Answer:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
Answer:
2 in front of water and 1 in front of oxygen
Explanation:
This question is describing balancing a chemical reaction. A balanced chemical reaction has the same number of atoms of each elements on both the reactant and product side. According to the question, the reactants contains 4 atoms of oxygen. The reactants give rise to water (H20) and O2 in the products side.
This reaction is most likely the decomposition of hydrogen peroxide (H2O2) as follows:
H2O2 (l) ----> H2O (l) + O2(g)
Based on the description, H2O2 will be 2H2O2 as it is said to contain four atoms of oxygen. This means that, in order to have a balanced equation, we must place coefficient 2 in front of water and coefficient 1 in front of oxygen. That is;
2H2O2 (l) ----> 2H2O (l) + O2(g)
Answer:
54 × 5/18 = 15m/s
Explanation:
to convert km/hr to m/s you multiply by 5/18
Answer:
physical change is the temporary change or riversible change here the physical properties r only changed
for example when water is cooled its get freezed Nd becomes ice similarly wen ice is heated again then it becomes water so here it's not changed permanently
I hope my ans is comprehensive
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Answer:
The decibel of the remaining pigs is 51.5 dB.
Explanation:
Decibel (dB) is a unit of measure of the intensity of a given sound.
Number of pigs = 199, noise level = 74.3 dB.
Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:
I 
N
I = kN
where k is the constant of proportionality.
⇒ k = 
= 
k = 0.3734
When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.
Thus, the becibel level (I) of the remaining pigs can be determined by:
I = kN
= 0.3734 × 138
= 51.53 dB
The becibel level (I) of the remaining pigs is 51.53 dB.
