Answer:
6.91 Hz
Explanation:
Volume of the cube
= (1.5 x 10⁻² )³m³
= 3.375 x 10⁻⁶ m³
mass of the cube
= 3.375 x 10⁻⁶ x 8920 ( 8.92 g / cm³ = 8920 kg/m³ )
m = 30.105 x 10⁻³ kg
Spring stretches by 2.65 x 10⁻² m due to a force of 1.5 N.
If k be the force constant
k x = F
K x2.65 x 10⁻² = 1.5
k = .566 x 10² N / m
Now frequency of oscillation for spring - mass system is given by
n = 
=
n = 6.91 Hz
I think i used calulater and it gives me 47.5
Answer:
The unknown substance is Aluminum.
Explanation:
We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:
Initial temperature (T₁) = 25 ⁰C
Final temperature (T₂) = 100 ⁰C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 100 – 25
ΔT = 75 ⁰C
Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:
Change in temperature (ΔT) = 75 ⁰C
Mass of the substance (M) = 135 g
Heat (Q) gained = 9133 J
Specific heat capacity (C) of substance =?
Q = MCΔT
9133 = 135 × C × 75
9133 = 10125 × C
Divide both side by 10125
C = 9133 / 10125
C = 0.902 J/gºC
Thus, the specific heat capacity of substance is 0.902 J/gºC
Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.
Answer:
A) Propagation of pressure fluctuations in a medium
B) air is the medium in which the wave is transported,
Explanation:
Part A.
A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.
The most correct answer is:
* Propagation of pressure fluctuations in a medium
Part b
air is the medium in which the wave is transported, otherwise it cannot propagate
