Answer:
low, low
Explanation:
Longer wavelengths will have lower frequencies, and shorter wavelengths will have higher frequencies.
Large amplitude waves contain more energy. The other is frequency, which is the number of waves that pass by each second. If more waves( or more wiggly lines) pass by, more energy is transferred each second
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
Answer:
Resultant displacement = 1222.3 m
Angle is 88.3 degree from +X axis.
Explanation:
A = 550 m north
B = 500 m north east
C = 450 m north west
Write in the vector form
A = 550 j
B = 500 (cos 45 i + sin 45 j ) = 353.6 i + 353.6 j
C = 450 ( - cos 45 i + sin 45 j ) = - 318.2 i + 318.2 j
Net displacement is given by
R = (353.6 - 318.2) i + (550 + 353.6 + 318.2) j
R = 35.4 i + 1221.8 j
The magnitude is
The direction is given by
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes
Now after applying brakes the distance covered by the car before it stops is given by kinematics equation
here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have
So total distance after which car will stop is
So car will not stop before the intersection as it is at distance 20 m
