+52° = 104°
Answer:
L = 40 cm
Explanation:
A microscope is an optical instrument built by two lenses in such a way that the image of the first is formed within the distance of the other (eyepiece), so that the latter creates an enlarged virtual image of the object, for which the magnification of the microscope is the same to the multiplication of the magnification of each lens
M = - L / f₀ (25 cm /
)
where fo and fe are the focal lengths of the objective and eyepiece, 25 cm is the near vision distance and L is the length of the microscope
L = - M f_{o} f_{e} / 25
let's calculate
L = - (-100) 2 5/25
L = 40 cm
Answer:
D) Roger is incorrect. Only a chemical change is taking place as evidenced by the light and heat of the burning candles.
Explanation:
- A physical change is a change in which there is no formation of new substances. Examples of physical changes are the melting or the evaporation of a substance (all phase transitions are examples of physical changes): in such cases, there is no formation of new substances.
- A chemical change is a change in which new substances form. Examples of chemical changes are the chemical reactions: for instance, when a candle burns, a reaction is taken place (oxygen is burnt, transforming into carbon dioxide + heat + light, so this is an example of chemical change).
Therefore, the correct answer is
D) Roger is incorrect. Only a chemical change is taking place as evidenced by the light and heat of the burning candles.
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y =
t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch