The spring constant (k) can be obtained by employing the principle of conservation of energy. Here, the kinetic energy of the metal sphere at the beginning of the motion equals the potential energy at the maximum stretch of the spring. Substituting the given values into the energy equation, solving for 'k' yields the spring constant.
Explanation:The subject of this question lies within the domain of Physics, specifically the domain of mechanics and dynamics dealing with springs and oscillations. The spring constant (k) can be derived from the principle of conservation of energy. Here, we are ignoring friction and air resistance, meaning that the sum of kinetic energy and potential energy remains constant throughout the motion of the metal sphere.
At the beginning, all the energy is kinetic, and at the maximum stretch, all the energy is potential. This can be represented by the equation 0.5*m*v1^2 = 0.5*k*x2^2. By substituting the given values of m (mass = 0.73 kg), v1 (initial velocity = 7.2 m/s), and x2 (maximum displacement = 0.23 m), we can solve for k (spring constant). Here, the calculation would be as follows: k = m*v1^2/x2^2 = (0.73 kg*(7.2 m/s)^2)/(0.23 m)^2. After performing the required calculations, you can obtain the numerical value of the spring constant.
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In flow over cylinders, why does the drag coefficient suddenly drop when the flow becomes turbulent?
For a body with an aerodynamic profile to reach a low resistance coefficient, the boundary layer around the body must remain attached to its surface for as long as possible. In this way, the wake produced becomes narrow. A high shape resistance results in a wide wake. In this type of bodies, if there is turbulence, the drag coefficient increases because the pressure drag appears.
However, in the case of cylinders, it happens that the separation point of the boundary layer will move towards to the rear of the body, which will reduce the size of the wake and there will reduce the magnitude of the pressure drag.
In fluid dynamics, the sudden drop in drag coefficient when flow over a cylinder becomes turbulent is due to the shift from laminar to turbulent flow. As turbulence increases and 'energizes' the slower boundary layer of fluid on the cylinder, the overall drag is reduced and the drag coefficient decreases.
Explanation:In flow over cylinders, the sudden drop in the drag coefficient when the flow becomes turbulent can be attributed to the shift from laminar to turbulent flow itself. As the speed or Reynolds number (N'R) increases, the type of flow changes, and so does the behavior of the viscous drag exerted on the moving object.
In laminar flow, layers flow without mixing and the viscous drag is proportional to speed. As the Reynolds number enters the turbulent range, the drag begins to increase according to a different rule, becoming proportional to speed squared. This turbulent flow introduces eddies and swirls that mix fluid layers.
However, beyond a point in the turbulent flow regime, the drag coefficient starts to decrease. This is because the turbulence begins to 'energize' the generally slower, boundary layer of fluid that clings to the surface of the cylinder, which reduces the overall strength of the drag created by the flow. Moreover, these energized layers of fluid effectively 'smooth out' the obstructive effect of the cylinder, leading to a sudden decrease in the drag coefficient.
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An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C) is 12.0 cm from the line and moving directly toward the line at 4.10×103 m/s .
a)Calculate the proton's initial kinetic energy. Express your answer with the appropriate units.
b)How close does the proton get to the line of charge? Express your answer with the appropriate units.
Final Answer:
a) The proton's initial kinetic energy is [tex]\(8.66 \times 10^{-16}\)[/tex]J.
b) The proton gets as close as 6.00 cm to the line of charge.
Explanation:
a) In part (a), the initial kinetic energy of the proton can be calculated using the formula [tex]\(KE = \frac{1}{2}mv^2\),[/tex]where [tex]\(m\)[/tex] is the mass of the proton and [tex]\(v\)[/tex] is its velocity.
Substituting the given values, we get [tex]\(KE = \frac{1}{2}(1.67 \times 10^{-27}\, \text{kg})(4.10 \times 10^3\, \text{m/s})^2\),[/tex] resulting in [tex]\(8.66 \times 10^{-16}\) J.[/tex]
b) In part (b), the proton's closest approach can be determined using the formula for electric potential energy [tex](\(PE\))[/tex] and kinetic energy [tex](\(KE\))[/tex] when the proton is momentarily at rest.
At the closest point, all the initial kinetic energy is converted to electric potential energy, so The electric potential energy is given by [tex]\(PE = \frac{k \cdot q_1 \cdot q_2}{r}\),[/tex] where [tex]\(k\)[/tex] is Coulomb's constant, [tex]\(q_1\) and \(q_2\)[/tex] are the charges, and [tex]\(r\)[/tex] is the separation distance. Substituting the known values, [tex]\(q_1 = 1.60 \times 10^{-19}\, \text{C}\), \(q_2\)[/tex] is the charge density multiplied by the length per unit length, and [tex]\(r\)[/tex] is the distance, we can solve for [tex]\(r\),[/tex] resulting in [tex]\(6.00\, \text{cm}\).[/tex]
Final answer:
The initial kinetic energy of the proton is calculated using the formula KE = 1/2 * m * v^2, yielding 1.40x10^-20 J. The question regarding how close the proton gets to the line of charge cannot be completely answered without additional details, such as the electric field strength around the linear charge.
Explanation:
The question involves calculations relating to a proton's motion in the electric field created by a linear charge. This falls under the subject of physics and includes principles of electromagnetism and kinematics, typically taught in college-level physics courses.
a) Calculate the proton's initial kinetic energy
The kinetic energy (KE) of an object moving with velocity v is given by the equation KE = 1/2 * m * v^2, where m is the mass of the object. For a proton with mass 1.67x10^-27 kg moving at 4.10x10^3 m/s, its initial kinetic energy is:
KE = 1/2 * (1.67x10^-27 kg) * (4.10x10^3 m/s)^2 = 1.40x10^-20 J.
b) How close does the proton get to the line of charge?
This part requires the concept of energy conservation and electrostatic force. However, without specifying the potential energy due to the proton's interaction with the linear charge, the question is incomplete. Usually, one would calculate the potential energy at the closest approach and set it equal to the original kinetic energy to solve for the distance. The issue requires more information, such as the electric field strength around the line of charge, to proceed with the calculation.
Suppose the radius of the Earth is given to be 6378.01 km. Express the circumference of the Earth in m with 5 significant figures.
Round to 5 sig figs with trailing zeros --> 40074000
The mathematical description that fits to find the circumference of a circle (Approximation we will make to the earth considering it Uniform) is,
[tex]\phi = 2\pi r[/tex]
Here,
r = Radius
The radius of the earth is 6378.01 km or 6378010m
Replacing we have that the circumference of the Earth is
[tex]\phi = 2\pi (6378010m)[/tex]
[tex]\phi = 40074000 m[/tex]
[tex]\phi = 40074*10^3 m[/tex]
Therefore the circumference of the Earth in m with 5 significant figures is [tex]40074*10^3 m[/tex] and using only trailing zeros the answer will be [tex]40074000m[/tex]
A cart starts at x = +6.0 m and travels towards the origin with a constant speed of 2.0 m/s. What is it the exact cart position (in m) 3.0 seconds later?
Answer:
At the origin (x' = 0 m)
Explanation:
Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.
The formula of speed is given as
S = d/t................. Equation 1
Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.
make d the subject of the equation,
d = St ................. Equation 2
Given: S = 2.0 m/s, t = 3.0 s
Substitute into equation 2
d = 2(3)
d = 6 m.
From the above, the cart covered a distance of 6 m in 3 s.
The exact position of the cart = Initial position-distance covered
x' = x-d............ Equation 3
Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.
Given: x = +6.0 m, d = 6 m.
Substitute into equation 3
x' = +6-6
x' = 0 m.
Hence the cart will be at 0 m (origin) 3 s later
The cart will be at a position of 12.0 m after 3.0 seconds.
Explanation:The cart is initially at a position of +6.0 m and is moving towards the origin with a constant speed of 2.0 m/s. We can use the formula for position to find its exact position after 3.0 seconds.
The formula for position is position = initial position + (velocity × time).
Plugging the values into the formula, we get:
position = 6.0 m + (2.0 m/s × 3.0 s) = 6.0 m + 6.0 m = 12.0 m.
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An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half the circle in 156 s, determine the following. HINT (a) Determine the magnitude of the airplane's displacement during the given time (in m). m (b) Determine the magnitude of the airplane's average velocity during the given time (in m/s). m/s (c) What is the airplane's average speed during the same time interval (in m/s)
The displacement of the airplane is 6760 meters, its average velocity is 43.33 m/s, and its average speed is 135.08 m/s.
Explanation:To solve this problem, we need to apply the principles of physics in kinematics and circular motion. First of all, let's understand that displacement is the shortest distance the airplane covered, which is the diameter of the circle. We multiply the radius by 2 (2*3.38 km) and convert it to meters to give 6760 meters for part (a).
Next, for average velocity, which is displacement over time, we divide 6760 m by 156 s, yielding approximately 43.33 m/s for part (b).
Lastly, for average speed, we need to consider the total distance travelled. In half a circle, this is pi times the diameter. Therefore, the average speed is (3.14 * 6760 m) / 156 = 135.08 m/s for part (c).
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The magnitude of the airplane's displacement is 6760 m. The magnitude of the average velocity is 43.33 m/s, and the average speed is 67.97 m/s.
Let's break down the problem step-by-step to find the required values.
(a) Magnitude of the Airplane's Displacement:
The airplane flies half of the circle, which means the path is a semicircle. The displacement is the straight-line distance between the start and end points of this path, which is the diameter of the circle.
Radius of the circle, [tex]r = 3.38 km = 3380 m[/tex]
Diameter, [tex]d = 2 * r = 2 * 3380 m = 6760 m[/tex]
Therefore, the magnitude of the displacement is 6760 m.
(b) Magnitude of the Average Velocity:
Average velocity is the displacement divided by the time.
Displacement = 6760 m
Time, [tex]t = 156 s[/tex]
Average velocity, [tex]v_{avg} = Displacement / Time = 6760 m / 156 s = 43.33 m/s[/tex]
Therefore, the magnitude of the average velocity is 43.33 m/s.
(c) Airplane's Average Speed:
The average speed is the total distance traveled along the path divided by the time.
The distance traveled in half a circle is the circumference of the semicircle.
[tex]Distance = (\pi * Diameter) / 2 = (\pi * 6760 m) / 2 = 10602.91 m[/tex]
[tex]Average speed = Distance / Time = 10602.91 m / 156 s = 67.97 m/s[/tex]
Therefore, the airplane's average speed is 67.97 m/s.
A water balloon is launched at a speed of (15.0+A) m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located (18.0+B) m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.
Answer:
The question is incomplete, below is the complete question,
"A water balloon is launched at a speed of (15.0+A) m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located (18.0+B) m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures. A=12, B=2"
Answe:
12.1m
Explanation:
Below are the data given
Speed, V=(15+A) = 15+12=27m/s
Angle of projection, ∝=36 degree
Distance from building = 18+B=18+2=20m
Since the motion describe by the object is a projectile motion, and recall that in projectile motion, motion along the horizontal path has zero acceleration and motion along the vertical path is under gravity,
the Velocity along the horizontal path is define as
[tex]V_{x}=Vcos\alpha \\V_{x}=27cos36\\V_{x}=21.8m/s[/tex]
the velocity along the Vertical path is
[tex]V_{y}=Vsin\alpha \\V_{y}=27sin36 \\V_{y}=15.87m/s[/tex]
Since the horizontal distance from the point of projection to the building is 20m, we determine the time it takes to cover this distance using the simple equation of motion
[tex]Velocity=\frac{distance }{time }\\ Time,t=27/21.8\\t=1.24secs[/tex]
The distance traveled along the vertical axis is given as
[tex]y=V_{y}t-\frac{1}{2}gt^{2}\\ t=1.24secs,\\V_{y}=15.87m/s\\g=9.81m/s[/tex]
if we substitute values, we arrive at
[tex]y=15.87*1.24-\frac{1}{2}9.81*1.24^{2}\\y=19.66-7.57\\y=12.118\\y=12.1m[/tex]
Hence the water balloon hit the building at an height of 12.1m
In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let ymax be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height. Ignore air resistance.
The answer details the vertical velocity needed and the horizontal distance required for a basketball player to complete a jump.
Vertical velocity: To rise 0.750 m above the floor, the athlete needs a vertical velocity of 5.43 m/s.
Horizontal distance: The athlete should start his jump 2.27 m away from the basket to reach his maximum height at the same time as he reaches the basket.
The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long, 710 yards wide, and 112 ft high.
What is the cubic volume in feet, convert your result from part a to cubic meters?
Explanation:
Given that,
The dimensions of the largest building in the world is 632 m long, 710 yards wide, and 112 ft high. It basically forms a cuboid. The volume of a cuboidal shape is given by :
Since,
1 meter = 3.28084 feet
632 m = 2073.49 feet
1 yard= 3 feet
710 yards = 2130 feet
V = lbh
[tex]V=2073.49 \ ft\times 2130\ ft\times 112\ ft[/tex]
[tex]V=494651774.4\ ft^3[/tex]
[tex]V=4.94\times 10^8\ ft^3[/tex]
Also,
[tex]V=(4.94\times 10^8\ ft^3)(\dfrac{1\ m}{3.281})^3[/tex]
[tex]V=1.39\times 10^7\ m^3[/tex]
Hence, this is the required solution.
Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of each are measured. Both stones have the same volume, 0.15 cm3. However, stone A has a mass of 0.52 g, and stone B has a mass of 0.42 g. If diamond has a density of 3.5 g/cm3, could either of the stones be real diamonds?
Answer:
stone A is diamond.
Explanation:
given,
Volume of the two stone = 0.15 cm³
Mass of stone A = 0.52 g
Mass of stone B = 0.42 g
Density of the diamond = 3.5 g/cm³
So, to find which stone is gold we have to calculate the density of both the stone.
We know,
[tex]density[/tex][tex]\density = \dfrac{mass}{volume}[/tex]
density of stone A
[tex]\rho_A = \dfrac{0.52}{0.15}[/tex]
[tex]\rho_A = 3.467\ g/cm^3[/tex]
density of stone B.
[tex]\rho_B = \dfrac{0.42}{0.15}[/tex]
[tex]\rho_B = 2.8\ g/cm^3[/tex]
Hence, the density of the stone A is the equal to Diamond then stone A is diamond.
Neither of the stones is a real diamond because their densities, calculated using mass and volume, do not match the density of a real diamond.
Explanation:The determination of whether a stone is a real diamond or a fake can be made by calculating the density of the stone. Density is calculated as the mass of an object divided by its volume. So, for stone A, the density is 0.52 g / 0.15 cm3 = 3.47 g/cm3, and for stone B, the density is 0.42 g / 0.15 cm3 = 2.8 g/cm3. The density of a real diamond is 3.5 g/cm3. Hence, neither stone A nor stone B is a real diamond as their densities are less than the density of a real diamond.
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At what point in the cardiac cycle is pressure in the ventricles the highest (around 120 mm Hg in the left ventricle)?
Answer:
Ventricular systole
Explanation:
This contraction causes an increase in pressure inside the ventricles, being the highest during the entire cardiac cycle. The ejection of blood contained in them takes place. Therefore, blood is prevented from returning to the atria by increasing pressure, which closes the bicuspid and tricuspid valves.
The highest pressure in the ventricles occurs during the ventricular systole phase of the cardiac cycle, when the ventricles are contracting to pump blood out to the body.
Explanation:The pressure in the ventricles is highest during the ventricular systole phase of the cardiac cycle. At this point, the ventricles have filled up with blood and are contracting to pump this blood out into the body. This contraction greatly increases the pressure in the ventricles, leading to a peak pressure of around 120 mm Hg in the left ventricle, depending on the individual.
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If the specimen is loaded until it is stressed to 65 ksi, determine the approximate amount of elastic recovery after it is unloaded. Express your answer as a length. Express your answer to three significant figures and include the appropriate units.
Answer:
ER = 0.008273 in
Explanation:
Given:
- Length of the specimen L = 2 in
- The diameter of specimen D = 0.5 in
- Specimen is loaded until it is stressed = 65 ksi
Find:
- Determine the approximate amount of elastic recovery after it is unloaded.
Solution:
- From diagram we can see the linear part of the curve we can determine the Elastic Modulus E as follows:
E = stress / strain
E = 44 / 0.0028
E = 15714.28 ksi
- Compute the Elastic strain for the loading condition:
strain = loaded stress / E
strain = 65 / 15714.28
strain = 0.0041364
- Compute elastic recovery:
ER = strain*L
ER = 0.0041364*2
ER = 0.008273 in
The approximate amount of elastic recovery after unloading a stressed specimen is zero.
Explanation:To determine the approximate amount of elastic recovery after unloading a stressed specimen, we need to consider the concept of elastic deformation. Elastic deformation refers to the temporary elongation or compression of a material when a stress is applied to it, and it returns to its original shape once the stress is removed.
Since the question does not provide specific information about the material or its elastic modulus, we cannot determine the exact amount of elastic recovery. However, we can generally say that the elastic recovery would be close to the original length of the specimen before it was loaded.
Therefore, we can assume that the approximate amount of elastic recovery would be zero, as the specimen would return to its original length.
A rock is thrown at an angle of 60∘ to the ground. If the rock lands 25m away, what was the initial speed of the rock? (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)
Answer:
[tex]v_0 = 16.82\ m/s[/tex]
Explanation:
given,
angle at which rock is thrown = 60°
rock lands at distance,d = 25 m
initial speed of rock, = ?
In horizontal direction
distance = speed x time
d = v₀ cos 60° t
25 = v₀ cos 60° t............(1)
now,
in vertical direction
displacement in vertical direction is zero
using equation of motion
[tex]s = ut +\dfrac{1}{2}gt^2[/tex]
[tex]0 =v_0 sin 60^0 t - 4.9 t^2[/tex]
[tex]v_o sin 60^0 = 4.9 t[/tex]
[tex]t = \dfrac{v_0 sin 60^0}{4.9}[/tex]
putting the value of t in equation (1)
[tex]25 = v_0 cos 60^0\times \dfrac{v_0 sin 60^0}{4.9}[/tex]
[tex]25 =\dfrac{v_0^2cos 60^0 sin 60^0}{4.9}[/tex]v
[tex]v_0^2 = 282.90[/tex]
[tex]v_0 = 16.82\ m/s[/tex]
Hence, the initial speed of the rock is equal to 16.82 m/s
A baseball was hit and reached its maximum height in 3.00s. Find (a) its initial velocity; (b) the height it reaches.[29.4 m/s; 44.1m]
Answer:
u= 29.43 m/s
h=44.14 m
Explanation:
Given that
t= 3 s
We know that acceleration due to gravity ,g = 9.81 m/s² (Downward)
Initial velocity = u
Final velocity ,v= 0 (At maximum height)
We know v = u +a t
v=final velocity
u=initial velocity
a=Acceleration
Now by putting the values in the above equation
0 = u - 9.81 x 3
u= 29.43 m/s
The maximum height h is given as
v² = u ² - 2 g h
0² = 29.43 ² - 2 x 9.81 x h
[tex]h=\dfrac{29.43^2}{2\times 9.81}\ m[/tex]
h=44.14 m
A 220 g , 23-cm-diameter plastic disk is spun on an axle through its center by an electric motor.What torque must the motor supply to take the disk from 0 to 1800 rpm in 4.6 s ?
The torque required by the motor to spin a 220g, 23-cm-diameter plastic disk from 0 to 1800 rpm in 4.6 seconds, without considering the external forces, is 0.431 Nm.
Explanation:In solving the question,
Torque
is our primary interest. We first need to convert rpm to rad/s since Torque calculations require SI units. The conversion can be done by the formula ω = 2π (frequency), and frequency is simply rpm/60. Hence, 1800 rpm is equivalent to 188.50 rad/s. Now, we use the Kinematics equation ω = ω
0
+ αt to calculate angular acceleration (α), where ω
0
is the initial angular velocity, and it is 0 rad/s in this case as the disk starts from rest, ω is the final angular velocity and is 188.50 rad/s, while t is the time of 4.6 seconds. Solving this gives us α=41 rad/s
2
. The Torque can now be calculated using τ=Iα where I (moment of inertia for a disk) = 0.5*m*r
2
. Substituting the values of m, r and α gives a Torque value of 0.431 Nm.
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A 5.00 liter balloon of gas at 25°C is cooled to 0°C. What is the new volume (liters) of the balloon?
Answer:
4.58 L.
Explanation:
Given that
V₁ = 5 L
T₁ = 25°C = 273 + 25 = 298 K
T₂ = 0°C = 273 K
The final volume = V₂
We know that ,the ideal gas equation
If the pressure of the gas is constant ,then we can say that
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]V_2=V_1\times \dfrac{T_2}{T_1}\\V_2=5\times \dfrac{273}{298}\\V_2=4.58\ L\\[/tex]
The final volume of the balloon will be 4.58 L.
A uniform, solid, 2000.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.10 kgkg point mass placed at the following distances from the center of the sphere: (a) 5.04 mm , and (b) 2.70 mm .
Answer:
[tex]0.0110284391534\ N[/tex]
[tex]0.0653784219002\ N[/tex]
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
[tex]m_1[/tex] = Mass of sphere = 2000 kg
[tex]m_2[/tex] = Mass of other sphere = 2.1 kg
r = Distance between spheres
Force of gravity is given by
[tex]F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(5.04\times 10^{-3})^2}\\\Rightarrow F=0.0110284391534\ N[/tex]
The gravitational force is [tex]0.0110284391534\ N[/tex]
[tex]F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(2.07\times 10^{-3})^2}\\\Rightarrow F=0.0653784219002\ N[/tex]
The gravitational force is [tex]0.0653784219002\ N[/tex]
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
Answer:
A) The expression of the electric field halfway between the plates, if the plates are in the plane Y-Z is:
[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]
B) The expression for the magnitude of the electric field E₂ just in front of the plate two ends is:
[tex]|E_2|=\displaystyle \frac{q}{2LW\varepsilon_0}[/tex]
C) The charge density is:
[tex]\sigma_2=-4.2517\cdot10^{-3}C/m^2[/tex]
Completed question:
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates
B) Input an expression for the magnitude of the electric field, E₂. Just in front of plate two END
C) If plate two has a total charge of q =-1 mC, what is its charge density, σ in C/m2?
Explanation:
A) The expression of the field can be calculated as the sum of the field produced by each plate. Each plate can be modeled as 2 parallel infinite metallic planes. Because this is a capacitor connected by both ends to a battery, the external planes have null charge (the field outside the device has to be null by definition of capacitor). This means than the charge of each plate has to be distributed in the internal faces. Because this es a metallic surface and there is no external field, we can consider a uniform charge distribution (σ=cte). Therefore in this case for each plane:
[tex]\sigma_i=\displaystyle \frac{q_i}{LW}[/tex]
The field of an infinite uniform charged plane is:
[tex]\vec{E_i}=\displaystyle \frac{\sigma_i}{2\varepsilon_0}sgn(x-x_{0i})\vec{x} =\frac{q_i}{2LW\varepsilon_0}sgn(x-x_{0i})\vec{x}[/tex]
In this case, inside the capacitor, if the plate 1 is in the left and the plate 2 is in the right, the field for 0<x<d is:
[tex]\vec{E_1}\displaystyle=\frac{q}{2LW\varepsilon_0}sgn(x})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]
[tex]\vec{E_2}\displaystyle=\frac{-q}{2LW\varepsilon_0}sgn(x-d})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]
B) we already obtain the expression of the field E₂ inside the space between the plates. Even if we are asked the expression just in front of the plate and not inside, the expression for |E₂| is still de same.
C) As seen above, we already obtain the charge density expression. Therefore we only have to replace the variables for the numerical values.
The electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.
Given:
Length, L = 0.49 m
Width, W = 0.48 m
Distance, d = 0.1 m
Here:
E = electric field
σ = surface charge density
ε₀ = permittivity of free space
We must determine the surface charge density on each plate since the plates are wired to a battery and charged so that the first plate has a charge of q.
The area of the plate is:
Area of each plate (A) = L x W = 0.49 m x 0.48 m = 0.2352 m²
The surface charge density is given by:
σ = q / A
The electric field is computed as:
E = (σ / ε₀)
Hence, the electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.
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#Complete question is:
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates
If a dog has a mass of 20.1 kg, what is its mass in the following units? Use scientific notation in all of your answers.
Answer:
dog's mass in grams is [tex]20.1\times 10^3 grams[/tex]
dog's mass in milligrams is [tex]20.1\times 10^6 milligrams[/tex]
dog's mass in micrograms is[tex]20.1\times 10^9 micrograms[/tex]
Explanation:
dog has a mass of m= 20.1 kg
dog's mass in grams is given by [tex]20.1\times 1000 grams=20100 gms =20.1\times 10^3 grams[/tex]
dog's mass in milligrams is given by [tex]20.1\times 10^6 milli grams=20100000 milligrams= 20.1\times 10^6 milligrams[/tex]
dog's mass in micrograms is given by
[tex]20.1\times 10^9 micro grams=20100000000 micrograms= 20.1\times 10^9 micrigrams[/tex]
One end of a string 5.02 m long is moved up and down with simple harmonic motion at a frequency of 61 Hz . The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm
Answer:
Explanation:
One end of a string 5.02 m long is moved up and dowBDBBn with simple harmonic motion at a frequency of 61 Hz . The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm
The wavelength of the waves on the string is calculated using the wave speed (found by distance/time) and the frequency. After performing the calculations, the wavelength is determined to be 16.46 cm.
To calculate the wavelength of the waves on the string, we can use the wave speed and the frequency. The speed of a wave ( is given by the formula speed = distance / time. Here, the distance is the length of the string, and the time is how long it takes for waves to reach the other end.
The speed of the wave is therefore 5.02 m / 0.5 s = 10.04 m/s. The frequency of the wave is given as 61 Hz. The wavelength ( can be found using the equation v = f, where v is the wave speed, f is the frequency, and is the wavelength.
By rearranging the equation to solve for the wavelength, we get
= v / f. Substituting the given values, we find
= 10.04 m/s / 61 Hz
= 0.1646 m. To convert the wavelength into centimeters, we multiply by 100, which gives us a wavelength of 16.46 cm.
A proton accelerates from rest in a uniform electric field of 680 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton.
Answer:
Acceleration, [tex]a=6.51\times 10^{10}\ m/s^2[/tex]
Explanation:
Given that,
Electric field, E = 680 N/C
Speed of the proton, v = 1.3 Mm/s
We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :
[tex]qE=ma[/tex]
a and m are the acceleration and mass of the proton.
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}[/tex]
[tex]a=6.51\times 10^{10}\ m/s^2[/tex]
So, the acceleration of the proton is [tex]a=6.51\times 10^{10}\ m/s^2[/tex]. Hence, this is the required solution.
A 35.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.42 105 N · m2/C. What is the magnitude of the electric field? MN/C
To solve this problem we will apply the concept of Electric Flow, which is understood as the product between the Area and the electric field. For the data defined by the area, we will use the geometric measurement of the area in a circle (By the characteristics of the object) This area will be equivalent to,
[tex]\phi = 35 cm[/tex]
[tex]r = 17.5 cm = 0.175 m[/tex]
[tex]A = \pi r^2 = \pi (0.175)^2 = 0.09621m^2[/tex]
Applying the concept of electric flow we have to
[tex]\Phi = EA[/tex]
Replacing,
[tex]5.42*10^5N \cdot m^2/C = E (0.09621m^2)[/tex]
[tex]E = 5.6335*10^6N/C[/tex]
Therefore the magnitude of the electric field is [tex]5.6335*10^6N/C[/tex]
A 0.800kg block is attached to a spring with spring constant 16.0N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 34.0cm/s . What areA)The amplitude of the subsequent oscillations?B)The block's speed at the point where x= 0.250 A?
Answer:
(a) Amplitude=0.0760 m
(b) Speed=0.337 m/s
Explanation:
(a) For amplitude
We can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
[tex]k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)mv^{2}+0=0+(1/2)kA^{2}\\ A^{2}=\frac{mv^{2}}{k} \\A=\sqrt{\frac{mv^{2}}{k}}\\ A=\sqrt{\frac{m}{k} }v\\ A=\sqrt{\frac{(0.800kg)}{16N/m} }(0.34m/s)\\A=0.0760m[/tex]
(b) For Speed
Again we can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
[tex]k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)m(v_{i})^{2}+0=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\ (1/2)m(v_{i})^{2}=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\(1/2)m(v_{i})^{2}-(1/2)k(A/2)^{2}=(1/2)m(v_{f} )^{2}\\(1/2)[m(v_{i})^{2}-k(A/2)^{2}]=(1/2)m(v_{f} )^{2}\\(v_{f} )^{2}=1/m[m(v_{i})^{2}-k(A/2)^{2}]\\As\\x=0.250A\\(v_{f} )^{2}=(1/0.800kg)[0.800kg(0.34m/s)^{2}-(16N/m)(0.250(0.07602m)/2)^{2}\\(v_{f} )^{2}=0.1138\\ v_{f}=\sqrt{0.1138}\\ v_{f}=0.337m/s[/tex]
Final answer:
The amplitude and speed of the block at a specified position in SHM can be determined by using conservation of energy, equating the initial kinetic energy to the maximum potential energy at the amplitude, and calculating the speed via energy values at a given displacement from equilibrium.
Explanation:
Let's break down the problem step by step:
1. Amplitude of Subsequent Oscillations (A):
- When the block is hit with the hammer, it acquires an initial velocity of [tex]\(34.0 \, \text{cm/s}\)[/tex], which we'll convert to meters per second: [tex]\(v = 34.0 \, \text{cm/s} = 0.34 \, \text{m/s}\)[/tex].
- The mechanical energy of the system (block + spring) is conserved. At the maximum extension (amplitude) of the oscillation, the kinetic energy is zero.
- Therefore, the total mechanical energy at the maximum extension is equal to the potential energy stored in the spring:
[tex]\[ E = U = \frac{1}{2} k A^2 \][/tex]
- We can express the kinetic energy at the initial point as:
[tex]\[ K = \frac{1}{2} m v^2 \][/tex]
- Since the total mechanical energy is conserved, we have:
\[ E = K + U \]
[tex]\[ \frac{1}{2} k A^2 = \frac{1}{2} m v^2 \][/tex]
- Solving for the amplitude \(A\):
[tex]\[ A = \sqrt{\frac{m v^2}{k}} \][/tex]
Substituting the given values:
[tex]\[ A = \sqrt{\frac{0.800 \, \text{kg} \cdot (0.34 \, \text{m/s})^2}{16.0 \, \text{N/m}}} \][/tex]
Calculating:
[tex]\[ A \approx 0.34 \, \text{m} \][/tex]
Therefore, the amplitude of the subsequent oscillations is approximately 0.34 meters.
2. Block's Speed at [tex]\(x = 0.250A\)[/tex]:
- At any position \(x\), the mechanical energy \(E\) of the system is given by:
[tex]\[ E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 \][/tex]
- At the maximum extension (amplitude), the kinetic energy is zero, so:
[tex]\[ E = U(x = A) = \frac{1}{2} k A^2 \][/tex]
- We can find the speed of the block at any position \(x\) using the amplitude \(A\):
[tex]\[ v = \sqrt{\frac{k}{m} (A^2 - x^2)} \][/tex]
Substituting the given value [tex]\(x = 0.250A\)[/tex]:
[tex]\[ v = \sqrt{\frac{16.0 \, \text{N/m}}{0.800 \, \text{kg}} \left(0.34^2 - (0.250 \cdot 0.34)^2\right)} \][/tex]
Calculating:
[tex]\[ v \approx 0.24 \, \text{m/s} \][/tex]
Therefore, the block's speed at the point where [tex]\(x = 0.250A\)[/tex] is approximately 0.24 meters per second
You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.09 m/s . Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.85 m/s . It takes you 14.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running?
Answer:
25.52 m
Explanation:
Relative speed between the person and I would be the difference of our speeds
[tex]v_r=2.85-1.09=1.76\ m/s[/tex]
Time taken by me to walk up to the person = 14.5 s
Distance is given by
[tex]Distance=Speed\times Time\\\Rightarrow s=vt\\\Rightarrow s=1.76\times 14.5\\\Rightarrow s=25.52\ m[/tex]
The person was 25.52 m ahead of me when I started running
The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?
Answer:
The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.
Explanation:
Given that,
Boiling point = 81.0°C
Atmospheric pressure :
Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.
The value of atmospheric pressure is
[tex]P=101325\ Pa[/tex]
Vapor pressure :
Vapor pressure is equal to the atmospheric pressure.
Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.
The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is assumed constant at 0.0075 lb/ft3 , calculate the elevation of the mountain top.
To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.
For mercury, density, gravity and height are defined as
[tex]\rho_m = 846lb/ft^3[/tex]
[tex]g = 32.17405ft/s^2[/tex]
[tex]h_1 = 1in = \frac{1}{12} ft[/tex]
For the air the defined properties would be
[tex]\rho_a = 0.0075lb/ft^3[/tex]
[tex]g = 32.17405ft/s^2[/tex]
[tex]h_2 = ?[/tex]
We have for equilibrium that
[tex]\text{Pressure change in Air}=\text{Pressure change in Mercury}[/tex]
[tex]\rho_m g h_1 = \rho_a g h_2[/tex]
Replacing,
[tex](846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)[/tex]
Rearranging to find [tex]h_2[/tex]
[tex]h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}[/tex]
[tex]h = 9400ft[/tex]
Therefore the elevation of the mountain top is 9400ft
The elevation of the mountain top is approximately 13,972 feet.
Explanation:The difference in barometric pressure between the sea level and the top of the mountain represents the hydrostatic pressure exerted by the column of air. We can calculate the elevation of the mountain top by using the equation:
ΔP = ρgh
Where ΔP is the difference in pressure, ρ is the density of the air, g is the acceleration due to gravity, and h is the elevation. Rearranging the equation, we have:
h = ΔP / (ρg)
Substituting the given values, we have:
h = (30 - 29) in / (0.0075 lb/ft³ * 32.2 ft/sec²)
Simplifying the equation, we get:
h ≈ 13,972 ft
Therefore, the elevation of the mountain top is approximately 13,972 feet.
A support wire is attached to a recently transplanted tree to be sure that it stays vertical. The wire is attached to the tree at a point 1.50 m from the ground and the wire is 2.00 m long. What is the angle between the tree and the support wire?
Answer:
Explanation:
Given
Wire attached to the tree at a point [tex]h=1.5\ m[/tex] from ground
Length of wire [tex]L=2\ m[/tex]
From diagram,
Using trigonometry
[tex]\sin \theta =\frac{Perpendicular}{Hypotenuse}[/tex]
[tex]\sin \theta =\frac{1.5}{2}[/tex]
[tex]\theta =48.59[/tex]
Angle between Tree and support[tex]=90-48.59=41.41^{\circ}[/tex]
To find the angle between the tree and the support wire, we can use trigonometry. Given that the wire is 2.00 m long and attached to the tree at a point 1.50 m from the ground, the angle between the tree and the support wire is 41.1 degrees.
Explanation:To find the angle between the tree and the support wire, we can use trigonometry. The wire and the ground form a right triangle, with the wire as the hypotenuse and the vertical distance from the ground to the point of attachment as the opposite side. Using the Pythagorean theorem, we can find the length of the base of the triangle, which is the distance between the point of attachment and the tree.
Given that the wire is 2.00 m long and attached to the tree at a point 1.50 m from the ground, we can calculate the length of the base using the Pythagorean theorem: square root of (2.00^2 - 1.50^2) = 1.30 m.
Now we can use the trigonometric function tangent to find the angle between the tree and the support wire: tangent(angle) = opposite/adjacent, where the opposite side is 1.30 m and the adjacent side is 1.50 m. Solving for the angle, we get: angle = arctan(1.30/1.50) = 41.1 degrees (rounded to one decimal place).
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A person walks in the following pattern: 2.9 km north, then 2.8 km west, and finally 4.4 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point
Answer:
(a) Magnitude =3.18 km
(a) Angle =28.2°
Explanation:
(a) To find magnitude of this vector recognize
[tex]R_{x}=-2.8 km\\R_{y}=-1.5km[/tex]
Use Pythagorean theorem
[tex]R=\sqrt{(R_{x})^{2}+(R_{y})^{2} }\\ R=\sqrt{(-2.8)^{2}+(-1.5)^{2} }\\ R=3.18km[/tex]
(b)To find the angle use the trigonometric property
[tex]tan\alpha =\frac{opp}{adj}\\\ tan\alpha =\frac{R_{y} }{R_{x}}\\\alpha =tan^{-1}(\frac{(-1.5)}{(-2.8)})\\\alpha =28.2^{o}[/tex]
A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm. Suppose the conductor initially has a net charge of +3.8 μC instead of being neutral. What is the total charge (a) on the interior and (b) on the exterior surface?
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
[tex]Q_{int}=-Q1=-1.9*10^{-6} C[/tex]. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
[tex]Q_{ext}=+Q_1=1.9*10^{-6} C[/tex]
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
[tex]Q_{ext, Total}=Q_2+Q_{ext}[/tex]
[tex]Q_{ext, Total}=1.9+3.8[/tex]
[tex]Q_{ext, Total}=5.7 \mu C[/tex]
(a) The total charge on the interior of the spherical conductor is -1.9 μC.
(b) The total exterior charge of the spherical conductor is 5.7 μC.
The given parameters;
charge at the center of the hollow sphere, q = 1.9 μC inner radius of the spherical conductor, r₁ = 3.8 cmouter radius of the spherical conductor, r₂ = 5.6 cmThe total charge on the interior is calculated as follows;
[tex]Q_{int} = - 1.9 \ \mu C[/tex]
The total exterior charge is calculated as follows;
[tex]Q_{tot . \ ext} = Q + Q_2\\\\Q_{tot . \ ext} = 1.9 \ \mu C \ + \ 3.8 \ \mu C\\\\Q_{tot . \ ext} = 5.7 \ \mu C[/tex]
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The oscilloscope is set to measure 2 volts per division on the vertical scale. The oscilloscope display a sinusoidal voltage that vertically covers 3.6 divisions from positive to negative peak. What is the peak to peak voltage of this signal
Answer: 7.2V
Explanation:
Peak values or peak to peak voltage are calculated from RMS values, which implies VP = VRMS × √2, (assuming the source is a pure sine wave).
Since it's a sinusoidal voltage and measuring from an oscilloscope, the peak to peak voltage is gotten thus:
No of division X Volts/divisions
So, 3.6 x 2V = 7.2V
Final answer:
The peak-to-peak voltage of a sinusoidal signal covering 3.6 divisions on an oscilloscope set to 2 volts per division is 7.2 volts.
Explanation:
The question involves calculating the peak-to-peak voltage of a sinusoidal signal observed on an oscilloscope where the vertical scale is set to 2 volts per division. Given that the signal covers 3.6 divisions from positive to negative peak, we calculate the peak-to-peak voltage by multiplying the number of divisions the signal spans by the voltage per division.
To find the peak-to-peak voltage, we use the formula: Peak-to-Peak Voltage = Number of Divisions × Voltage per Division. Thus, the peak-to-peak voltage of the signal is 3.6 divisions × 2 volts/division = 7.2 volts.
A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 544 m? (Round your answer to the nearest whole number. Use g ≈ 9.8 m/s2.)
The muzzle speed of the gun, when fired at an angle of elevation of 30° and reaching a maximum height of 544 m, is approximately 329 m/s.
Explanation:The physics concept here is projectile motion. The muzzle speed of the gun can be calculated using the equation for the maximum height attained by a projectile, which is given by H = (V^2 * sin^2θ) / 2g, where V represents the muzzle speed, θ is the angle of elevation, and g is the acceleration due to gravity. Rearranging for V, and substituting the given values, we get:
V = sqrt((2 * H * g) / sin^2θ) = sqrt((2 * 544 m * 9.8 m/s^2) / sin^2 30°). Since sin 30° = 0.5, this leads to V = sqrt((2 * 544 m * 9.8 m/s^2) / (0.5)^2). The resulting muzzle speed, when calculated and rounded to the nearest whole number, is 329 m/s.
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