Answer:
915 Hz
Explanation:
The observed frequency from a sound source is given as
f₀ = f [(v + v₀)/(v+vₛ)]
where
f₀ = observed frequency of the sound by the observer = ?
f = actual frequency of the sound wave = 983 Hz
v = actual velocity of the sound waves = 343 m/s
vₛ = velocity of the source of the sound waves = 55.9 m/s
v₀ = velocity of the observer = 28.4 m/s
f₀ = 983 [(343+28.4)/(343+55.9)]
f₀ = 915.2 Hz = 915 Hz
For t1:
t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec
For t2:
t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec
Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)
d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m
Where:
d = hor. distance
ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25
The answer is 3.19 : 2.25
Answer:
33.6 Ns backward.
Explanation:
Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.
From Newton's second law of motion,
Impulse = change in momentum
I = mΔv................................. Equation 1
Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.
Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)
Substituting into equation 1
I = 28(-1.2)
I = -33.6 Ns
Thus the impulse = 33.6 Ns backward.
Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.
Answer:
a. E = 122.4 N/C
b. E = 58.2 N/C
c. E = 0
Explanation:
The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.
In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.
A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.
B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.
C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.
