Answer:
-36.71%
Explanation:
Using the Generalized Work Energy Principle, the puck is brought to rest by an external force and the system has no kinetic energy
#Denote the stopping distance and force required with a prime and observe from (i) that:
#If stopping distance increases by a factor of 79/50;
So F decreases by 36.71% . we expect force to reduce since the same amount of work is repeatedly done on the system over a long distance.
When conducting a search to identify a text's credibility and reliability, you have to check the following characteristics:
- Sources: where the information is obtained and it is supposed to be true. It is said that you need at least three different sources that explain the same information for it to be validated.
- Is the article is current?: articles could change all the time, specially scientific ones because new discoveries can change what was discovered before, so it is important to check if the text you are reading is current or not, and if it is not, you need to check if something has changed during all those years.
<h3>Article's credibility</h3>
In this exercise, you have to present an article and describe the purpose of the source and if the article is current or not.
For example, let's select an article called "How using social media affects teenagers".
The purpose of the sources in this article is to demonstrate how social media affects teenagers using different surveys.
Check more information about sources here brainly.com/question/24708478
Answer:
a. d₁/d₂ = 1.09 b. 0.054 mW
Explanation:
a. What is the ratio of the diameter of the first student's eardrum to that of the second student?
We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²
So, I ∝ I/d²
I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.
Given that I₂ = 1.18I₁
I₂/I₁ = 1.18
Since I₁/I₂ = d₂²/d₁²
√(I₁/I₂) = d₂/d₁
d₁/d₂ = √(I₂/I₁)
d₁/d₂ = √1.18
d₁/d₂ = 1.09
So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09
b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?
We know intensity, I = P/A where P = acoustic power and A = area = πd²/4
Now, P = IA
= I₂A₂
= I₂πd₂²/4
= 1.18I₁πd₂²/4
Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m
So, P = 1.18I₁πd₂²/4
= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4
= 0.691244π × 10⁻⁴ W/4 =
2.172 × 10⁻⁴ W/4
= 0.543 × 10⁻⁴ W
= 0.0543 × 10⁻³ W
= 0.0543 mW
≅ 0.054 mW
Answer:
Explanation:
given,
moles of air compressed, n = 1.70 mol
initial temperature, T₁ = 390 K
Power supply by the compressor, P = 7.5 kW
Heat removed = 1.3 kW
Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.
Time of compression = time of the hay revolution
=
=
=
=0.273 s
Using first law of thermodynamics
U = Q - W
now,
Power supplied 
= 7.5 kW
heat removed 
= 1.3 kW
now,
we know,
C_v for air = 5 cal/° mol
= 5 x 4.186 J/mol°C = 20.93 J/mol°C
now,
the temperature change per compression stroke is equal to 47.57°C.
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>
Explanation:
When the charge of 2e is placed in between the plates .
The force applied on this charge by plates is = q E
here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C
and E is the magnitude of electric field intensity
The work done = Force x displacement
Thus W = q E x S
here S is displacement
Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8
= 1.02 x 10⁻¹⁷ J
This work will be converted into the kinetic energy of charge .
Thus K.E = 1.02 x 10⁻¹⁷ J
