Answer:
The surface charge density on the plate, [tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]
Explanation:
It is given that,
Area of parallel plate capacitor, [tex]A=2\times 10^{-3}\ m^2[/tex]
Separation between the plates, [tex]d=5\times 10^{-2}\ mm[/tex]
Electric field between the plates, [tex]E=8.5\times 10^{6}\ V/m[/tex]
We need to find the surface charge density on the plates. The formula for electric field is given by :
[tex]E=\dfrac{\sigma}{\epsilon}[/tex]
Where
[tex]\sigma[/tex] = surface charge density
[tex]\sigma=E\times \epsilon[/tex]
[tex]\sigma=8.5\times 10^{6}\ V/m\times 8.85\times 10^{-12}\ C^2/Nm^2[/tex]
[tex]\sigma=0.000075\ C/m^2[/tex]
[tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]
Hence, this is the required solution.
Final answer:
The surface charge density on the plates of a parallel-plate air-filled capacitor 7.5225 x [tex]10^{-5}[/tex] [tex]C/m^2[/tex].
Explanation:
To determine the surface charge density on the plates of the parallel-plate air-filled capacitor, we can use the relationship between the electric field (E), the permittivity of free space
[tex](\sigma\))[/tex]. The electric field is defined as
[tex](E = \frac{\sigma}{\varepsilon_0}).[/tex]
Given that the electric field (E) is
[tex](8.5 \times 10^6 V/m)[/tex] and the permittivity of free space [tex]((\varepsilon_0))[/tex] is
[tex](8.85 \times 10^{-12} C^2/N \cdot m^2)[/tex], we can rearrange the formula to solve for the surface charge density [tex]((\sigma)):[/tex]
[tex](\sigma = E \cdot \varepsilon_0 = (8.5 \times 10^6 V/m) \times (8.85 \times 10^{-12} C^2/N \cdot m^2) = 7.5225 \times 10^{-5} C/m^2).[/tex]
Thus, the surface charge density on the plates is
[tex](7.5225 \times 10^{-5} C/m^2).[/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
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
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.
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 cm2 is rotated in 0.040 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 6.0 x 10-5 T. What is the total magnetic flux through the coil before it is rotated?
Final answer:
The total magnetic flux through the coil before rotation is calculated using the formula Φ = B × A × cos(θ), with B being the Earth's magnetic field strength, A the area of the coil, and θ the angle between the field and the normal to the coil, resulting in a flux of 7.2 × 10⁻⁸ Tm².
Explanation:
The total magnetic flux Φ through the coil before it is rotated is given by the formula Φ = B × A × cos(θ), where B is the magnetic field strength, A is the area enclosed by the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil. Before the rotation, the plane of the coil is perpendicular to Earth's magnetic field, making θ = 0° (or cos(θ) = 1), so the cos(θ) term does not affect the calculation.
To calculate the magnetic flux, convert the area from cm² to m² by multiplying by 10⁻⁴, and use the given values:
B = 6.0 × 10⁻⁵ T (Earth's magnetic field)A = 12 cm² × 10⁻⁴ = 1.2 × 10⁻³ m² (area of the coil)Φ = B × A × cos(0°) = 6.0 × 10⁻⁵ T × 1.2 × 10⁻³ m² × 1 = 7.2 × 10⁻⁸ Tm²Thus, the total magnetic flux through the coil before it is rotated is 7.2 × 10⁻⁸ Tm².
. 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)
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.
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]
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.
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]
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 string is wrapped around a pulley of radius 3.60 cm , and a weight hangs from the other end. The weight falls with a constant acceleration 3.00 m/s^2 What's the angular acceleration of the pulley?
Answer:
83.3 rad/s^2
Explanation:
r = 3.6 cm = 0.036 m
a = 3 m/s^2
the relation between the linear acceleration and angular acceleration is given by
a = r α
where, α is angular acceleration, a be the linear acceleration.
α = a / r
α = 3 / 0.036 = 83.3 rad/s^2
Two identical loudspeakers are some distance apart. A person stands 5.20 m from one speaker and 4.10 m from the other. What is the third lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 344 m/s.
Answer:
125.09 Hz
Explanation:
Given : A person stands 5.20 from one speaker and 4.10 m away from the other speaker.
Distance between the speakers is 5.20 - 4.10 =1.10 m
We know that
For destructive interferences
n . λ / 2
where n =1,3,5,7....
Therefore difference between the speakers is
5.10 - 4.20 = 1 X 0.5 λ
λ = 2.2
Given velocity of sound is V = 344 m/s
Therefore frequency, f = [tex]\frac{v}{\lambda }[/tex]
= [tex]\frac{344}{2.2 }[/tex]
= 156.36 Hz
Now, the third lowest frequency is given by
λ = (5.20-4.10) X 5 X 0.5
= 2.75 m
Therefore frequency, f = [tex]\frac{v}{\lambda}[/tex]
= [tex]\frac{344}{2.75}[/tex]
= 125.09 Hz
Therefore third lowest frequency is 125.09 Hz
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
A 1500-kg car accelerates from 0 to 30 m/s in 6.0 s. What is the minimum average power delivered by the engine?
Answer:
Power, P = 112500 watts
Explanation:
It is given that,
Mass of the car, m = 1500 kg
Initial velocity of the car, u = 0
Final velocity of the car, v = 30 m/s
Time taken, t = 6 s
The minimum average power delivered by the engine is given by :
P = W/t
Where
W = work done
t = time taken
Work done = change in kinetic energy
So, [tex]P=\dfrac{\Delta K}{t}=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (30\ m/s)^2}{6\ s}[/tex]
P = 112500 watts
So, the minimum average power delivered by the engine is 112500 watts. Hence, this is the required solution.
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]
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
(a) What net external force is exerted on a 1100.0-kg artillery shell fired from a battleship if the shell is accelerated at 2.40 × 10 4 m/s 2 ? (b) What is the magnitude of the force exerted on the ship by the artillery shell, and why?
Explanation:
It is given that,
Mass of the artillery shell, m = 1100 kg
Acceleration of he shell, [tex]a=2.4\times 10^4\ m/s^2[/tex]
(a) The magnitude of force exerted on the shell can be calculated using Newton's second law of motion as :
F = ma
[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]
F = 26400000 N
or
[tex]F=2.6\times 10^7\ N[/tex]
(b) The magnitude of the force exerted on the ship by the artillery shell will be same but the direction is opposite as per Newton's third law of motion. i.e [tex]2.6\times 10^7\ N[/tex]
The magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.
(a) Net External Force on the Artillery Shell:
We can use Newton's second law of motion (ΣF = ma) to solve this problem. Here, ΣF represents the net external force acting on the object (artillery shell), m is the mass of the shell (1100.0 kg), and a is the acceleration of the shell (2.40 × 10^4 m/s²).
Force (F) = mass (m) × acceleration (a)
F = 1100.0 kg × 2.40 × 10^4 m/s²
F = 2.64 × 10^7 N
Therefore, the net external force exerted on the artillery shell is 2.64 × 10^7 N.
(b) Force on the Ship and Explanation:
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the force exerted on the artillery shell to accelerate it (calculated in part (a)) is equal in magnitude but opposite in direction to the force exerted by the shell on the ship.
Therefore, the magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.
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
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.
Learn more about Vector Addition here:https://brainly.com/question/35874968
#SPJ3
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]
For more information, refer the link given below
https://brainly.com/question/12446100?referrer=searchResults
A length change 0.08 m will occur for an object that is L= 56 m long. If the coefficient of thermal expansion is5.3 x 10 /C and if the original temperature is 253 C, find the increase in temperature.
Answer:
Increase in temperature = 269.54 °C
Explanation:
We have equation for thermal expansion
ΔL = LαΔT
Change in length, ΔL = 0.08 m
Length, L = 56 m
Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1
Change in temperature, ΔT = T - 253
Substituting
0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)
(T - 253) = 269.54
T = 522.54 °C
Increase in temperature = 269.54 °C
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 %.
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) 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
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 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]
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]
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.)
The Large Magellanic Cloud is a small galaxy that orbits the Milky Way. It is currently orbiting the Milky Way at a distance of roughly 160000 light-years from the galactic center at a velocity of about 300 km/s.
1) Use these values in the orbital velocity law to get an estimate of the Milky Way's mass within 160000 light-years from the center. (The value you obtain is a fairly rough estimate because the orbit of the Large Magellanic Cloud is not circular.)
Express your answer using one significant figure.
Answer: [tex]2(10)^{42}kg [/tex]
Explanation:
Approaching the orbit of the Large Magellanic Cloud around the Milky Way to a circular orbit, we can use the equation of velocity in the case of uniform circular motion:
[tex]V=\sqrt{G\frac{M}{r}}[/tex] (1)
Where:
[tex]V=300km/s=3(10)^{5}m/s[/tex] is the velocity of the Large Magellanic Cloud's orbit, which is assumed as constant.
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M[/tex] is the mass of the Milky Way
[tex]r=160000ly=1.51376(10)^{21}m[/tex] is the radius of the orbit, which is the distance from the center of the Milky Way to the Large Magellanic Cloud.
Now, if we want to know the estimated mass of the Milky Way, we have to find [tex]M[/tex] from (1):
[tex]M=\frac{V^{2} r}{G}[/tex] (2)
Substituting the known values:
[tex]M=\frac{(3(10)^{5}m/s)^{2}(1.51376(10)^{21}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (3)
[tex]M=\frac{1.362384(10)^{32}\frac{m^{3}}{s^{2}}}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex]
Finally:
[tex]M=2.0416(10)^{42}kg\approx 2(10)^{42}kg[/tex] >>>This is the estimated mass of the Milky Way
To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.
Explanation:To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The formula for this law is v = sqrt(GM/r), where v is the velocity, G is the gravitational constant, M is the Milky Way's mass, and r is the distance from the center. Rearranging the equation to solve for M, we get M = v^2 * r / G.
Using the given values of v = 300 km/s and r = 160,000 light-years, we need to convert them to appropriate units. 1 light-year is approximately 9.461 × 10^15 meters. After performing the necessary conversions and calculations, the estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.
Learn more about Estimating the Milky Way's mass within a given distance here:https://brainly.com/question/32474051
#SPJ3