Answer:
3.6 x 10⁻⁷ H
Explanation:
C = capacitance of the capacitor = 6.57 x 10⁻¹² F
L = inductance of the inductor = ?
f = frequency of broadcasting system = 103.9 MHz = 103.9 x 10⁶ Hz
frequency is given as
[tex]f = \frac{1}{2\pi \sqrt{LC}}[/tex]
[tex]103.6\times 10^{6} = \frac{1}{2(3.14)\sqrt{(6.57\times 10^{-12})L}}[/tex]
Inserting the values
L = 3.6 x 10⁻⁷ H
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.
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]
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
(a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 µT. (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.
Answer:
(a) 1.3 x 10^6 Hz
(b) 76.73 cm
Explanation:
(a)
the formula for the frequency is given by
f = B q / 2 π m
where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.
B = 46.7 micro tesla = 46.7 x 10^-6 T
q = 1.6 x 10^-19 C
m = 9.1 x 10^-31 kg
f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz
(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J
K = 1/2 mv^2
182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 6.3 x 10^6 m/s
r = m v / B q
Where, r be the radius of circular path
r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)
r = 0.7673 m = 76.73 cm
A small charged bead has a mass of 1.0 g. It is held in a uniform electric field of magnitude E = 200,000 N/C, directed upward. When the bead is released, it accelerates upward with an acceleration of 20 m/s2. What is the charge on the bead?
Answer:
10^-7 C
Explanation:
m = 1 g = 10^-3 kg, E = 200,000 N/C, a = 20 m/s^2, u = 0
Let q be the charge on bead
Force = m a = q E
a = q E / m
q = m a / E = (10^-3 x 20) / 200000 = 10^-7 C
A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 20.0 \rm kW of electric power (generated by doing work at the rate 20.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 20.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 7.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K=QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 300 hours per month during the winter. What does one month's heating cost in the home with a 16.0 kW electric heater?
What does one month's heating cost in the home of a neighbor who uses a heat pump to provide the same amount of heating?
Answer:
a) 2.85 kW
b) $ 432
c) $ 76.95
Explanation:
Average price of electricity = 1 $/40 MJ
Q = 20 kW
Heat energy production = 20.0 KJ/s
Coefficient of performance, K = 7
also
K=(QH)/Win
Now,
Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7
where
P(in) is the input power
Thus,
P(in) = 20/7 = 2.85 kW
b) Cost = Energy consumed × charges
Cost = ($1/40000kWh) × (16kW × 300 × 3600s)
cost = $ 432
c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95
Final answer:
To heat a home for one month, an electric heater requires $432, while a heat pump with a coefficient of performance of 7 costs significantly less at $61.71, showcasing the efficiency and cost-effectiveness of using a heat pump for home heating.
Explanation:
To calculate the cost of heating for one month using a 16.0 kW electric heater and a heat pump with a coefficient of performance of 7.00, we start with understanding how a heat pump works. It essentially moves heat from the outdoors indoors, making it more efficient than directly converting electricity into heat. The first step is to calculate the total energy used by the electric heater and then compare it to the energy utilized efficiently by the heat pump.
Cost Calculation for Electric Heater:
Power usage of heater = 16.0 kW
Total hours run per month = 300 hours
Total energy used per month = 16.0 kW × 300 hours = 4800 kWh
Since 1 kWh equals 3.6 MJ, the total energy in MJ = 4800 × 3.6 = 17280 MJ
Average price for electricity = 40 MJ per dollar
Cost = 17280 MJ / (40 MJ/dollar) = $432
Cost Calculation for Heat Pump:
The coefficient of performance (COP) of the heat pump = 7.00
Efficiency implies that for every unit of energy (work) used, 7 units of heat are transferred inside.
To provide the same heat as the electric heater, the energy consumed by the heat pump = 4800 kWh / 7 = 685.71 kWh
Total energy in MJ = 685.71 × 3.6 = 2468.56 MJ
Cost = 2468.56 MJ / (40 MJ/dollar) = $61.71
Thus, using a heat pump is significantly more cost-efficient for home heating compared to using a direct electric heater.
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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The work W done in moving an object from (0,0)(0,0) to (13,4)(13,4) subject to a constant force FF is W=F ⋅rW=F ⋅r, where rr is the vector with its head at (13,4)(13,4) and tail at (0,0)(0,0). The units are feet and pounds. (a) Suppose the force F=8cosθi+8sinθjF=8cosθi+8sinθj. Find WW in terms of θθ. (b) Suppose the force FF has magnitude of 55 lb and makes an angle of π6π6 rad with the horizontal, pointing right. Find WW in foot-pounds. (Use symbolic notation and fractions where needed.)
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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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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Two small plastic spheres each have a mass of 2 grams and a charge of -50 nC. They are placed 2 cm apart (center to center). A) What is the magnitude of the electric force on each sphere? B) If one sphere is held in place what is the resulting acceleration magnitude of the other?
Answer:
Part a)
F = 0.056 N
Part b)
a = 28.13 m/s/s
Explanation:
Part a)
Electrostatic force between two charged balls is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we will have
[tex]q_1 = q_2 = 50 nC[/tex]
r = 2 cm
now we will have
[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.02^2}[/tex]
so here we have
[tex]F = 0.056 N[/tex]
Part b)
Now due to above force the acceleration of one ball which is released is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{0.056}{2 \times 10^{-3}}[/tex]
[tex]a = 28.13 m/s^2[/tex]
A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm to their eye, what should be the focal length of the prescribed lens? Assume the lens will be 2 cm from the eye.
Answer:
[tex]f = 8.89 cm[/tex]
Explanation:
As we know that Far sighted person has near point shifted to 80 cm distance
so he is able to see the object 80 cm
now the distance of lens from eye is 2 cm
and the person want to see the objects at distance 10 cm
so here the image distance from lens is 80 cm and the object distance from lens is 8 cm
now from lens formula we have
[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]
[tex]-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}[/tex]
[tex]f = 8.89 cm[/tex]
What is the pressure exerted on a 100 cm2 plate by a weight of 200 N resting on the plate? 0 A) 20 MPa O B) 20,000 kPa C) 2 Pa ○ D) .05 Pa 0 E) 20 kPa Save
Answer:
The answer is E) 20kPa
Explanation:
Here we use the definition of Pressure: Pressure is a physical quantity that measures the projection of force in the perpendicular direction per unit area. In this case, wehave a 200N Force over a 100cm^2 Area
In order to have the pressure in international units we need to convert those cm^2 to m^2 this way:
We know 100cm is equal to 1m
[tex](100cm^2)*((1m^2)/(100cm)^2)[/tex]
In this case, the area will be equal to 0,01m^2
With this information we use the Pressure equation, which based on the definition will be:
[tex]P=F/A[/tex]
Solving:
[tex]P=(200N)/(0,01m^2)[/tex]
[tex]P=20000Pa[/tex]
Then P is the same as 20kPa
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.
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.
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.
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
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
15. Determine the zeros for and the end behavior of f(x) = x(x - 4)(x + 2)
Answer:
[tex]x=0\\x=4\\x=-2[/tex]
Explanation:
The zeros of a function are all the values of x for which f (x) = 0.
Therefore to find the zeros of the function I must equal f(x) to zero and solve for x.
[tex]f(x) = x(x - 4)(x + 2)=0[/tex]
[tex]x(x - 4)(x + 2)=0[/tex]
We have the multiplication of 3 factors x, (x-4) and (x + 2)
Then the function will be equal to zero when one of the factors is equal to zero, that is:
[tex]x = 0\\(x-4) = 0,\ x = 4\\(x + 2) = 0,\ x = -2[/tex]
Note that [tex]f(x) = x (x - 4) (x + 2)[/tex] is a cubic function of positive principal coefficient, the graph starts from [tex]-\infty[/tex] and cuts to the x-axis at [tex]x = -2[/tex], then decreases and cuts by second once to the x-axis at [tex]x = 0[/tex], it finally cuts the x-axis for the third time at [tex]x = 4[/tex] and then tends to [tex]\infty[/tex]
The zeros of the function f(x) = x(x - 4)(x + 2) are 0, 4, and -2. The end behavior of the function indicates that as x approaches infinity, f(x) approaches infinity, and as x approaches negative infinity, f(x) approaches negative infinity.
To determine the zeros of the function f(x) = x(x - 4)(x + 2), we need to find the values of x that make f(x) equal to zero. The function will be zero when any of the factors are zero. Therefore, the zeros of f(x) are 0, 4, and -2.
The end behavior of the polynomial function is determined by the leading term. Since our function has a leading term x3 with a positive coefficient, as x approaches positive infinity, f(x) also approaches positive infinity, and as x approaches negative infinity, f(x) approaches negative infinity. This is characteristic of an odd-degree polynomial with a positive leading coefficient.
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.
Learn more about Elastic Collision here:https://brainly.com/question/33268757
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Two narrow slits 84 mum apart are illuminated with light of wavelength 650 nm. What is the angle of the m = 3 bright fringe in radians What is the angle in degrees
Answer:
The angle is 1.33°.
Explanation:
Given that,
Distance [tex]d=84\times10^{-6}\ m[/tex]
Wavelength = 650 nm
Number of fringe = 3
We need to calculate the angle
Using formula of angle for brightness
[tex]d\sin\theta=m\lambda[/tex]
[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
Where, d = distance
[tex]\lambda[/tex] =wavelength
m = number of fringe
Put the value into the formula
[tex]\theta=\sin^{-1}\dfrac{3\times650\times10^{-9}}{84\times10^{-6}}[/tex]
[tex]\theta=1.33^{\circ}[/tex]
Hence, The angle is 1.33°.
A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.44 m/s2 and β = 5.00×10−2 m/s3
a. Calculate the average velocity of the car for the time interval t=0 to t1 = 1.95 s .
b. Calculate the average velocity of the car for the time interval t=0 to t2 = 3.96 s .
c. Calculate the average velocity of the car for the time interval t1 = 1.95 s to t2 = 3.96 s .
Answer:
a) Average velocity of the car for the time interval t=0 to t = 1.95 s is 2.62 m/s.
b) Average velocity of the car for the time interval t=0 to t = 3.96 s is 4.92 m/s.
c) Average velocity of the car for the time interval t=1.95 to t = 3.96 s is 7.15 m/s.
Explanation:
We have x(t)= α t²− β t³
That is x(t)= 1.44 t²− 5 x 10⁻² t³
Average velocity is ratio of distance traveled to time.
a)Average velocity of the car for the time interval t=0 to t = 1.95 s
x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m
x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m
Time difference = 1.95 - 0 = 1.95 s
Average velocity [tex]=\frac{5.10}{1.95}=2.62m/s[/tex]
b)Average velocity of the car for the time interval t=0 to t = 3.96 s
x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m
x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m
Time difference = 3.96 - 0 = 3.96 s
Average velocity [tex]=\frac{19.48}{3.96}=4.92m/s[/tex]
c)Average velocity of the car for the time interval t=0 to t = 3.96 s
x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m
x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m
Time difference = 3.96 - 1.95 = 2.01 s
Average velocity [tex]=\frac{19.48-5.10}{2.01}=7.15m/s[/tex]
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.
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
From a lake, water is pumped at a rate of 67 L/s to a storage tank positioned 14 m above while consuming 16.4 kW of electrical power. (Assuming no frictional losses or kinetic energy change taking place), determine the overall efficiency of the pump-motor unit. (Take gravitational acceleration as 10 m/s2).
Answer:
57 %
Explanation:
input power = 16.4 kW = 16.4 x 10^3 W = 16400 W
Water pumped per second = 67 L/s
Mass of water pumped per second, m = Volume of water pumped epr second x density of water
m = 67 x 10^-3 x 1000 = 67 kg/s
height raised, h = 14 m
Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W
efficiency = output power / input power = 9380 / 16400 = 0.57
% efficiency = 57 %
thus, the efficiency of the pump is 57 %.
. In an elastic collision, what happens to the change in kinetic energy? A) It is transformed into heat and also used to deform colliding objects. B) It is converted into potential energy. C) It is transformed into momentum such that momentum is conserved. D) All of the above. E) None of the above.
Explanation:
There are two types of collision. First one is elastic and other one is inelastic collision.
The linear momentum of two objects before and after the collision remains the same in case of elastic collision. Also, the kinetic energy in elastic collision is conserved. But in case of inelastic collision, the momentum of two objects before and after the collision remains constant but the kinetic energy is in conversed. It changes form one form of energy to another.
Hence, the correct option is (E) "None of the above".
C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)
In an ideal elastic collision, the following happens:
Total Kinetic Energy is Conserved: The total kinetic energy of the system (colliding objects) before the collision is equal to the total kinetic energy after the collision. No energy is lost or gained in terms of kinetic energy.
Here's why the other options are incorrect:
A) Heat and Deformation: While real collisions might involve some energy loss due to heat and deformation, an ideal elastic collision is considered perfectly elastic, meaning no energy is lost in these ways.
B) Potential Energy: The collision doesn't necessarily involve a change in potential energy of the objects.
C) Momentum: Momentum is always conserved in any collision, elastic or inelastic.
D) All of the Above: Only the conservation of total kinetic energy applies to an ideal elastic collision.
E) None of the Above: In an ideal elastic collision, kinetic energy is conserved.
Therefore, the correct answer is:
C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)
Electric Field of a Point Charge Suppose the electric field 0.500 meters from a positive point charge is 1.25 x 105 Newtons per Coulomb. What is the charge? 0 A. 1.67 ?C (-) B. 3.48 ?C ° C. 5.14 ?C ? D. 7.29
Answer:
Charge, q = 3.48 μC
Explanation:
It is given that,
Electric field, [tex]E=1.25\times 10^5\ N/C[/tex]
Distance form positive charge, r = 0.5 meters
We need to find the charge. It is given by the formula as follows :
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]q=\dfrac{Er^2}{k}[/tex]
k = electrostatic constant
[tex]q=\dfrac{1.25\times 10^5\ N/C\times (0.5\ m)^2}{9\times 10^9}[/tex]
q = 0.00000347 C
or
[tex]q=3.48\times 10^{-6}\ C[/tex]
[tex]q=3.48\ \mu C[/tex]
So, the charge is 3.48 μC. Hence, this is the required solution.
A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
Answer:
v = 14.35 m/s
Explanation:
As we know that crate is placed on rough bed
so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate
So here we have
[tex]\mu mg = \frac{mv^2}{R}[/tex]
here we have
[tex]\mu g = \frac{v^2}{R}[/tex]
now we know that
[tex]v = \sqrt{\mu Rg}[/tex]
here we have
[tex]\mu = 0.600[/tex]
R = 35 m
g = 9.81 m/s/s
now plug in all values in above equation
[tex]v = \sqrt{(0.600)(35)(9.81)}[/tex]
[tex]v = 14.35 m/s[/tex]
On a day when the temperature is 8 degrees Celcius, what is the speed of sound in air, in units of m/s.
Answer:
v = 335.7 m/s
Explanation:
As we know that speed of sound in air is given by the formula
[tex]v = \sqrt{\frac{\gamma RT}{M}}[/tex]
now we have
[tex]\gamma = 1.4[/tex] For air
M = 29 g/mol = 0.029 kg/mol
T = 8 degree Celcius = 273 + 8 = 281 K
R = 8.31 J/mol K
now from above formula we have
[tex]v = \sqrt{\frac{(1.4)(8.31)(281)}{0.029}}[/tex]
[tex]v = 335.7 m/s[/tex]