Answer:
speed when the block had slid 3.40 m is 2.68 m/s
Explanation:
given data
distance = 6.80 m
speed = 3.80 m/s
to find out
speed when the block had slid 3.40 m
solution
we will apply here equation of motion that is
v²-u² = 2×a×s ..............1
here s is distance, a is acceleration and v is speed and u is initial speed that is 0
so put here all value in equation 1 to get a
v²-u² = 2×a×s
3.80²-0 = 2×a×6.80
a = 1.06 m/s²
so
speed when distance 3.40 m
from equation 1 put value
v²-u² = 2×a×s
v²-0 = 2×1.06×3.40
v² = 7.208
v = 2.68
so speed when the block had slid 3.40 m is 2.68 m/s
Answer:
(A)8,421.1 V/m
(B)
(C)
(D)
Explanation:
area (A) = 7.6 cm^{2} = 0.00076 m^{2}
distance (d) = 1.9 mm = 0.0019 m
potential difference (v) = 16 V
(A) electric field (E) = Δv / d = 16 / 0.0019 = 8,421.1 V/m
(B) capacitance = (∈₀A)/d
where ∈₀ = permitivity of free space =
capacitance = 
capacitance (C) =
(C) charge (q) = C x v =
=
(D) surface charge density = charge (q) / area =
/ 0.00076 =
Answer:
The correct option is (c) "periodic wave"
Explanation:
The relationship between the angular frequency and the radial frequency is :

f is the radial frequency
Frequency, wavelength, velocity of propagation, and related variables do not depend on the amplitude of wave. Travelling waves are also known as periodic wave if their waveform repeats every time interval T. Frequency, wavelength and the speed of wave are the factors that defines traveling waves.
So, the correct option is (c) "periodic wave".