Answer:
a) Acceleration is zero
, c) Speed is cero
Explanation:
a) the equation that governs the simple harmonic motion is
x = A cos (wt +φφ)
Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition
Body acceleration is
a = d²x / dt²
Let's look for the derivatives
dx / dt = - A w sin (wt + φ)
a = d²x / dt² = - A w² cos (wt + φ)
In the instant when it is not stretched x = 0
As the spring is released at maximum elongation, φ = 0
0 = A cos wt
Cos wt = 0 wt = π / 2
Acceleration is valid for this angle
a = -A w² cos π/2 = 0
Acceleration is zero
b)
c) When the spring is compressed x = A
Speed is
v = dx / dt
v = - A w sin wt
We look for time
A = A cos wt
cos wt = 1 wt = 0, π
For this time the speedy vouchers
v = -A w sin 0 = 0
Speed is cero
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
The answer is Rarefaction
Answer:
Explanation:
Given that,
Two resistor has resistance in the ratio 2:3
Then,
R1 : R2 = 2:3
R1 / R2 =⅔
3 •R1 = 2• R2
Let R2 = R
Then,
R1 = ⅔R2 = 2/3 R
So, if the resistor are connected in series
Let know the current that will flow in the circuit
Series connection will have a equivalent resistance of
Req = R1 + R2
Req = R + ⅔ R = 5/3 R
Req = 5R / 3
Let a voltage V be connect across then, the current that flows can be calculated using ohms law
V = iR
I = V/Req
I = V / (5R /3)
I = 3V / 5R
This the current that flows in the two resistors since the same current flows in series connection
Now, using ohms law again to calculated voltage in each resistor
V= iR
For R1 = ⅔R
V1 =i•R1
V1 = 3V / 5R × 2R / 3
V1 = 3V × 2R / 5R × 3
V1 = 2V / 5
For R2 = R
V2 = i•R2
V2 = 3V / 5R × R
V2 = 3V × R / 5R
V2 = 3V / 5
Then,
Ratio of voltage 1 to voltage 2
V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5
V1 : V2 = 2V / 5 × 5 / 3V.
V1 : V2 =2 / 3
V1:V2 = 2:3
The ratio of their voltages is also 2:3
<span>In order to calculate an average, we should sum all numbers and divide them by quantity.
Let’s work with qualifications first. Let’s say you got a 10 in 1 exam, then an 8 in 2 exams and a 4 in 2 exams. Your average will be:
= (10*1+8*2+4*2) / 5 = 6.8
If 6 is the minimum, you will pass.
There is another way to calculate this average: applying distributive property.
= 10*1/5+8*2/5+4*2/5 = 6.8
Remember you can convert the fractions into equivalent fractions: 1/5 = 20/100; 2/5 = 40/100
= 10*20/100+8*20/100+4*20/100 = 6.8
We actually don’t have the number of atoms of each mass… we have the percentage instead! So we need to learn this last method for atoms.
Let’s go back to our atoms problem:
73.71 % of atoms have a mass of 27.98 u
14.93 % of atoms have a mass of 28.98 u
11.36 % of atoms have a mass of 29.97 u
So let’s put that in the formula:
Average mass = 27.98 u*73.71 /100 + 28.98 u*14.93 /100 + 29.97u*11.36 /100
So what you have to know is that a percentage can be converted into a fraction, and you should work that fraction in order to find the average. We can make the calculus shorter putting 100 as the common denominator:
Average mass = (27.98 u*73.71 + 28.98 u*14.93 + 29.97u*11.36)/100
So actually we are taking the percentage as if it was the quantity, and 100 as if it was the total (the total of all percentages is always 100). Maybe we don’t have 100 atoms, but it will be the same proportion anyway, whatever number we have! And here it is the result:
Average mass = 28,36u
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