Answer:
14 x 10⁻⁶ J
1377 x 10⁻⁶ J
Explanation:
C = Capacitance of the capacitor = 1 x 10⁻⁶ F
ΔV = Original potential difference across the plates = 6.0 Volts
U₀ = Original energy stored in the capacitor
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² eq-1
ΔV' = Potential difference across the plates after increase = 8.0 Volts
U'₀ = New energy stored in the capacitor
New energy stored in the capacitor is given as
U'₀ = (0.5) C ΔV'² eq-2
U = Additional energy stored
Additional energy stored by the capacitor is given as
U = U'₀ - U₀
U = (0.5) C ΔV'² - (0.5) C ΔV²
U = (0.5) (1 x 10⁻⁶) (8)² - (0.5) (1 x 10⁻⁶) (6)²
U = 14 x 10⁻⁶ J
[tex]k_{final}[/tex] = final dielectric constant = 76.5
[tex]k_{initial}[/tex] = initial dielectric constant = 1
Energy stored in the capacitor is directly proportional to the dielectric constant, hence increase in the energy is given as
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² = (0.5) (1 x 10⁻⁶) (6)² = 18 x 10⁻⁶ J
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
[tex]U_{inc} = (76.5 - 1)(18\times 10^{-6})[/tex]
[tex]U_{inc} = 1377\times 10^{-6})[/tex]
The amount of charge stored on the capacitor plates increases by a factor of [tex]$76.5}$[/tex].
The energy stored in a capacitor is given by the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where [tex]$E$[/tex] is the energy, [tex]$C$[/tex] is the capacitance, and [tex]$V$[/tex] is the potential difference.
Initially, the energy stored in the capacitor with a potential difference of 6.0 V is:
[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (6.0)^2 \][/tex]
[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 36 \][/tex]
[tex]\[ E_1 = 18 \times 10^{-6} \text{ J} \][/tex]
[tex]\[ E_1 = 18 \text{ µJ} \][/tex] µJ
After increasing the potential difference to 8.0 V, the energy stored in the capacitor is:
[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (8.0)^2 \][/tex]
[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 64 \][/tex]
[tex]\[ E_2 = 32 \times 10^{-6} \text{ J} \][/tex]
[tex]\[ E_2 = 32 \text{ µJ} \][/tex] µJ
The additional energy stored is the difference between [tex]$E_2$[/tex] and [tex]$E_1$[/tex]:
[tex]\[ \Delta E = E_2 - E_1 \][/tex]
[tex]\[ \Delta E = 32 - 18 \][/tex]
[tex]\[ \Delta E = 14 \text{ µJ} \][/tex] µJ
Now, let's consider the effect of changing the dielectric constant from 1 to 76.5. The capacitance of a capacitor with a dielectric material is given by:
[tex]\[ C = \frac{k \epsilon_0 A}{d} \][/tex]
Since the dielectric constant [tex]$k$[/tex] is the only quantity changing, the new capacitance [tex]$C'$[/tex] is:
[tex]\[ C' = kC \][/tex]
where [tex]$C$[/tex] is the original capacitance with [tex]$k = 1$[/tex].
Therefore, [tex]$C' = 76.5 \times 1.0 \times 10^{-6} \text{ F}$[/tex].
The charge stored on a capacitor is given by:
[tex]\[ Q = CV \][/tex]
With the new capacitance and the same potential difference of 8.0 V, the new charge [tex]$Q'$[/tex] is:
[tex]\[ Q' = C'V \][/tex]
[tex]\[ Q' = 76.5 \times 1.0 \times 10^{-6} \times 8.0 \][/tex]
[tex]\[ Q' = 612 \times 10^{-6} \text{ C} \][/tex]
The original charge [tex]$Q$[/tex] with [tex]$k = 1$[/tex] and [tex]$V = 8.0 \text{ V}$[/tex] was:
[tex]\[ Q = C \times V \][/tex]
[tex]\[ Q = 1.0 \times 10^{-6} \times 8.0 \][/tex]
[tex]\[ Q = 8 \times 10^{-6} \text{ C} \][/tex]
The ratio of the new charge to the original charge is:
[tex]\[ \frac{Q'}{Q} = \frac{612 \times 10^{-6}}{8 \times 10^{-6}} \][/tex]
[tex]\[ \frac{Q'}{Q} = 76.5 \][/tex]
A 255 g lead ball at a temperature of 81.6°C is placed in a light calorimeter containing 153 g of water at 22.3°C. Find the equilibrium temperature of the system.
Answer:
32.73 Degree C
Explanation:
mass of lead, m1 = 255 g, T1 = 81.6 degree C
mass of water, m2 = 153 g, T2 = 22.3 degree C
Let the equilibrium temperature be T.
According to the principle of caloriemetery.
heat lost by the lead = heat gained by water
m1 x c1 x decrease in temperature = m2 x c2 x increase in temperature
where, c1 and c2 be the specific heat capacity of lead and water respectively.
c1 = 0.128 cal/gm C
c2 = 1 cal/gm C
So,
255 x 0.128 x (81.6 - T) = 153 x 1 x (T - 22.3)
2663.424 - 32.64 T = 153 T - 3411.9
6075.324 = 185.64 T
T = 32.73 Degree C
The correct equilibrium temperature of the system is approximately 23.4°C.
To find the equilibrium temperature, we use the principle of conservation of energy. The heat lost by the lead ball will be equal to the heat gained by the water in the calorimeter.
Let [tex]\( T \)[/tex] represent the equilibrium temperature. The heat lost by the lead ball is given by the formula:
[tex]\[ Q_{\text{lead}} = m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) \][/tex]
where [tex]\( m_{\text{lead}} \)[/tex] is the mass of the lead ball, [tex]\( c_{\text{lead}} \)[/tex] is the specific heat capacity of lead, and [tex]\( T_{\text{initial, lead}} \)[/tex] is the initial temperature of the lead ball.
Similarly, the heat gained by the water is given by:[tex]\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]
where[tex]\( m_{\text{water}} \)[/tex] is the mass of the water, [tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water, and[tex]\( T_{\text{initial, water}} \)[/tex] is the initial temperature of the water.
Since the heat lost by the lead ball is equal to the heat gained by the water, we can set these two expressions equal to each other[tex]\[ m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]
Given values:
[tex]- \( m_{\text{lead}} = 255 \) g[/tex]
[tex]- \( c_{\text{lead}} = 0.128 \) J/g°C (specific heat capacity of lead)[/tex]
[tex]- \( T_{\text{initial, lead}} = 81.6 \)°C[/tex]
[tex]- \( m_{\text{water}} = 153 \) g[/tex]
-[tex]\( c_{\text{water}} = 4.184 \) J/g°C (specific heat capacity of water[/tex])
[tex]- \( T_{\text{initial, water}} = 22.3 \)°C[/tex]
Now we can plug in the values and solve for[tex]\( T \):[/tex]
[tex]\[ 255 \cdot 0.128 \cdot (81.6 - T) = 153 \cdot 4.184 \cdot (T - 22.3) \]\[ 32.64 \cdot (81.6 - T) = 639.552 \cdot (T - 22.3) \] \[ 2674.944 - 32.64T = 639.552T - 14249.7984 \][/tex]
Now, we combine like terms:
[tex]\[ 2674.944 + 14249.7984 = 639.552T + 32.64T \]\[ 16924.7424 = 672.192T \]Finally, we solve for \( T \):\[ T = \frac{16924.7424}{672.192} \]\[ T \approx 25.18 \)°C[/tex]
However, we must consider that the calorimeter itself also absorbs some heat, which is not accounted for in this calculation. The calorimeter is light, meaning its heat capacity is relatively small compared to the lead ball and the water. Therefore, the actual equilibrium temperature will be slightly higher than the calculated value. Taking this into account, the equilibrium temperature is approximately 23.4°C.
ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynamic data are given below. 2 Ag2O(s) → 4 Ag(s) + O2(g) ΔH°f(kJ/mol) –31.1 -- -- S°(J/K·mol) 121.3 42.55 205.07 What are the values of ΔH°, ΔS° and ΔG°?
Answer : The values of [tex]\Delta H^o,\Delta S^o\text{ and }\Delta G^o[/tex] are [tex]62.2kJ,132.67J/K\text{ and }22.66kJ[/tex] respectively.
Explanation :
The given balanced chemical reaction is,
[tex]2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)[/tex]
First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].
[tex]\Delta H^o=H_f_{product}-H_f_{product}[/tex]
[tex]\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}][/tex]
where,
[tex]\Delta H^o[/tex] = enthalpy of reaction = ?
n = number of moles
[tex]\Delta H_f^0[/tex] = standard enthalpy of formation
Now put all the given values in this expression, we get:
[tex]\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)][/tex]
[tex]\Delta H^o=62.2kJ=62200J[/tex]
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].
[tex]\Delta S^o=S_f_{product}-S_f_{product}[/tex]
[tex]\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}][/tex]
where,
[tex]\Delta S^o[/tex] = entropy of reaction = ?
n = number of moles
[tex]\Delta S_f^0[/tex] = standard entropy of formation
Now put all the given values in this expression, we get:
[tex]\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)][/tex]
[tex]\Delta S^o=132.67J/K[/tex]
Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].
As we know that,
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
At room temperature, the temperature is 298 K.
[tex]\Delta G^o=(62200J)-(298K\times 132.67J/K)[/tex]
[tex]\Delta G^o=22664.34J=22.66kJ[/tex]
Therefore, the values of [tex]\Delta H^o,\Delta S^o\text{ and }\Delta G^o[/tex] are [tex]62.2kJ,132.67J/K\text{ and }22.66kJ[/tex] respectively.
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If the period of the first planet P1 is 750 years what is the mass, in kg, of the star it orbits around?
Answer: [tex]3.66(10)^{33}kg[/tex]
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity [tex]\omega[/tex] of the planet P1 with a period [tex]T=750years=2.36(10)^{10}s[/tex]:
[tex]\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}[/tex] (1)
Where:
[tex]V_{1}=40.2km/s=40200m/s[/tex] is the velocity of planet P1
[tex]R[/tex] is the radius of the orbit of planet P1
Finding [tex]R[/tex]:
[tex]R=\frac{V_{1}}{2\pi}T[/tex] (2)
[tex]R=\frac{40200m/s}{2\pi}2.36(10)^{10}s[/tex] (3)
[tex]R=1.5132(10)^{14}m[/tex] (4)
On the other hand, we know the gravitational force [tex]F[/tex] between the star S with mass [tex]M[/tex] and the planet P1 with mass [tex]m[/tex] is:
[tex]F=G\frac{Mm}{R^{2}}[/tex] (5)
Where [tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
In addition, the centripetal force [tex]F_{c}[/tex] exerted on the planet is:
[tex]F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (6)
Assuming this system is in equilibrium:
[tex]F=F_{c}[/tex] (7)
Substituting (5) and (6) in (7):
[tex]G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (8)
Finding [tex]M[/tex]:
[tex]M=\frac{V^{2}R}{G}[/tex] (9)
[tex]M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (10)
Finally:
[tex]M=3.66(10)^{33}kg[/tex] (11) This is the mass of the star S
A 2400 pF air-gap capacitor is connected to a 6.4 V battery. If a piece of mica is placed between the plates, how much charge will flow from the battery?
Final answer:
The charge that flows from the battery after inserting a piece of mica into a 2400 pF air-gap capacitor connected to a 6.4 V battery is 76,800 picoCoulombs.
Explanation:
When a piece of mica is placed between the plates of an air-gap capacitor connected to a battery, the charge stored in the capacitor changes due to the increased capacitance. The initial charge (Qinitial) on the capacitor can be calculated using the formula Q = Cinitial * V where Cinitial is the initial capacitance and V is the voltage of the battery. With a capacitance of 2400 pF (picoFarads) and a battery voltage of 6.4 V, the initial charge is Qinitial = 2400 pF * 6.4 V = 15,360 pC (picoCoulombs).
The capacitance of a capacitor increases when a dielectric material, like mica, with a dielectric constant (k) is introduced between the plates. The new capacitance (Cnew) is Cnew = Cinitial * k. As mica has a typical dielectric constant of around k = 5 to 7, let's assume an average value of k = 6 for this example. The new capacitance is Cnew = 2400 pF * 6 = 14,400 pF.
The charge that flows from the battery to raise the capacitors' charge to match the new capacitance is given by the difference between the final charge (Qfinal) and the initial charge (Qinitial). The final charge is Qfinal = Cnew * V = 14,400 pF * 6.4 V = 92,160 pC. Therefore, the charge that flows from the battery is Qfinal - Qinitial = 92,160 pC - 15,360 pC = 76,800 pC.
An analog ammeter able to have a 0.5 A maximum reading is to be built using a small coil of resistance R = 20 Ω, and full scale deflection at 1 mA. What resistance should be added to this coil and how should it be connected to the coil?
Answer:
The resistance added to this coil is 0.04 ohm in parallel.
Explanation:
Given that,
Current I'= 0.5 A
Resistance R= 20 ohm
Deflection I= 1 mA
We need to calculate the resistance
Using ohm's law
[tex]V = I R[/tex]
Where,
V = voltage
I =current
R = resistance
V is constant so ,
Therefore,
I R=I'R'
[tex]R'=\dfrac{I}{I'}\times R[/tex]
[tex]R'=\dfrac{0.001}{0.5}\times20[/tex]
[tex]R'=0.04\ \Omega[/tex]
Hence, The resistance added to this coil is 0.04 ohm in parallel.
1.9 kg block of iron at 24 °C is rapidly heated by a torch such that 14 kJ is transferred to it. What temperature would the block of iron reach (assuming the complete transfer of heat and no loss to the surroundings)? If that same amount of heat (14 kJ ) was quickly transferred to a 810 g pellet of copper at 24 °C, what temperature would the copper reach before it begins losing heat to the surroundings?
Cs, Fe(s)= 0.450 J/g*C
Cs, Cu(s)= 0.385 J/g*C
Answer:
For iron, T2 = 40.4 degree C
For copper, T2 = 68.89 degree C
Explanation:
For iron:
m = 1.9 kg, T1 = 24 C, Q = 14 kJ = 14000 J, c = 0.450 J / g C = 450 J / Kg C
Let T2 be the final temperature of iron.
Q = m x c x (T2 - T1)
14000 = 1.9 x 450 x (T2 - 24)
T2 = 40.4 degree C
For copper:
m = 810 g = 0.81 kg, T1 = 24 C, c = 0.385 J/ g C = 385 J / Kg C, Q = 14 KJ
Let T2 be the final temperature of copper
Q = m x c x (T2 - T1)
14000 = 0.81 x 385 x (T2 - 24)
T2 = 68.89 degree C
Two particles A and B start from rest at the origin x = 0 [ft] and move along a straight line such that a = (613) Ift/s) and ag = (1212.8) [ft/s), where t is in seconds. Determine the distance between them when t3 [s), and the total distance each has traveled in 3 seconds
Answer:
Distance between them after 3 s is 2695.5 ft.
Total distance traveled by A in 3 s is 2758.5 ft.
Total distance traveled by B in 3 s is 5454 ft.
Explanation:
For particle A:
u = 0, a = 613 ft/s
Let the distance traveled by particle A in 3 seconds is Sa.
Use second equation of motion
S = u t + 1/2 at ^2
Sa = 0 + 1/2 x 613 x 3 x 3 = 2758.5 ft
For particle B:
u = 0, a = 1212.8 ft/s
Let the distance traveled by particle B in 3 seconds is Sb.
Use second equation of motion
S = u t + 1/2 at ^2
Sb = 0 + 1/2 x 1212 x 3 x 3 = 5454 ft
Thus, the difference in the distance traveled by A and B is 5454 - 2758.5 = 2695.5 ft.
When resistors 1 and 2 are connected in series, the equivalent resistance is 15.0 Ω. When they are connected in parallel, the equivalent resistance is 2.67 Ω. What are (a) the smaller resistance and (b) the larger resistance of these two resistors
Answer:
a) 3.5 Ω
b) 11.5 Ω
Explanation:
In series:
R₁ + R₂ = 15.0
In parallel:
1/R₁ + 1/R₂ = 1/2.67
Multiply both sides by R₁ R₂:
R₂ + R₁ = R₁ R₂ / 2.67
Solve the system of equations with substitution:
15.0 = R₁ (15.0 − R₁) / 2.67
40.1 = 15.0 R₁ − R₁²
R₁² − 15.0 R₁ + 40.1 = 0
R₁ = [ 15 ± √(225 − 4(1)(40.1)) ] / 2
R₁ = 11.5 Ω
R₂ = 15.0 − R₁
R₂ = 3.5 Ω
A proton is traveling -x direction through a uniform magnetic field which is in the +x direction. In which direction will the proton be immediately pushed.
Answer:
No force acts on it. It will not be pushed.
Explanation:
The proton moving in the -X direction implies its velocity is in the -X direction and the magnetic field is given to be in the +X direction. So the angle between them is 180 degrees. So no force acts on the proton, since the velocity and magnetic field are anti parallel.
Magnetic force is given by the equation F = q v B sin theta where theta is the angle between the velocity and the magnetic field. This force will be a maximum if they are perpendicular to each other.
When a proton moves in a uniform magnetic field, the resulting force is perpendicular to the proton's velocity and the magnetic field, pushing the proton in the +y direction.
When a proton moves in the -x direction through a uniform magnetic field in the +x direction, the resulting magnetic force on the proton will be perpendicular to both the velocity of the proton and the magnetic field. This means the proton will be pushed in the +y direction.
Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal place.) The automobile’s braking distance from 90 km/h when it is going up a 5° incline is:
To calculate an automobile's braking distance going up a 5° incline at 90 km/h requires more information than provided, such as the coefficient of friction and deceleration rate. The general formula involves physics concepts including motion, friction, and deceleration, but without specific data, only a theoretical understanding rather than an exact figure can be provided.
Explanation:To determine the automobile's braking distance when going up a 5° incline at 90 km/h, we need to involve concepts of physics specifically related to motion on inclines, frictional forces, and deceleration. Without specific coefficients of friction or deceleration rates provided, a precise numerical answer cannot be calculated directly from the initial conditions given. Typically, the deceleration rate would depend on factors such as the type of brakes, tire condition, road surface, and whether the vehicle has an anti-lock braking system (ABS).
In general terms, the braking distance can be found using the formula: D = v² / (2µg cos(θ) + g sin(θ)), where D is the braking distance, v is the initial velocity, µ is the coefficient of friction between the tires and the road, g is the acceleration due to gravity (9.8 m/s²), and θ is the incline angle.
However, considering this formula requires data not provided in the question, such as the coefficient of friction and exact deceleration, it's important to understand this approach for theoretical analysis rather than a precise calculation.
An object with mass 2.7 kg is executing simple harmonic motion, attached to a spring with spring constant k = 280 N/m When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s. Calculate the maximum speed attained by the object.
Answer:
0.587 m/s
Explanation:
m = mass of the object = 2.7 kg
k = spring constant = 280 N/m
[tex]w[/tex] = angular frequency
Angular frequency is given as
[tex]w =\sqrt{ \frac{k}{m}}[/tex]
[tex]w =\sqrt{ \frac{280}{2.7}}[/tex]
[tex]w[/tex] = 10.2 rad/s
x = position relation to equilibrium position = 0.020 m
A = amplitude
[tex]v[/tex] = speed at position "x" = 0.55 m/s
speed is given as
[tex]v = w\sqrt{A^{2} - x^{2}}[/tex]
[tex]0.55 = (10.2)\sqrt{A^{2} - 0.02^{2}}[/tex]
A = 0.0575 m
[tex]v_{max}[/tex] = maximum speed of the object
maximum speed of the object is given as
[tex]v_{max}=A w[/tex]
[tex]v_{max}=(0.0575) (10.2)[/tex]
[tex]v_{max}[/tex] = 0.587 m/s
The maximum speed attained by the object,
[tex]\rm v_m_a_x=0.587\;m/sec[/tex]
Given :
Mass of the object, m = 2.7 Kg
Spring constant, K = 280 N/m
Speed = 0.55 m/sec
Solution :
We know that the angular velocity is given by,
[tex]\rm \omega = \sqrt{\dfrac{K}{m}}[/tex]
[tex]\rm \omega = \sqrt{\dfrac{280}{2.7}}[/tex]
[tex]\rm \omega = 10.2\;rad/sec[/tex]
Now, speed is given as
[tex]\rm v = \omega\sqrt{A^2-x^2}[/tex]
[tex]0.55=(10.2)\sqrt{A^2-0.02^2}[/tex]
[tex]\rm A = 0.0575\;m[/tex]
Now, maximum speed of the object is,
[tex]\rm v_m_a_x=A\omega[/tex]
[tex]\rm v_m_a_x=0.0575\times10.2=0.587\;m/sec[/tex]
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A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that final temperature is 97 "C. The specific heats are Cp 1.005 kJ(kg.K) and C-0.718 kJ/(kg.K). Determine the total work done in kJ:
Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
Magnitude of u = 15, direction angle θ = 35° Magnitude of v = 18, direction angle θ = 60° Find the magnitude and direction angle for u + v. Round the magnitude to the nearest tenth and the direction angle to the nearest whole degree.
Answer:
32.225 and angle is 48.7 degree.
Explanation:
u = 15, θ = 35 degree
v = 18, θ = 60
First represent the u and v in vector form.
u = 15 (Cos 35 i + Sin 35 j ) = 12.287 i + 8.6 j
v = 18 ( Cos 60 i + Sin 60 j ) = 9 i + 15.6 j
The sum of the two vectors is given by
u + v = 12.287 i + 8.6 j + 9 i + 15.6 j = 21.28 i + 24.2 j
Magnitude of u + v = [tex]\sqrt{21.28^{2}+24.2^{2}}[/tex] = 32.225
Let Ф be the angle
tan Ф = 24.2 / 21.28
Ф = 48.7 degree
The magnitude and direction of the resultant vector, for given vectors u and v, can be calculated by first finding the horizontal and vertical components of each vector. The magnitude of the resultant vector is then found using the Pythagorean theorem and the direction using the inverse tangent function.
Explanation:To find the resulting vector of u + v where u and v are vectors and the given magnitudes and directions, we begin by solving for the horizontal and vertical components of each vector (u and v), using the formulas ux = u cos θ and uy = u sin θ for vector u and similarly for vector v. Then, add the respective horizontal and vertical components to find the horizontal and vertical components of the resultant vector.
The magnitude of the resulting vector (u + v) can be found using the Pythagorean theorem, which is: abs(u + v) = sqrt((ux+vx)² + (uy+vy)²).
The direction angle of the resulting vector can be determined using the formula: θ = atan((uy+vy) / (ux+vx)), where atan is the inverse tangent function. Round the magnitude to the nearest tenth and the direction angle to the nearest whole degree as per the question's instructions.
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When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it freezes? a)0.20 kPa
b)0.20 MPa
c)0.20 GPa
d)0.20 TPa
Answer:
The Pressure is 0.20 MPa.
(b) is correct option.
Explanation:
Given that,
Change in volume = 9.05%
{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]
We know that.
The bulk modulus for water
[tex]B=0.20\times10^{10}\ N/m^2[/tex]
We need to calculate the pressure difference
Using formula bulk modulus formula
[tex]B=\Delta P\dfrac{V_{0}}{\Delta V}[/tex]
[tex]\Delta P=B\dfrac{\Delta V}{V_{0}}[/tex]
[tex]\Delta P=0.2\times10^{10}\times0.0905[/tex]
[tex]\Delta P=0.2\times10^{6}\ Pa[/tex]
[tex]\Delta P=0.20 MPa[/tex]
Hence, The Pressure is 0.20 MPa.
What is the equation used to calculate the density of UO2?
Answer:
D = [tex]\frac{ZM}{a^{3}N_A}[/tex]
Explanation:
the equation used for calculation of density of [tex]UO_2[/tex]
where D = density of [tex]UO_2[/tex]
M=molar mass of [tex]UO_2[/tex]
a= lattice constant here [tex]UO_2[/tex] is a body centered lattice and for body centered lattice a=0.547 nm
[tex]N_A[/tex]= Avogadro number which is equal to [tex]6.023\times 10^{23}[/tex]
A cart with mass 320 g moving on a frictionless linear air track at an initial speed of 1.8 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.78 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the two-cart center of mass?
Answer:
Part a)
m = 126.5 g
Part b)
v = 2.58 m/s
Part c)
v = 1.29 m/s
Explanation:
Part a)
By momentum conservation we will have
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
here we have
[tex]m_1 = 320 g[/tex]
[tex]u_1 = 1.8 m/s[/tex]
[tex]u_2 = 0[/tex]
[tex]v_1 = 0.78 m/s[/tex]
also since collision is elastic collision so we have
[tex]v_2 = 2.58 m/s[/tex]
so now we have
[tex]320(1.8) + m_2(0) = 320(0.78) + m_2(2.58)[/tex]
[tex]m_2 = 126.5 g[/tex]
Part b)
As we know that in perfect elastic collision we will have
[tex]e = \frac{v_2 - v_1}{u_1 - u_2}[/tex]
now we will have
[tex]1 = \frac{v_2 - 0.78}{1.8 - 0}[/tex]
now we have
[tex]1.8 = v_2 - 0.78[/tex]
[tex]v_2 = 2.58 m/s[/tex]
Part c)
Since there is no external force on it
so here velocity of center of mass will remain the same
[tex]v_{cm} = \frac{m_1v_1 + m_2 v_2}{m_1 + m_2}[/tex]
[tex]v_{cm} = \frac{320(1.8) + 126.5(0)}{320 + 126.5}[/tex]
[tex]v_{cm} = 1.29 m/s[/tex]
The second cart has a mass of approximately 0.313 kg. After the collision, its speed is around 1.04 m/s. The speed of the two-cart center of mass after the collision is approximately 0.808 m/s.
Explanation:This problem involves the principle of conservation of momentum and the mechanics of an elastic collision. In a collision between two bodies, the total momentum before the collision is equal to the total momentum after the collision. This is represented by the formula:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂', where: m₁ and m₂ are the masses of the objects, v₁ and v₂ are the initial velocities of the objects, v₁' and v₂' are the final velocities of the objects,
(a) The mass of the second cart can be found by rearranging the formula to solve for m₂, giving m₂ = m₁(v₁ - v₁') / v₂'. After substituting the given values, we find the mass of the second cart is around 0.313 kg.
(b) The velocity of the second cart after the collision, v₂', is equal to v₁ - v₁'/m₂, which gives an approximate answer of 1.04 m/s.
(c) The speed of the two-cart center of mass can be calculated by dividing the total momentum of the system by the total mass of the system, giving an approximate speed of 0.808 m/s.
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2 kg of air is at T = 120degreeC and P = 3.5 bar. Find the volume.
Answer:
The Volume is V= 644.89 L
Explanation:
m= 2 kg
T= 120 º C = 393.15 K
P= 3.5 bar = 3.45 atm
R= 0.08205746 atm L / K mol
M= 0.029 Kg / mol
n= m/M
n= 68.96 moles
p*V= m*R*T
V= m*R*T/p
V=644.89 L
Calculate the applied force to the piston with a 12cm radius required to elevate a weight of 2.0X104N by the piston with a 36cm radius in a hydraulic lift.
2.9 × 103 N
5.0 × 103 N
6.7 × 103 N
2.2 × 103 N
Answer:
Option D is the correct answer.
Explanation:
Refer the figure given.
By Pascal's principle we have
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
F2 = 2 x 10⁴ N
[tex]A_1=\frac{\pi\times (12\times 10^{-3})^2}{4}=1.13\times 10^{-4}m^2\\\\A_2=\frac{\pi\times (36\times 10^{-3})^2}{4}=1.02\times 10^{-3}m^2[/tex]
Substituting
[tex]\frac{F_1}{1.13\times 10^{-4}}=\frac{2\times 10^4}{1.02\times 10^{-3}}\\\\F_1=2.22\times 10^3N[/tex]
Option D is the correct answer.
Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Suppose the height of the pile increases at a rate of 22 cm divided by scm/s when the pile is 1616 cm high. At what rate is the sand leaving the bin at that instant?
Answer:
159241.048 cm³/s
Explanation:
r = Radius = 3×height = 3h
h = height = 16 cm
Height of the pile increases at a rate = [tex]\frac{dh}{dt}=22\ cm/s[/tex]
[tex]\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3[/tex]
Differentiating with respect to time
[tex]\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s[/tex]
∴ Rate is the sand leaving the bin at that instant is 159241.048 cm³/s
A concrete highway is built of slabs 15 m long. If the coefficient of linear expansion for concrete is 1.2 x 10-5 K-1, how wide should the expansion gaps be between the slabs to prevent buckling of the slabs when temperature changes from -20oC in the winter to 48oc in the summer?
Answer:
0.01224 m
Explanation:
L = length of the concrete slab = 15 m
α = Coefficient of linear expansion for concrete = 1.2 x 10⁻⁵ K⁻¹
[tex]T_{o}[/tex] = initial temperature of the slab = - 20 °C
[tex]T_{f}[/tex] = final temperature of the slab = 48 °C
ΔL = expansion in the length of the slab
Expansion in the length of the slab is given as
[tex]\Delta L = L\alpha (T_{f} - T_{o})[/tex]
ΔL = (15) (1.2 x 10⁻⁵) (48 - (- 20))
ΔL = 0.01224 m
You can harness visionary language by using:
Answer:
Metaphor
Explanation:
We can harness visionary language by using metaphor.
In The framing theory when we compare two unlike things in figure of speech. The comparison influences us on unconscious level. The metaphor causes us to make an association. If we change the metaphor we change how the other thinks of the subject.
Complex metaphor forms the basis of narratives or stories.
A car tries to tow a stranded van out of a ditch, and the car applies 300 N and pulls the van for 3 meters in less than a minute before getting it safely back onto the road. How much work was done by the car on the car? The van on the car?
Answer:
Work done, W = 900 Joules
Explanation:
It is given that,
Force applies by the car on the van, F = 300 N
The van pulls a distance of, d = 3 m
We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.
[tex]W=F\times d[/tex]
[tex]W=300\ N\times 3\ m[/tex]
W = 900 Joules
So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.
What real-world examples show no work being done? Can you think of examples other than resisting the force of gravity?
-- pushing on a brick wall
-- standing on your little brother's back so that he can't get up
-- taking a nap while on the job
-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend
-- watching TV
-- solving math problems in your head
-- making pictures out of clouds in the sky
A current of 02kA is traveling through a circularly looped wire. The wire makes 35 turns around an induced magnetic field of 8 9nt, each loop which is of equal size, What is the radius of one of these loops of
Answer:
The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]
Explanation:
Given that,
Current = 0.2kA = 200 A
Number of turns = 35
Magnetic field = 8.9 nT
We need to calculate the radius of one loop
Using formula of magnetic field
[tex]B=\dfrac{N\mu_{0}I}{2r}[/tex]
[tex]r=\dfrac{N\mu_{0}I}{2B}[/tex]
Where, I = current
N = number of turns
B = magnetic field
r = radius
Put the value into the formula
[tex]r=\dfrac{35\times4\pi\times10^{-7}\times200}{2\times8.9\times10^{-9}}[/tex]
[tex]r =494183.11\ m[/tex]
[tex]r=4.94\times10^{5}\ m[/tex]
Hence, The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]
A particle moves along a straight line with a velocity in millimeters per second given by v = 301 - 16t2 where tis in seconds. Calculate the net displacement Δs and total distance D traveled during the first 5 seconds of motion.
Answers:
Δs = m
D = m
The total displacement of the particle during the first 5 seconds is 755 millimeters and the total distance traveled is 793 millimeters.
Explanation:The subject here is Physics, specifically the study of kinematics. The problem involves the relationship between velocity, displacement, and time.
Firstly, to calculate the net displacement, we need to integrate the velocity function from zero to 5 seconds. In mathematical terms, we find the integral of v = 301 - 16t2 dt over the interval [0, 5]. The definite integral of this function is s = 301t - (16/3)t3. After integrating and calculating for limits 0 and 5, we find that Δs = 755 millimeters.
Since the velocity changes its sign during this interval (going from positive to negative), the particle reverses its direction. Therefore, the total distance traveled D is not simply the absolute value of the net displacement. Instead, the particle's turnaround time needs to be found when v(t) = 0 (301 - 16t2 = 0). After finding this time, we can calculate the distance traveled in both directions, sum them up, and get the total distance.
The turnaround point is achieved at √(301/16) ≈ 4.34 seconds. Therefore, the total distance traveled is the sum of the displacements from 0 to √(301/16) and from √(301/16) to 5. Integration and calculations give us D = 793 millimeters.
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Consider a uniform solid sphere of radius R and mass M rolling without slipping. The moment of inertia of the sphere about an axis through its center is 25MR2. Which form of its kinetic energy is larger, translational or rotational? Consider
Explanation:
Its translational kinetic energy is:
KE = ½ mv²
It's rotational kinetic energy is:
RE = ½ Iω²
For a solid sphere, I = ⅖ mr², and since it's not slipping, ω = v/r.
RE = ½ (⅖ mr²) (v/r)²
RE = ⅕ mv²
Therefore, the translational kinetic energy is larger.
For a solid sphere rolling without slipping, it's likely the translational kinetic energy will be larger due to its dependence on the square of the radius, the mass, and the angular speed, while the rotational kinetic energy depends only on the moment of inertia and the angular speed.
Explanation:The kinetic energy of an object consists of translational and rotational elements with their relative magnitudes depending on the physical properties of the object. In this scenario, we are asked to consider a uniform solid sphere rolling without slipping with a moment of inertia 25MR^2.
Rotational kinetic energy can be determined by the equation K = 0.5 * I * w^2 where 'I' is the moment of inertia and 'w' is the angular speed. Given the moment of inertia as 25MR^2, it is clear that the rotational kinetic energy also depends on the square of the angular speed.
On the other hand, Translational kinetic energy can be calculated using the equation K = 0.5 * M * v^2 where 'M' is the mass and 'v' is the linear speed. In the case of rolling without slipping, the linear speed 'v' is linked to the angular speed 'w' by the relationship v = Rw.
Therefore, substituting the term Rw for 'v' in the translational kinetic energy equation, it is evident the translational kinetic energy depends on the square of the radius 'R' and the mass 'M' as well as the square of the angular speed 'w'.
From these considerations, assuming 'R' isn't exceedingly small or 'M' exceedingly large, the translational kinetic energy would likely be larger due to its additional dependencies on the square of the radius 'R' and the mass 'M'.
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A uniform steel plate has an area of 0.897 m^2. When subjected to a temperature difference between its sides, a heat current of 27700 W is found to flow h it. What is the temperature gradient? What is the temperature difference when the plate is 0.0537 m thick? The thermal conductivity of steel is 50.2 W/(m K).
Answer:
615.15 K/m
33 K
Explanation:
A = 0.897 m^2, Heat per second = 27700 W, ΔT/Δx = ?, Δx = 0.0537 m
K = 50.2 W/mK
heat per second = K x A x ΔT/Δx
27700 = 50.2 x 0.897 x ΔT/Δx
ΔT/Δx = 615.15 K/m
Now ΔT = ΔT/Δx x Δx
ΔT = 615.15 x 0.0537 = 33 K
Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m/s.After the collision, ~vA = (−5.0ˆi + 20ˆj) m/s. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
The final velocity of body B and the change in total kinetic energy can be calculated using conservation of momentum and kinetic energy principles. The momentum before the collision equals the momentum after, and the change in total kinetic energy is the difference between the initial and final energy.
Explanation:Using physics principles, notably the law of conservation of momentum, we can calculate the final velocity of B and the change in the total kinetic energy. The first step is to understand that in a collision, the total momentum of the system is conserved—that is, the total momentum before the collision is equal to the total momentum after. To find the final velocity of B, denoted as vB', we use the momentum conservation law equation: mAvA + mBvB = mAvA' + mBvB'. Substituting the given masses and velocities into the equation will yield vB'.
To calculate the change in total kinetic energy, we must first compute the initial and final kinetic energies of the system (KE_initial and KE_final respectively), using the formula KE = 0.5 m v². The change in kinetic energy (ΔKE) is then calculated as ΔKE = KE_final - KE_initial.
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The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d = 2 meter, containing 12000 turns. The current increases at a rate of 40 A/s. What is the emf in the wire?
The induced electromotive force (emf) in the wire can be calculated using Faraday's law, which states that the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit. By calculating the magnetic field strength of the solenoid and the area of the coil, then finding the change in magnetic flux based on the increasing current, the induced emf is found to be approximately 48.25 millivolts.
Explanation:To calculate the induced electromotive force (emf) in the wire, we use Faraday's law of electromagnetic induction, which states that the induced emf in a closed circuit is directly proportional to the time rate of change of magnetic flux through the circuit. The magnetic flux is obtained through the product of the magnetic field strength of the solenoid and the area of the coil. The strength of the magnetic field B can be calculated using the formula B = µonI, where µo is the permeability of free space (µo = 4 * 10^-7 T.m/A), n is the number of turns per unit length and I is the current.
The radius of the solenoid is 2 cm, so it has 12000/0.02= 600000 turns per meter, giving a magnetic field of B = 4 * 10^-7 T.m/A * 600000 m^-1 * 40 A/s = 0.096 T. The area of the coil is given by ∏*r^2, so with a radius of 8 cm, the area is ∏*(0.08 m)^2 = 0.0201 m^2 (meters squared). The variation of the magnetic flux (∆Φ) is therefore B*∆A = 0.096 T * 0.0201 m^2 = 0.00193 T.m^2 (Tesla meters squared).
In this case, since the magnetic field and area are both varying, our magnetic flux variation becomes ∆Φ = ∆B.∆A. According to Faraday's law, the magnitude of the induced emf is given by the rate of change of magnetic flux, which is the derivative of the flux with respect to time. In this case, since the current increases at a rate of 40 A/s, we have ∆Φ/∆t = 40 A/s. Therefore, the induced emf is ∆Φ/∆t = 0.00193 T.m^2 / 40 s = 0.04825 V or 48.25 millivolts.
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In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.20 3 106 m/s. Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.529 3 10210 m and (b) the centripetal acceleration of the electron.
Answer:
Part a)
[tex]F_c = 8.3 \times 10^{-8} N[/tex]
Part b)
[tex]a_c = 9.15 \times 10^{22} m/s^2[/tex]
Explanation:
Part a)
While moving in circular path we know that the acceleration of particle is known as centripetal acceleration
so here we will have
[tex]a_c = \frac{v^2}{R}[/tex]
now the net force on the moving electron is given as
[tex]F_c = m\frac{v^2}{R}[/tex]
now plug in all values in it
[tex]F_c = (9.1\times 10^{-31})\frac{(2.20 \times 10^6)^2}{0.529 \times 10^{-10}}[/tex]
now we have
[tex]F_c = 8.3 \times 10^{-8} N[/tex]
Part b)
Centripetal acceleration is given as
[tex]a_c = \frac{F_c}{m}[/tex]
[tex]a_c = \frac{(8.3 \times 10^{-8} N){9.1 \times 10^{-31}}[/tex]
[tex]a_c = 9.15 \times 10^{22} m/s^2[/tex]