Answer:
[tex]K = 0.076 W/m K[/tex]
Explanation:
Heat required to melt the complete ice is given as
[tex]Q = mL[/tex]
here we have
m = 5.0 kg
[tex]L = 3.35 \times 10^5 J/kg[/tex]
now we have
[tex]Q = (5 kg)(3.35 \times 10^5)[/tex]
[tex]Q = 1.67 \times 10^6 J[/tex]
now the power required to melt ice in 8.4 hours is
[tex]P = \frac{Q}{t} = \frac{1.67\times 10^6}{8.4 \times 3600 s}[/tex]
[tex]P = 55.4 Watt[/tex]
now by formula of conduction we know
[tex]P = \frac{KA(\Delta T)}{x}[/tex]
now we have
[tex]55.4 = \frac{K(0.73 m^2)(25 - 5)}{0.02}[/tex]
[tex]K = 0.076 W/m K[/tex]
A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?
Answer:
2 kg
Explanation:
Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.
Sum of the moments at the balance point:
-(Mg)(L/3) + (mg)(L/2 − L/3) = 0
(Mg)(L/3) = (mg)(L/2 − L/3)
(Mg)(L/3) = (mg)(L/6)
2M = m
M = 1 kg, so m = 2 kg.
The mass of the rod is 2 kg.
By equating the torques due to the ball and the rod, we determine the mass of the rod is 6 kg.
To determine the mass of the rod, we need to use the concept of torque balance. Here, we have a 1 kg ball hanging at the end of a 1 m long rod.
The system balances at a point one-third of the distance from the end holding the mass. This means the pivot point is located 1/3 meter from the end with the ball.
The torque due to the ball:
Torque_ball = (1 kg) * (9.8 m/s²) * (1 m)The torque due to the rod:
The rod can be considered as having its mass concentrated at its center of mass, which is 0.5 m from either end. Since the pivot is 1/3 m from the ball end, the rod's center of mass is (0.5 m - 1/3 m) = 1/6 m from the pivot point. Thus, the torque due to the rod is:Torque_rod = (mass of the rod) * (9.8 m/s²) * (1/6 m)Since the system is in balance, the torques must be equal:
Torque_ball = Torque_rod
(1 kg) * (9.8 m/s²) * (1 m) = (mass of the rod) * (9.8 m/s²) * (1/6 m)
Solving for the mass of the rod:
(1 kg) * (9.8 m/s²) * (1 m) = (mass of the rod) * (9.8 m/s²) * (1/6)
mass of the rod = 6 kg
Thus, the mass of the rod is 6 kg.
What determines how much induced current will flow through a conductor?
Answer:
Rate of change of magnetic flux
Explanation:
The induced current is equal to the ratio of induced emf to the resistance of the conductor.
According to the Faraday's law of electromagnetic induction, the induced emf is proportional to the rate of change of magnetic flux.
The magnetic field due to a 2-A current flowing in a long, straight wire is 8 μT at a point P, a certain distance away from the wire. How far is this point from the center of the wire?
Answer:
Distance from the center of wire is 0.05 meters.
Explanation:
It is given that,
Current flowing in the wire, I = 2 A
Magnetic field, [tex]B=8\ \mu T=8\times 10^{-6}\ T[/tex]
Let d is the distance from the center of the wire. The magnetic field at a distance d from the wire is given by :
[tex]B=\dfrac{\mu_oI}{2\pi d}[/tex]
[tex]d=\dfrac{\mu_oI}{2\pi B}[/tex]
[tex]d=\dfrac{4\pi \times 10^{-7}\times 2\ A}{2\pi \times 8\times 10^{-6}\ T}[/tex]
d = 0.05 meters
So, the distance from the wire is 0.05 meters. Hence, this is the required solution.
To find the distance from the wire to point P where the magnetic field is 8 μT due to a 2-A current, the formula B = μ₀I/(2πR) is used, and the calculation reveals that R = 0.1 mm.
To, calculating the distance from the center of the wire to a point P where the magnetic field due to a 2-A current flowing in a long, straight wire is 8 μT.
To find this distance, we use the formula for the magnetic field around a long straight wire, given by B = μ₀I/(2πR), where B is the magnetic field strength, μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A), I is the current, and R is the distance from the wire.
Plugging the given values into this equation, we get:
8 x 10⁻⁶ T = (4π x 10⁻⁷ Tm/A)(2A) / (2πR)
Simplifying, we find that R = (4π x 10⁻⁷ Tm/A)(2A) / (2π x 8 x 10⁻⁶ T) = 0.0001 m or 0.1 mm.
An infinite plane of charge has surface charge density 7.2 μC/m^2. How far apart are the equipotential surfaces whose potentials differ by 100 V?
Answer:
so the distance between two points are
[tex]d = 0.246 \times 10^{-3} m[/tex]
Explanation:
Surface charge density of the charged plane is given as
[tex]\sigma = 7.2 \mu C/m^2[/tex]
now we have electric field due to charged planed is given as
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
now we have
[tex]E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}[/tex]
[tex]E = 4.07 \times 10^5 N/C[/tex]
now for the potential difference of 100 Volts we can have the relation as
[tex]E.d = \Delta V[/tex]
[tex]4.07 \times 10^5 (d) = 100[/tex]
[tex]d = \frac{100}{4.07 \times 10^5}[/tex]
[tex]d = 0.246 \times 10^{-3} m[/tex]
Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a charge of 0.785 µC, and the right-hand sphere carries a charge of 1.47 µC. What is the equilibrium separation between the centers of the two spheres?
Answer:
Explanation:
Answer:
0.632 m
Explanation:
let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.
According to the diagram,
d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)
Let T be the tension in the string.
Resolve the components of T.
T Sinθ = k q1 q2 / d^2
T Sinθ = k q1 q2 / (2LSinθ)² .....(2)
T Cosθ = mg .....(3)
Dividing equation (2) by equation (3), we get
tanθ = k q1 q2 / (4 L² Sin²θ x mg)
tan θ Sin²θ = k q1 q2 / (4 L² m g)
For small value of θ, tan θ = Sin θ
So,
Sin³θ = k q1 q2 / (4 L² m g)
Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)
Sin³θ = 0.2523
Sinθ = 0.632
θ = 39.2 degree
So, the separation between the two charges, d = 2 x L x Sin θ
d = 2 x 0.5 x 0.632 = 0.632 m
Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density of 0.1 lbm/ft3 , exit through a circular cross section with a radius of 1 ft at 1450 ft/s relative to the engine. Find the mass of fuel which is supplied to the engine each minute.
Answer:
1387908 lbm/h
Explanation:
Air flowing into jet engine = 70 lbm/s
ρ = Exhaust gas density = 0.1 lbm/ft³
r = Radius of exit with a circular cross section = 1 ft
v = Exhaust gas velocity = 1450 ft/s
Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel
Q = (70+x) lbm/s
Area of exit with a circular cross section = π×r² = π×1²= π m²
Now from energy balance
Q = ρ×A×v
⇒70+x = 0.1×π×1450
⇒70+x = 455.53
⇒ x = 455.53-70
⇒ x = 385.53 lbm/s
∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h
To find the mass of fuel supplied to the engine per minute, calculate the exhaust gas volume flow rate, then the mass flow rate using gas density, and subtract the air mass flow rate from it. Convert the resulting fuel mass flow rate from seconds to minutes by multiplying by 60.
Explanation:The student's question is about calculating the mass flow rate of fuel supplied to the engine of a jet, using the conservation of mass and given data about the exhaust gases. The density of the exhaust gas, the velocity of the exhaust, and the cross-sectional area of the outlet are provided.
First, calculate the volume flow rate of exhaust gases by multiplying the area of the outlet by the velocity of the exhaust. Since the radius is given, the area (A) can be calculated with the formula A = πr^2, where r is the radius. Then, the volume flow rate (V) is found with the formula V = A × exhaust velocity.
Next, multiply the volume flow rate by the exhaust gas density to find the mass flow rate of the exhaust gas. Finally, the difference between the mass flow rate of the incoming air and the mass flow rate of the exhaust gas will give the mass flow rate of the fuel injected into the engine. Since the question asks for the mass of fuel per minute, this rate should be converted from seconds to minutes by multiplying by 60.
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point located 6.00 cm from the axis of rotation of the centrifuge with a speed of 99 m/s when the centrifuge is at full speed moves (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up? (b) How many revolutions does the centrifuge make as it goes from rest to its final angular speed?
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
.A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.
Final answer:
A system can sustain a nonzero velocity with zero net external force in accordance with Newton's first law of motion, such as a hockey puck moving at a constant velocity on an ice surface without friction.
Explanation:
A system can definitely have a nonzero velocity even when the net external force acting upon it is zero. This situation aligns with Newton's first law of motion, also known as the law of inertia, which states that an object will maintain its state of rest or uniform motion unless acted upon by a net external force.
An example of this scenario can be seen when an object is moving at a constant velocity on a frictionless surface, where no external forces are acting to accelerate or decelerate it. In real-world terms, consider a hockey puck gliding across a smooth ice surface with no additional forces applied to it; it will continue to move at the same speed and in the same direction.
A cylindrical specimen of some metal alloy having an elastic modulus of 102 GPa and an original cross-sectional diameter of 3.8 mm will experience only elastic deformation when a tensile load of 2440 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.47 mm.
Answer:
[tex]l=222.803mm[/tex]
Explanation:
Given:
Elastic modulus, E = 102 GPa
Diameter, d = 3.8mm = 0.0038 m
Applied tensile load = 2440N
Maximum allowable elongation, = 0.47mm = 0.00047
Now,
The cross-sectional area of the specimen,[tex]A_o=\frac{\pi d^2}{4}[/tex]
substituting the values in the above equation we get
[tex]A_o=\frac{\pi 0.0038^2}{4}[/tex]
or
[tex]A_o=1.134\times 10^{-5}[/tex]
now
the stress (σ) is given as:
[tex]\sigma=\frac{Force}{Area}[/tex]
and[tex]E=\frac{\sigma}{\epsilon}[/tex]
where,
[tex]\epsilon =\ Strain[/tex]
also,
[tex]\epsilon=\frac{\Delta l}{l}[/tex]
where,
[tex]l=initial \ length[/tex]
thus,
[tex]E=\frac{\frac{F}{A_o}}{\frac{\Delta l}{l}}[/tex]
or on rearranging we get,
[tex]l=\frac{E\times \Delta l\times A}{F}[/tex]
substituting the values in the above equation we get
[tex]l=\frac{102\times 10^9\times 0.00047\times 1.134\times 10^{-5}}{2440}[/tex]
or
[tex]l=0.222803m[/tex]
or
[tex]l=222.803mm[/tex]
A ball of mass 2 kg fallsfrom a rof that is 5 m above the ground. How much gravitational potential energy did the ball lose when it reached the ground, in units of joules? give your answer as a positive value
Answer:
98.1 J
Explanation:
GPE = mass × g × height
= 2 × 9.81 × 5
= 98.1 J
Charge 1 of +5 micro-coulombs is placed at the origin, charge 2 of +24 micro-coulombs is placed at x = +0.23 m, y = -0.69 m, charge 3 of -5 micro-coulombs is placed at x = -0.27 m, y = 0 m. What is the magnitude of the total electric force on charge 1 in Newtons?
Answer:
[tex]F_{net} = 4.22 N[/tex]
Explanation:
Since charge 1 and charge 2 are positive in nature so here we will have repulsion type of force between them
It is given as
[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]
[tex]F_{12} = \frac{(9\times 10^9)(5 \mu C)(24 \mu C)}{0.23^2 + 0.69^2}\frac{-0.23\hat i + 0.69 \hat j}{\sqrt{0.23^2 + 0.69^2}}[/tex]
[tex]F_{12} = 2.81(-0.23\hat i + 0.69\hat j)[/tex]
Since charge three is a negative charge so the force between charge 1 and charge 3 is attraction type of force
[tex]F_{13} = \frac{(9\times 10^9)(5 \mu C)(5 \mu C)}{0.27^2 + 0^2} (-\hat i)[/tex]
[tex]F_{13} = 3.1(- \hat i)[/tex]
Now we will have net force on charge 1 as
[tex]F_{net} = F_{12} + F_{13}[/tex]
[tex]F_{net} = (-0.65 \hat i + 1.94 \hat j) + (-3.1 \hat i)[/tex]
[tex]F_{net} = (-3.75 \hat i + 1.94 \hat j)[/tex]
now magnitude of total force on the charge is given as
[tex]F_{net} = 4.22 N[/tex]
Be sure to answer all parts. Calculate the mass of each of the following: (a) a sphere of gold with a radius of 12.5 cm. (The volume of a sphere with a radius r is V = (4/3)πr3; the density of gold is 19.3 g/cm3.) × 10 gEnter your answer in scientific notation. (b) a cube of platinum of edge length 0.049 mm (density = 21.4 g/cm3). × 10 gEnter your answer in scientific notation. (c) 67.1 mL of ethanol (density = 0.798 g/mL). g
The mass of a sphere of gold with a radius of 12.5 cm is 1.5789752 x 10^5 g in scientific notation. A cube of platinum with an edge length of 0.049 mm has a mass of 2.517269 x 10^-3 g. The mass of 67.1 mL of ethanol with a density of 0.798 g/mL is 53.5458 g.
Explanation:To calculate the mass of a sphere of gold with a radius of 12.5 cm using the density formula (density = mass/volume), first we need to find the volume of the sphere using the formula V = (4/3)πr^3. Then we can multiply the volume by the density of gold, which is 19.3 g/cm^3.
(a) Volume of gold sphere = (4/3)π(12.5 cm)^3 = 8,181.229 cm^3
Mass of gold sphere = volume × density = 8,181.229 cm^3 × 19.3 g/cm^3 = 157,897.52 g
The mass in scientific notation is 1.5789752 × 105 g.
(b) Volume of platinum cube = edge^3 = (0.049 cm)^3 = 1.17649 × 10^-4 cm^3
Mass of platinum cube = volume × density
= 1.17649 × 10^-4 cm^3 × 21.4 g/cm^3 = 2.517269 × 10^-3 g
(c) Mass of ethanol = volume × density
= 67.1 mL × 0.798 g/mL = 53.5458 g
What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
The weight of an 8 kg substance can be calculated in various units using the weight equation w = mg and the appropriate conversion factors. The weight is 78.4 N, 0.0784 kN, 78.4 kg·m/s², 8 kgf, 17.64 lbf, and 10.83 lbm·ft/s².
Explanation:To calculate the weight of an object in different units, we need to use the equation for weight: w = mg, where m is the mass of the object and g is the acceleration due to gravity. In this case, the mass (m) of the substance is given as 8 kg, and the value of g on Earth is approximately 9.80 m/s².
Therefore, the weight of the substance in various units is:
Newtons (N): w = mg = (8 kg)(9.80 m/s²) = 78.4 N.KiloNewtons (kN): 78.4 N = 0.0784 kN (as 1kN = 1000 N).kg·m/s²: This is just another term for Newton, so the weight is 78.4 kg·m/s².Kilogram-force (kgf): Here, 1 kgf equals the gravitational force exerted on a 1 kg mass, so 8 kgf.Pound-force (lbf): Since 1 N = 0.225 lbf, the weight in lbf is 78.4 N * 0.225 lb/N = 17.64 lbf.Pound-mass feet per second squared (lbm·ft/s²): We can use the conversion factor 1 lbm·ft/s² = 0.13825 N, so the weight is 78.4 N * 0.13825 lbm·ft/s²/N = 10.83 lbm·ft/s².Learn more about Weight Conversion here:https://brainly.com/question/11429990
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A counterclockwise current runs through a square wire loop in the xy plane centered at the origin. The length of each side is d. What is the magnetic field on the z axis?
Answer:
[tex]B_{net} = \frac{2\sqrt2 \mu_0 i}{\pi d}[/tex]
Explanation:
Magnetic field due to straight current carrying wire is given by the formula
[tex]B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)[/tex]
now we will have for one side of the square at its center position given as
[tex]B = \frac{\mu_0 i}{4\pi (\frac{d}{2})}(sin45 + sin45)[/tex]
[tex]B = \frac{2\sqrt2 \mu_0 i}{4 \pi d}[/tex]
now for the we have for complete square loop it will become 4 times of the one side
[tex]B_{net} = 4 B[/tex]
[tex]B_{net} = 4 \frac{2\sqrt2 \mu_0 i}{4 \pi d}[/tex]
[tex]B_{net} = \frac{2\sqrt2 \mu_0 i}{\pi d}[/tex]
In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.
Answer:
9.74 x 10^7 m/s
Explanation:
V = 27000 V
energy of electrons = e x V
K = 1.6 x 10^-19 x 27000 = 43200 x 10^-19 J
Energy = 1/2 m v^2
43200 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v^2 = 9.495 x 10^15
v = 9.74 x 10^7 m/s
One piece of copper jewelry at 111°C has exactly twice the mass of another piece, which is at 28°C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter (c of copper = 0.387 J/g·K)?
Answer:
83.33 C
Explanation:
T1 = 111 C, m1 = 2m
T2 = 28 C, m2 = m
c = 0.387 J/gK
Let the final temperature inside the calorimeter of T.
Use the principle of calorimetery
heat lost by hot body = heat gained by cold body
m1 x c x (T1 - T) = m2 x c x (T - T2)
2m x c X (111 - T) = m x c x (T - 28)
2 (111 - T) = (T - 28)
222 - 2T = T - 28
3T = 250
T = 83.33 C
Thus, the final temperature inside calorimeter is 83.33 C.
A particular wire has a resistivity of 6.47×10-8 Ωm and a cross-sectional area of 2.32 mm2. A length of this wire is to be used as a resistor that will develop 130 W of power when connected to a 9.00 V battery. What length of wire is required?
Using the formula for power:
P = V^2 / R
130 W = (9.00 V)^2 / R
Solve for r:
R = 81/130
R = 0.623 ohms
Now solve for the length of wire:
R = rho L / A
A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2
Now you have:
0.623 = (6.47x10^-8) L / (2.32x10^-6)
L = 22.34 m (Round answer as needed.)
Answer:
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An object is oscillating on a spring with a period of 4.60 s. At time t = 0.00 s the object has zero speed and is at x = 8.30 cm. What is the acceleration of the object at t = 2.50 s?
The acceleration of the object at t = 2.50 s in simple harmonic motion can be found using the equation a = -ω²x, where ω is the angular frequency and x is the displacement from the equilibrium position.
Explanation:The acceleration of the object at t = 2.50 s can be found using the equation for simple harmonic motion:
a = -ω²x
where ω is the angular frequency and x is the displacement from the equilibrium position.
The period of the oscillation is related to the angular frequency by the equation:
T = 2π/ω
Substituting the given period (T = 4.60 s) into the equation and solving for ω, we get:
ω = 2π/T = 2π/4.60 s
Now, substituting the values we have, ω = 2π/4.60 s and x = 8.30 cm, into the acceleration equation:
a = -ω²x = -(2π/4.60 s)² * 8.30 cm
Calculate the value of a to find the acceleration of the object at t = 2.50 s using the given equation for acceleration.
Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter. The resistivity of copper is 1.68×10−8Ω⋅m and the resistivity of aluminum is 2.65×10−8Ω⋅m.
Answer: 0.258
Explanation:
The resistance [tex]R[/tex] of a wire is calculated by the following formula:
[tex]R=\rho\frac{l}{s}[/tex] (1)
Where:
[tex]\rho[/tex] is the resistivity of the material the wire is made of. For aluminium is [tex]\rho_{Al}=2.65(10)^{-8}m\Omega[/tex] and for copper is [tex]\rho_{Cu}=1.68(10)^{-8}m\Omega[/tex]
[tex]l[/tex] is the length of the wire, which in the case of aluminium is [tex]l_{Al}=12m[/tex], and in the case of copper is [tex]l_{Cu}=30m[/tex]
[tex]s[/tex] is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:
[tex]s=\pi{(\frac{d}{2})}^{2}[/tex] (2) Where [tex]d[/tex] is the diameter of the circumference.
For aluminium wire the diameter is [tex]d_{Al}=2.5mm=0.0025m[/tex] and for copper is [tex]d_{Cu}=1.6mm=0.0016m[/tex]
So, in this problem we have two transversal areas:
For aluminium:
[tex]s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}[/tex]
[tex]s_{Al}=0.000004908m^{2}[/tex] (3)
For copper:
[tex]s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}[/tex]
[tex]s_{Cu}=0.00000201m^{2}[/tex] (4)
Now we have to calculate the resistance for each wire:
Aluminium wire:
[tex]R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}[/tex] (5)
[tex]R_{Al}=0.0647\Omega[/tex] (6) Resistance of aluminium wire
Copper wire:
[tex]R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}[/tex] (6)
[tex]R_{Cu}=0.250\Omega[/tex] (7) Resistance of copper wire
At this point we are able to calculate the ratio of the resistance of both wires:
[tex]Ratio=\frac{R_{Al}}{R_{Cu}}[/tex] (8)
[tex]\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}[/tex] (9)
Finally:
[tex]\frac{R_{Al}}{R_{Cu}}=0.258[/tex] This is the ratio
Two point charges Q1 = +5.70 nC and Q2 = −2.30 nC are separated by 45.0 cm. (a) What is the electric potential at a point midway between the charges? 111.3 Incorrect: Your answer is incorrect. seenKey 136 U
Explanation:
It is given that,
Charge, [tex]Q_1=5.7\ nC=5.7\times 10^{-9}\ C[/tex]
Charge, [tex]Q_2=-2.3\ nC=-2.3\times 10^{-9}\ C[/tex]
Distance between charges, d = 45 cm = 0.45 m
We need to find the electric potential at a point midway between the charges. It is given by :
[tex]V=\dfrac{kQ_1}{r_1}+\dfrac{kQ_2}{r_2}[/tex]
[tex]V=k(\dfrac{Q_1}{r_1}+\dfrac{Q_2}{r_2})[/tex]
Midway between the charges means, r = 22.5 cm = 0.225 m
[tex]V=9\times 10^9\times (\dfrac{5.7\times 10^{-9}\ C}{0.225\ m}+\dfrac{-2.3\times 10^{-9}\ C}{0.225\ m})[/tex]
[tex]V=136\ volts[/tex]
So, the electric potential at a point midway between the charges is 136 volts. Hence, this is the required solution.
Final answer:
The electric field strength at the specified point is calculated by finding the vector sum of the electric fields due to each charge, using Coulomb's law and the superposition principle.
Explanation:
The electric field strength at a point due to a single point charge can be calculated using Coulomb's law. The electric field E is the force F per unit charge q, and it acts in the direction away from a positive charge or towards a negative charge. For two point charges, Q₁ and Q₂, each will create an electric field at the point of interest, and the net electric field is the vector sum of these two fields.
To find the electric field at a point that is 20 cm from Q₁ (a positive 3 nC charge) and 50 cm from Q₂ (a negative 4 nC charge), we calculate the electric field from each charge individually and then combine them. The formula for the electric field due to a point charge is E = k * |Q| / r², where k = 8.99 x 10⁹ Nm²/C² is the Coulomb's constant, Q is the charge, and r is the distance from the charge to the point of interest.
Since the charges are opposite in sign and different in magnitude, and the point is closer to Q₁, the electric fields due to each charge will not cancel out, and the net electric field will have a direction based on the vector addition of the individual fields.
Which of the following characterizes the earth’s revolution? a) it takes approximately 24 hours b) it is responsible for creating the day/night relationship c) it determines the timing of seasons and the length of the year d) it is clockwise when viewed from above the North Pole.
Explanation:
The revolutions of the Earth (also called translation movement), consist of the elliptical orbit that describes the Earth around the Sun.
In this sense, a complete revolution around the Sun occurs every 365 days, 5 hours, 48 minutes and about 46 seconds. It is thanks to this movement and that the Earth's axis is tilted with respect to the plane of its orbit about [tex]23\º[/tex], that the four seasons of the year exist.
For this reason, some regions receive different amounts of sunlight according to the seasons of the year. These variations are more evident near the poles and softer or imperceptible in the tropics (near the equator). Because near the equator the temperature tends to be more stable, with only two seasons: rain and drought.
The Earth's revolution around the Sun, taking approximately 365.24 days, determines the timing of seasons and the length of the year, and when viewed from the North Pole, the revolution is counterclockwise.
The revolution of the Earth around the Sun characterizes the Earth's journey through space as it orbits the Sun. This movement takes approximately 365.24 days, which equals one year.
The revolution is responsible for the timing of seasons and the length of the year. When observed from above the North Pole, the Earth's revolution around the Sun occurs counterclockwise, which is different from the rotation of the Earth on its own axis, the latter causing day and night cycles. Therefore, the correct answer to the student's question about what characterizes the Earth's revolution is (c) it determines the timing of seasons and the length of the year.
A proton is moving at 105 m/s at a point where the potential is 10 V. Later, it is at a place where the potential is 5 V. What is its speed there, assuming energy is conserved?
Answer:
The speed is [tex]7.07\times10^{4}\ m/s[/tex]
Explanation:
Given that,
Speed of proton [tex]v= 10^{5}\ m/s[/tex]
Final potential = 10 v
Initial potential = 5 V
We need to calculate the speed
Using formula of energy
[tex]\dfrac{1}{2}mv^2=eV[/tex]
[tex]v^2=\dfrac{2eV}{m}[/tex]
The speed of the particle is directly proportional to the potential.
[tex]v^2\propto V[/tex]
Put the value into the formula
[tex](10^{5})\propto 10[/tex]....(I)
For 5 V,
[tex]v^2\propto 5[/tex].....(II)
From equation (I) and (II)
[tex]\dfrac{(10^{5})^2}{v^2}=\dfrac{10}{5}[/tex]
[tex]v=70710.67\ m/s[/tex]
[tex]v=7.07\times10^{4}\ m/s[/tex]
Hence, The speed is [tex]7.07\times10^{4}\ m/s[/tex]
The speed of the proton in the second place is 74.3 m/s.
To calculate the speed of the proton in the second place, first, we need to find the mass of the proton.
Using,
P.E = mv²/2............ Equation 1Where:
P.E = potential energy of the protonm = mass of the protonv = speed of the proton.Make m the subject of the equation
m = 2P.E/v²............. Equation 2Given:
P.E = 10 Vv = 105 m/sSubstitute these values into equation 2
m = 2×10/(105²)m = 1.81×10⁻³ kg.Finally, to calculate the speed in the second place, we make v the subject of equation 1
v = √(2P.E/m)................. Equation 3Given:
P.E = 5 Vm = 1.81×10⁻³ kgSubstitute these values into equation 3
v = √[(2×5)/(1.81×10⁻³)]v = 74.3 m/sHence, The speed of the proton in second place is 74.3 m/s.
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A certain helium-neon laser pointer, emitting light with a wavelength of 632 nm, has a beam with an intensity of 825 W/m2 and a diameter of 2.80 mm. How many photons are emitted by the laser pointer every second?
Answer:
[tex]N = 1.62 \times 10^{15}[/tex]
Explanation:
Energy of each photon is given as
[tex]E = \frac{hc}{\lambda}[/tex]
here we will have
[tex]\lambda = 632 nm[/tex]
now we will have
[tex]E = \frac{(6.626\times 10^{-34})(3 \times 10^8)}{632 \times 10^{-9}}[/tex]
[tex]E = 3.14 \times 10^{-19} J[/tex]
now let say there is N number of photons per second
so power due to photons is
[tex]P = N(3.14 \times 10^{-19})[/tex]
now intensity is given as power received per unit area
so we have
[tex]I = \frac{P}{A}[/tex]
[tex]825 = \frac{N(3.14 \times 10^{-19})}{\pi (1.40 \times 10^{-3})^2}[/tex]
[tex]825 = N(5.11 \times 10^{-14})[/tex]
[tex]N = 1.62 \times 10^{15}[/tex]
The drift speed in a copper wire is 4.08 × 10-5 m/s for a typical electron current. Calculate the magnitude of the electric field inside the copper wire. The mobility of mobile electrons in copper is 4.5 × 10-3 (m/s)/(N/C). (Note that though the electric field in the wire is very small, it is adequate to push a sizable electron current through the copper wire.)
Answer:
9.07 x 10^-3 N/C
Explanation:
The mobility of electrons is defined as the ratio of drift velocity to the applied electric field.
Vd = 4.08 x 10^-5 m/s
Mobility = 4.5 x 10^-3 m/s
Let the electric field is E.
Mobility = Vd / E
E = Vd / mobility
E = 4.08 x 10^-5 / (4.5 x 10^-3)
E = 9.07 x 10^-3 N/C
The magnitude of the electric field inside the copper wire is approximately 9.067 — 10-3 N/C.
To calculate the magnitude of the electric field inside the copper wire, we can use the relationship between drift speed (v_d), mobility (μ), and electric field (E):
[tex]\[ v_d = \mu \cdot E \][/tex]
Given the drift speed (v_d) of 4.08 — 10^-5 m/s and the mobility (μ) of 4.5 — 10^-3 (m/s)/(N/C), we can rearrange the equation to solve for the electric field (E):
[tex]\[ E = \frac{v_d}{\mu} \][/tex]
Substituting the given values:
[tex]\[ E = \frac{4.08 \times 10^{-5} \text{ m/s}}{4.5 \times 10^{-3} \text{ (m/s)/(N/C)}} \][/tex]
[tex]\[ E = \frac{4.08}{4.5} \times 10^{-5 + 3} \text{ N/C} \][/tex]
[tex]\[ E = 0.9067 \times 10^{-2} \text{ N/C} \][/tex]
[tex]\[ E \approx 9.067 \times 10^{-3} \text{ N/C} \][/tex]
Therefore, the magnitude of the electric field inside the copper wire is approximately 9.067 — 10^-3 N/C.
High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 42 m/s. Find the speed of the golf ball just after impact. m/s
Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball, [tex]p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s[/tex]
After the collision, final momentum [tex]p_f=0.21\ kg\times 42\ m/s+0.046v[/tex]
Using the conservation of momentum as :
[tex]p_i=p_f[/tex]
[tex]11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v[/tex]
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
A jet turbine rotates at a velocity of 7,500 rpm. Calculate the stress acting on the turbine blades if the turbine disc radius is 70 cm and the cross-sectional area is 15 cm2. Take the length to be 10 cm and the alloy density to be 8.5 g/cm3.
Answer:
stress = 366515913.6 Pa
Explanation:
given data:
density of alloy = 8.5 g/cm^3 = 8500 kg/m^3
length turbine blade = 10 cm = 0.1 m
cross sectional area = 15 cm^2 = 15*10^-4 m^2
disc radius = 70 cm = 0.7 m
angular velocity = 7500 rpm = 7500/60 rotation per sec
we know that
stress = force/ area
force = m*a
where a_{c} is centripetal acceleration =
[tex]a_{c} =r*\omega ^{2}= r*(2*\pi*\omega)^{2}[/tex]
=[tex]0.70*(2*\pi*\frac{7500}{60})^{2}[/tex]
= 431795.19 m/s^2
mass = [tex]\rho* V[/tex]
Volume = area* length = 15*10^{-5} m^3
[tex]mass = m = \rho*V = 8500*15*10^{-5} kg[/tex]
force = m*a_{c}
[tex]=8500*15*10^{-5}*0.70*(2*\pi*\frac{7500}{60})^{2}[/tex]
force = 549773.87 N
stress = force/ area
= [tex]\frac{549773.87}{15*10^{-5}}[/tex]
stress = 366515913.6 Pa
Final answer:
The question requires the calculation of stress on jet turbine blades using physics principles involving centrifugal force and material stress.
Explanation:
The question involves calculating the stress on the turbine blades of a jet engine, which requires a knowledge of physics concepts, particularly mechanics and dynamics. The given data include the turbine's rotational velocity (7,500 rpm), radius of the turbine disc (70 cm), the cross-sectional area of the blades (15 cm2), the length of the blades (10 cm), and the alloy density (8.5 g/cm3). To solve this, one would need to calculate the centrifugal force acting on the blades due to rotation and then divide that force by the cross-sectional area to find the stress. However, the calculation involves steps and concepts not provided in the information above, so the direct calculation cannot be completed without additional physics formulae and explanation.
A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning.
To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.
The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:
F = Eq
F is the electric force, E is the electric field, and q is the charge
Let us adjust the equation for only the horizontal components of the above quantities:
Fx = (Ex)(q)
Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.
Given values:
F = 6.57×10⁻²N
q = 6.80×10³C
Plug in these values and solve for Ex:
6.57x10⁻² = Ex(6.80×10³)
Ex = 9.66×10⁻⁶N/C
Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.
We know the weight of the sphere is given by:
W = mg
W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.
We also know this equation:
F = Eq
Let us adjust for the vertical components:
Fy = (Ey)(q)
Set Fy equal to W and we get:
(Ey)(q) = mg
Given values:
q = 6.80×10³C
m = 0.018kg
g = 9.81m/s²
Plug in the values and solve for Ey:
(Ey)(6.80×10³) = 0.018(9.81)
Ey = 2.60×10⁻⁵N/C
Let's now use the Pythagorean theorem to find the total magnitude of the electric field:
E = [tex]\sqrt{Ex^{2}+Ey^{2}}[/tex]
E = 2.77×10⁻⁵N/C
The direction of the electric field is given by:
θ = tan⁻¹(Ey/Ex)
θ = 20.4° off the horizontal
An irrigation channel has a rectangular cross section of 1.5 ft deep x 11 ft wide on the input side. On the far end of the channel, the channel expands to 6 ft wide while maintaining the same depth. If the water flowing into the channel has a speed of 30 ft/sec, calculate the velocity of the water flow on the far end of the channel.
Answer:
55 ft/s
Explanation:
A₁ = Area of rectangular cross-section at input side = 1.5 x 11 = 16.5 ft²
A₂ = Area of rectangular cross-section at far end = 1.5 x 6 = 9 ft²
v₁ = speed of water at the input side of channel = 30 ft/s
v₂ = speed of water at the input side of channel = ?
Using equation of continuity
A₁ v₁ = A₂ v₂
(16.5) (30) = (9) v₂
v₂ = 55 ft/s
An airplane moves through the air at a constant speed. The engines’ thrust applies a force in the direction of motion, and this force is equal in magnitude and opposite in direction to the drag force. Reducing thrust will cause the plane to fly at a slower—but still constant—speed. Explain why this is so.
Explanation:
This is because the drag force suffered by the aircraft is proportional to the speed at which it travels. The thrust of the engines prints a speed to the plane and this speed prints a drag force, always reaching an equilibrium point of these two forces where the speed of the plane is constant and the acceleration is equal to zero.
Therefore, by reducing the thrust, the drag force is greater and the plane begins to decrease its speed, until it reaches the point where the new drag force is matched with the new thrust force, giving it a new final speed , without acceleration.
An airplane flying at a constant speed maintains a balance between engine thrust and drag force. Reducing thrust lessens this force, leading to a new equilibrium at a lower speed due to the drag force being proportional to the square of the velocity.
Explanation:When an airplane moves through air, its engines provide thrust to counteract drag, which is a force opposing the motion. The drag force increases with the speed of the airplane, following a relationship where the magnitude of the drag force is proportional to the square of its speed. According to this principle, if an airplane reduces its thrust, the drag force eventually balances the reduced thrust at a new, lower equilibrium speed, which allows the plane to continue flying at a new constant speed, but slower.
This balance between thrust and drag is a direct application of Newton's second law of motion, where the net force applied to an object is equal to its mass times its acceleration (F=ma). In the steady flight of an airplane, when thrust equals drag and lift equals weight, the acceleration is zero, resulting in constant velocity flight. Reducing the thrust results in an initial slowing down of the airplane until the drag force decreases to match the new lower thrust, and the plane achieves a new constant (lower) velocity.
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A certain gas occupies a volume of 3.7 L at a pressure of 0.91 atm and a temperature of 283 K. It is compressed adiabatically to a volume of 0.85 L. Determine (a) the final pressure and (b) the final temperature, assuming the gas to be an ideal gas for which γ = 1.4.
Answer:
(a)
P₂ = 7.13 atm
(b)
T₂ = 157.14 K
Explanation:
(a)
V₁ = initial volume = 3.7 L = 3.7 x 10⁻³ m³
V₂ = final volume = 0.85 L = 0.85 x 10⁻³ m³
P₁ = Initial Pressure of the gas = 0.91 atm = 0.91 x 101325 = 92205.75 Pa
P₂ = Final Pressure of the gas = ?
Using the equation
[tex]P_{1} V{_{1}}^{\gamma } = P_{2} V{_{2}}^{\gamma }[/tex]
[tex](92205.75) (3.7\times 10^{-3})^{1.4 } = P_{2} (0.85\times 10^{-3})^{1.4 }[/tex]
[tex]P_{2}[/tex] = 722860 Pa
[tex]P_{2}[/tex] = 7.13 atm
(b)
T₁ = initial temperature =283 K
T₂ = Final temperature = ?
using the equation
[tex]P{_{1}}^{1-\gamma } T{_{1}}^{\gamma } = P{_{2}}^{1-\gamma } T{_{2}}^{\gamma }[/tex]
[tex](92205.75)^{1-1.4 } (283)^{1.4 } = (722860)^{1-1.4 } T{_{2}}^{1.4 }[/tex]
T₂ = 157.14 K