Answer:
(a) convex mirror
(b) virtual and magnified
(c) 23.3 cm
Explanation:
The having mirror is convex mirror.
distance of object, u = - 20 cm
magnification, m = 1.4
(a) As the image is magnified and virtual , so the mirror is convex in nature.
(b) The image is virtual and magnified.
(c) Let the distance of image is v.
Use the formula of magnification.
Use the mirror equation, let the focal length is f.
Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm
Answer:
They don’t ‘represent’ anything, they are properties of the wave.
Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.
If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.
For many waves, we don’t perceive them at all (e.g. radio waves).
For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
Answer:
I'm sorry I don't have a answer but I like your pfp
I believe the correct response would be B. It would decrease.
