Explanation:
A) The distance between the two successive compressions (or rarefactions) is actually called the wavelength of the longitudinal waves.
B) Wavelengths of longitudinal and transverse waves are comparable in the fact that in a transverse wave, the particles move perpendicular to the direction the wave travels whereas in a longitudinal wave the particles are displaced along the direction to the direction the wave travels
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Answer:
0.647 nC
Explanation:
The force experienced by a charge due to the presence of an electric field is given by
where
q is the charge
E is the magnitude of the electric field
In this problem, each antenna is modelled as it was a single point charge, experiencing a force of
Therefore, if the electric field magnitude is
Then the charge on each antenna would be
I’m not completely sure but most likely is is the 10 mile bike ride, I hope I can help! (:
Answer:
The gravitational force exerted on the object is 75 N (answer D)
Explanation:
Hi there!
The gravitational force is calculated as follows:
F = m · g
Where:
F = force of gravity.
m = mass of the object.
g = acceleration due to gravity (unknown).
For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:
y = y0 + v0 · t + 1/2 · a · t²
Where:
y = position at time t.
y0 = initial position.
v0 = initial velocity.
t = time.
g = acceleration due to gravity.
Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.
Then, after 2 seconds, the height of the object will be -30 m:
y = y0 + v0 · t + 1/2 · g · t²
-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²
-30 m = 1/2 · g · 4 s²
-30 m = 2 s ² · g
-30 m/2 s² = g
g = -15 m/s²
Then, the magnitude of the gravitational force will be:
F = m · g
F = 5 kg · 15 m/s²
F = 75 N
The gravitational force exerted on the object is 75 N (answer D)
Have a nice day!
