Answer:
<em>Since I can see no choices, I answered it in my own understanding.</em>
Brian - amplitude and frequency
Marcia - amplitude and longitudinal wave
Explanation:
"Sound" and "sound waves" are essential part of a person's life. They can be used for<u> communicating</u> and <u>detecting some object</u>s.
Brian loves singing in the shower which means that he is using a greater amplitude. Amplitude refers to the<em> intensity of the sound </em>or the amount of energy that a sound carries. When one sings in the shower, the sound cannot travel very far. It bounces immediately back to the person singing thus, making the sound bigger. Brian is also using a <em>different range of </em><em>frequency</em><em> compared to his normal way of talking.</em> The frequency of a normal male voice is normally 85 to 180 Hz. A person singing may have a frequency as high as 1,500 Hz.
Marcia talks loudly on the phone. This means that she is also using a greater amplitude because the intensity of her voice is big. Since she is using the telephone, this means that her voice travels in a longitudinal wave through the telephone. This allows her voice to reach to the person on the other end of the line.
Answer:
The horizontal component is zero.
The vertical component is 
Explanation:
Given that,
The lizard climb 7m directly up on a tree.
We know that,
The horizontal component is
The vertical component is
If the lizard climb 7m directly up on a tree then,
We need to find the components
Using given data
The horizontal component of lizard is
The vertical component is
Hence, The horizontal component is zero.
The vertical component is
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Answer:
t = 5.05 s
Explanation:
This is a kinetic problem.
a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m
b) in this system the equations of motion are
y = v₀ t + ½ g t²
where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth
e) y = 0 + ½ g t²
t = √ (2y / g)
t = √(2 125 / 9.8)
t = 5.05 s
