Answer:
v = 7.5*10⁶ m/s
Explanation:
While accelerating through a potential difference of 160 V, the electron undergoes a change in the electric potential energy, as follows:
ΔUe = q*ΔV = (-e)*ΔV = (-1.6*10⁻¹⁹ C) * 160 V = -2.56*10⁻¹⁷ J (1)
Due to the principle of conservation of energy, in absence of non-conservative forces, this change in potential energy must be equal to the change in kinetic energy, ΔK:
ΔK = Kf -K₀
As the electron accelerates from rest, K₀ =0.
⇒ΔK =Kf = [tex]\frac{1}{2}*me*vf^{2}[/tex] (2)
From (1) and (2):
ΔK = -ΔUe = 2.56*10⁻¹⁷ J = [tex]\frac{1}{2}*me*vf^{2}[/tex]
where me = mass of the electron = 9.1*10⁻³¹ kg.
Solving for vf:
[tex]vf =\sqrt{\frac{2*(2.56e-17J)}{9.1e-31kg} } =7.5e6 m/s[/tex]
⇒ vf = 7.5*10⁶ m/s
Final answer:
The speed of an electron after accelerating from rest through a potential difference of 160 V can be calculated using the conservation of energy principle, resulting in a velocity of approximately 5.93 × 10^6 m/s.
Explanation:
To calculate the speed of an electron after it accelerates from rest through a potential difference, we use the concept of conservation of energy where the electrical potential energy converted into kinetic energy is expressed as qAV = ½mv². Here, q is the charge of the electron (-1.60 × 10^-19 C), V is the potential difference (160 V), m is the mass of the electron (9.11 × 10^-13 kg), and v is the velocity of the electron we need to find. Rearranging the equation to solve for v and substituting the given values gives:
v = sqrt(2 × q × V / m) = √(2 × (-1.60 × 10^-19 C) × 160 V / (9.11 × 10^-13kg))
Performing the calculation yields a velocity of approximately 5.93 × 10^6 m/s.
What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?
Answer:
elastic force and weight are related to the acceleration of the System.
Explanation:
The relationship between these two forces can be found with Newton's second law.
[tex]F_{e}[/tex] - W = m a
K x - m g = m a
We see that elastic force and weight are related to the acceleration of the System.
If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur
The relationship between the applied force of a hanging mass and the spring force of the spring is governed by Hooke's law, where the force is proportional to displacement. At equilibrium, the spring force balances the gravitational force, and any displacement results in harmonic oscillation.
Explanation:The relationship between the applied force of a hanging mass on a spring and the spring force is described by Hooke's law. When a mass is attached to a vertical spring, it stretches until the spring force equals the force of gravity on the mass. This point is called the equilibrium position. The spring force at this point is k times the displacement from the spring's unstretched length, where k is the spring constant. If the mass is displaced from this equilibrium position, a restoring force acts to return the mass to equilibrium, leading to harmonic oscillation.
In the context of springs, two scenarios arise. When the spring is relaxed, there is no force on the block, indicating an equilibrium situation. Upon stretching or compressing the spring, Hooke's law predicts the force on the block as being proportional to the displacement from its equilibrium position, x, with force F being -Kx. Here, -K represents the negative spring constant, signifying a restoring force opposite to the direction of displacement.
For a massive continuous spring with negligible gravitational effect compared to the spring force (KL » mg), when the top of the spring is driven up and down, the position of the bottom can be described as a function of time. This shows that both vertical and horizontal spring-mass systems follow similar dynamics and obey the laws governing harmonic motion.
An air traffic controller notices two signals from two planes on the radar monitor. One plane is at altitude 1162 m and a 10.1-km horizontal distance to the tower in a direction 34.2° south of west. The second plane is at altitude of 4162 m and its horizontal distance is 9.5 km directed 21.5° south of west. What is the distance between these planes in kilometers?
Answer:
[tex]|R|=4.373km[/tex]
Explanation:
Given data
For first plate let it be p₁
[tex]p_{z1}=1162 m\\ p_{x1}=10.1km\\\alpha _{1}=34.2^{o}[/tex]
For second plate let it be p₂
[tex]p_{z2}=4162 m\\p_{x2}=9.5km\\\alpha _{2}=21.5^{o}[/tex]
To find
Distance R between them
Solution
To find distance between two plates first we need to find p₁ and p₂
Finding p₁
According to vector algebra
[tex]p_{1}=p_{x1}i+p_{y1}j+p_{z1}k\\ as\\tan\alpha =tan(34.2^{o} )=(p_{y1}/p_{x1})\\p_{y1}=10.1tan(34.2^{o} )\\p_{y1}=6.864km[/tex]
So we get
[tex]p_{1}=-10.1i-6.864j+1.162k[/tex]
Now to find p₂
[tex]p_{2}=p_{x2}i+p_{y2}j+p_{z2}k\\ as\\tan\alpha =tan(21.5^{o} )=(p_{y2}/p_{x2})\\p_{y2}=9.5tan(21.5^{o} )\\p_{y2}=3.74km[/tex]
So we get
[tex]p_{2}=-9.5i-3.74j+4.162k[/tex]
Now for distance R
According to vector algebra the position vector R between p₁ and p₂
[tex]R=p_{1}-p_{2}\\ R=(p_{x1}-p_{x2})i+(p_{y1}-p_{y2})j+(p_{z1}-p_{z2})k\\R=(-10.1-(-9.5))i+(-6.864-(-3.74))j+(1.162-4.162)k\\R=-0.6i-3.124j-3k\\|R|=\sqrt{(0.6)^{2}+(3.124)^{2}+(3)^{2} }\\ |R|=4.373km[/tex]
To find the distance between two planes, calculate the horizontal distances using the given angles and distances. Use the Pythagorean theorem to find the distance by applying the formula. The distance between the two planes is approximately 14.1 km.
Explanation:To find the distance between the two planes, we can use the Pythagorean theorem. First, we calculate the horizontal distances using the given angles and distances. For the first plane, the horizontal distance is 10.1 km * cos(34.2°), and for the second plane, it is 9.5 km * cos(21.5°). Next, we use the Pythagorean theorem to find the distance between the two planes: d = √((9.5 km * cos(21.5°))^2 + (10.1 km * cos(34.2°))^2).
Calculating the values, we get d ≈ 14.1 km.
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A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door. Where is the pivot point and what is the maximum torque?
Answer:
Explanation:
Given
Force applied [tex]F=80\ N[/tex]
Door is [tex]d=3\ m[/tex] wide
for gate to revolve Properly Pivot Point must be at center i.e. 1.5 from either end
Torque applied is [tex]T=force\times distance[/tex]
Maximum torque
[tex]T_{max}=F\times \frac{d}{2}[/tex]
[tex]T_{max}=80\times \frac{3}{2}[/tex]
[tex]T_{max}=120\ N-m[/tex]
Final answer:
The pivot point of a revolving door is at its central axis, and the maximum torque exerted by the businesswoman on the revolving door is 120 Nm, calculated by multiplying the force (80 N) by the perpendicular distance from the pivot point to the point of application (1.5 m).
Explanation:
The question is asking about the concept of torque, which is a measure of the turning force on an object. In the scenario described, a businesswoman is applying a force of 80 N to a revolving door. The maximum torque can be calculated by multiplying the force applied by the perpendicular distance from the pivot point to the point of application. The pivot point of a revolving door is its central axis, which is the center of the door.
To calculate the maximum torque, we use the formula:
Torque = Force * Distance.
Here, Torque = 80 N * 1.5 m (since the force is applied at the edge of the 3 m wide door, the distance from the pivot point is half the width).
Thus, Torque = 120 Nm
This is the maximum torque exerted by the woman on the door relative to its central.
At the time t = 0, the boy throws a coconut upward (assume the coconut is directly below the monkey) at a speed v0. At the same instant, the monkey releases his grip, falling downward to catch the coconut. Assume the initial speed of the monkey is 0, and the cliff is high enough so that the monkey is able to catch the coconut before hitting the ground. The time taken for the monkey to reach the coconut is
Answer:
The time taken for the monkey to reach the coconut, t is = H/v₀
Explanation:
Let the coordinate of the boy's hand be y = 0. The height of the tree above the boy's hand be H.
Coordinate of where the monkey meets coconut = y
Using the equations of motion,
For the monkey, initial velocity = 0m/s, time to reach coconut = t secs and the height at which coconut is reached = H-y
For the coconut, g = -10 m/s², initial velocity = v₀, time to reach monkey = t secs and height at which coconut meets monkey = y
For monkey, H - y = ut + 0.5gt², but u = 0,
H - y = 0.5gt²..... eqn 1
For coconut, y = v₀t - 0.5gt² ....... eqn 2
Substituting for y in eqn 1
H - y = H - (v₀t - 0.5gt²) = 0.5gt²
At the point where the monkey meets coconut, t=t
H - v₀t + 0.5gt² = 0.5gt²
v₀t = H
t = H/v₀
Solved!
Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after accelerating through potential difference 4V?
Final answer:
When a charge is accelerated through a potential difference, its speed can be calculated using the equation v = √(2qV/m). In the given example, an electron is accelerated through a potential difference of 4V, resulting in a speed of approximately 290 mV.
Explanation:
When a charge is accelerated through a potential difference, it gains kinetic energy. The relationship between the potential difference and the speed of the charge can be calculated using the equation:
v = √(2qV/m)
Where v is the final speed of the charge, q is the charge of the particle, V is the potential difference, and m is the mass of the particle.
In the given example, an electron with a charge of -1.60 × 10-19 C and a mass of 9.11 × 10-31 kg is accelerated to a speed of 10 × 104 m/s through a potential difference. To calculate the potential difference, we can rearrange the equation to:
V = m(v²)/(2q)
Substituting the values:
V = (9.11 × 10-31 kg)(10 × 104 m/s)²/ (2)(-1.60 × 10-19 C)
V = 2.91 × 10-2 V
So, the potential difference is approximately 29 mV.
A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 52 cm . What is the spring constant of the spring?
To find the spring constant (k) of the spring, the potential energy of the falling safe is equated with the elastic potential energy of the spring at maximum compression. The calculated spring constant is approximately 20641 N/m.
To determine the spring constant of the spring, we can use the conservation of energy principle. Since the safe falls from a height onto the spring and compresses it, we can equate the potential energy of the safe at the height to the elastic potential energy stored in the spring at maximum compression.
The potential energy (PE) of the safe when it is above the spring is given by PE = mgh, where m is the mass (1100 kg), g is the acceleration due to gravity (9.81 m/s2), and h is the height (2.4 m). The elastic potential energy (EPE) stored in the spring when compressed is given by [tex]EPE = (1/2)kx^2[/tex], where k is the spring constant we need to find, and x is the compression distance (0.52 m).
Equating these two energies, [tex]mgh = (1/2)kx^2[/tex], we solve for the spring constant (k). Plugging in the values gives us:
[tex]1100 kg * 9.81 m/s^2 * 2.4 m = (1/2)k * (0.52 m)^2[/tex]
Solving for k, we get:
[tex]k = (1100 kg * 9.81 m/s^2 * 2.4 m) / (0.5 * (0.52 m)^2)[/tex]
After calculating, we find that the spring constant k is approximately 20641 N/m.
The spring constant k of the spring is 232614 N/m.
Use energy conservation: lost gravitational energy equals stored elastic potential energy in spring.
Gravitational Potential Energy (GPE) lost by the safe:
Height dropped by safe: [tex]\( h = 2.4 \, \text{m} + 0.52 \, \text{m} = 2.92 \, \text{m} \)[/tex]
Mass of safe (m): 1100 kg
Acceleration due to gravity [tex](\( g \)): \( 9.81 \, \text{m/s}^2 \)[/tex]
GPE lost = [tex]\( mgh = 1100 \times 9.81 \times 2.92 \)[/tex]
Elastic Potential Energy (EPE) stored in the spring:
Compression of spring [tex](\( x \)): 0.52 m (52 cm)[/tex]
EPE stored = [tex]\( \frac{1}{2} k x^2 \)[/tex]
Set lost gravitational potential energy equal to spring's stored elastic energy.
[tex]\[ mgh = \frac{1}{2} k x^2 \][/tex]
Rearrange the formula to solve for k:
[tex]\[ k = \frac{2mgh}{x^2} \][/tex]
Substitute the values:
[tex]\[ k = \frac{2 \times 1100 \times 9.81 \times 2.92}{(0.52)^2} \][/tex]
[tex]\[ k = \frac{2 \times 1100 \times 9.81 \times 2.92}{0.2704} \][/tex]
[tex]\[ k = \frac{62898.672}{0.2704} \][/tex]
[tex]\[ k \approx 232613.6 \, \text{N/m} \][/tex]
If a guitar string has a fundamental frequency of 500 Hz, what is the frequency of its second overtone?
You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?
Final answer:
The speed of the putty just before it strikes the ceiling is approximately 0.342 m/s, and it takes approximately 0.97 seconds to reach the ceiling.
Explanation:
To calculate the speed of the putty just before it strikes the ceiling, we can use the equation for vertical motion:
v = u + gt
Where:
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time
In this case, the object is thrown upwards, so the final velocity when it reaches the ceiling will be zero. The initial velocity is 9.50 m/s and the acceleration due to gravity is -9.8 m/s^2. Substitute these values into the equation and solve for t:
0 = 9.50 - 9.8t
t = 0.97 seconds
Therefore, the speed of the putty just before it strikes the ceiling is 9.50 - 9.8(0.97) = -0.342 m/s. Since speed cannot be negative, the actual speed is 0.342 m/s.
To calculate the time it takes for the putty to reach the ceiling, we can use the same equation and solve for t:
0 = 9.50 - 9.8t
t = 0.97 seconds
So, it takes the putty approximately 0.97 seconds to reach the ceiling.
Final answer:
The speed of the putty just before it strikes the ceiling is 9.50 m/s, the same as its initial speed due to conservation of energy. The time it takes for the putty to reach the ceiling can be calculated using kinematic equations.
Explanation:
The problem presented concerns the motion of a glob of putty thrown upward and involves calculations based on the principles of kinematics in physics.
Answer to Part (a): What is the speed of the putty just before it strikes the ceiling?
The speed of the putty just before it strikes the ceiling can be found by using the kinematic equation for velocity under constant acceleration (gravity). Since the putty is thrown upwards, it will slow down under the influence of gravity until it reaches its maximum height. At this point, it will start to fall back down, accelerating under gravity until it hits the ceiling. Assuming no energy loss, the speed of the putty just before it strikes the ceiling will be the same as its initial speed when it leaves the hand, which is 9.50 m/s.
Answer to Part (b): How much time from when it leaves your hand does it take the putty to reach the ceiling?
The time it takes for the putty to reach the ceiling can be calculated using the kinematic equation s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration due to gravity (approximately -9.81 m/s² when upwards is considered positive), and t is the time. Solving for t gives us the time taken for the putty to reach the height of 3.60 m.
A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve this acceleration is (a) 56 kW (b) 222 kW (c) 2.5 kW (d) 62 kW (e) 90 kW
To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this
Initial Velocity,
[tex]V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]V_i = 16.6667m/s[/tex]
Final Velocity,
[tex]V_f = 100km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]V_f = 27.7778m/s[/tex]
Now Power is defined as the change of Energy over the time,
[tex]P = \frac{E}{t}[/tex]
But Energy is equal to Kinetic Energy,
[tex]P = \frac{\frac{1}{2} m\Delta v^2}{t}[/tex]
[tex]P = \frac{\frac{1}{2} m(v_f^2-v_i^2)}{t}[/tex]
Replacing,
[tex]P = \frac{\frac{1}{2} (900)(27.7778^2-16.6667^2)}{4}[/tex]
[tex]P = 56kW[/tex]
Therefore the correct answer is A.
Final answer:
The additional power needed to achieve the acceleration is 62 kW.So,option (d) 62 kW is correct.
Explanation:
To calculate the additional power needed to achieve acceleration, we can use the formula for power: Power = Force x Velocity. We know the mass of the car (900 kg), the initial velocity (60 km/h), and the final velocity (100 km/h). We can convert the velocities to m/s and calculate the force required to accelerate the car. Then, we can multiply the force by the change in velocity to find the additional power needed.
In this case, the additional power needed is approximately 62 kW. Therefore, option (d) 62 kW is the correct answer.
A thin, square, conducting plate 40.0 cm on a side lies in the xy plane. A total charge of 4.70 10-8 C is placed on the plate. You may assume the charge density is uniform. (a) Find the charge density on each face of the plate. 1.47e-07 What is the definition of surface charge density? C/m2 (b) Find the electric field just above the plate. magnitude N/C direction (c) Find the electric field just below the plate. magnitude N/C direction
Answer:
(a) Surface charge density is the charge per unit area.
[tex]\sigma = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]
(b) [tex]\vec{E} = (+\^z)~8.34\times 10^3~{\rm N/C}[/tex]
(c) [tex]\vec{E} = (-\^z)~8.34\times 10^3~{\rm N/C}[/tex]
Explanation:
(a) Surface charge density is the charge per unit area. The area of the square plate can be calculated by its side length.
[tex]A = l^2 = (0.4)^2 = 0.16 ~{\rm m^2}[/tex]
Half of the total charge is distributed on one side and the other is distributed on the other side.
Therefore, surface charge density on each face of the plate is
[tex]\sigma = Q/A = \frac{2.35 \times 10^{-8}}{0.16} = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]
(b) To find the electric field just above the plate, Gauss' Law can be used. Normally, Gauss' Law can only be used in infinite sheet (considering the flat surfaces), but just above the surface can be considered that the distance from the surface is much much smaller than the length of the plate (x << l).
In order to apply Gauss' Law, we have to draw an imaginary cylinder with radius r. The cylinder has to stay perpendicular to the plane.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r^2 = \frac{\pi r^2 \sigma}{\epsilon_0}\\E = \frac{\sigma}{2\epsilon_0}\\E = \frac{1.46 \times 10^{-7}}{2\times 8.8\times 10^{-12}} = 8.34\times 10^3~{\rm N/C}[/tex]
The direction of the electric field is in the upwards direction.
(c) The magnitude of the electric field is the same as that of upper side. Only the direction is reversed, downward direction.
The pressure drop needed to force water through a horizontal 1-in.-diameter pipe is 0.55 psi for every 8-ft length of pipe. (a) Determine the shear stress on the pipe wall. Determine the shear stress at distances (b) 0.3 and (c) 0.5 in. away from the pipe wall.
Answer:
(a). The shear stress on the pipe wall is 0.2062 lb/ft²
(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²
(c). The shear stress at the distance 0.5 in away from the pipe wall is zero.
Explanation:
Given that,
Diameter = 1 in
Pressure = 0.55 psi
Length = 8 ft
We need to calculate the radius of the pipe
Using formula of radius
[tex]r=\dfrac{D}{2}[/tex]
Put the value into the formula
[tex]r=\dfrac{1}{2}[/tex]
[tex]r=0.5\ in[/tex]
(a). We need to calculate the shear stress on the pipe wall
Using formula of shear stress
[tex]\dfrac{\Delta p}{L}=\dfrac{2\tau}{r}[/tex]
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0.5}{2\times8\times12}[/tex]
[tex]\tau=0.2062\ lb/ft^2[/tex]
(b). We need to calculate the shear stress at the distance 0.3 in
Using formula of shear stress
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0.3}{2\times8\times12}[/tex]
[tex]\tau=0.12375\ lb/ft^2[/tex]
(c). We need to calculate the shear stress at the distance 0.5 in away from the pipe wall
r = 0.5-0.5 = 0
Using formula of shear stress
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0}{2\times8\times12}[/tex]
[tex]\tau=0[/tex]
Hence, (a). The shear stress on the pipe wall is 0.2062 lb/ft²
(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²
(c). The shear stress at the distance 0.5 in away from the pipe wall is zero.
A ball is thrown horizontally off a cliff. If the initial speed of the ball is (15.0 + A) m/s and the cliff is (25.0 + B) m high, how far from the base of the cliff will the ball land in the water below? Calculate the answer in meters (m) and round to three significant figures.A=4B=54
Answer:
76.3 m
Explanation:
We are given that
Initial speed of the ball,u=(15+A)m/s
Height of cliff,h=(25.9+B) m
We have to find the distance from the base of the cliff the ball will land in the water below.
A=4 and B=54
Distance=[tex]u\sqrt{\frac{2h}{g}}[/tex]
Using the formula and substitute the values
[tex]D=(15+4)\sqrt{\frac{2(25+54)}{9.8}}[/tex]
Because [tex]g=9.8m/s^2[/tex]
[tex]D=19\sqrt{\frac{158}{9.8}}[/tex]
D=76.3
Hence, the distance from the base of the cliff the ball will land in the water below=76.3 m
A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.
What is the electric potential at the center of the sphere?
Answer:
The electric potential at the center of the sphere is 36 V.
Explanation:
Given that,
Radius R= 0.600 m
Distance D = 1.20 m
Electric potential = 18.0 V
We need to calculate the electric potential
Using formula of electric potential
[tex]V=\dfrac{kq}{r}[/tex]
Put the value into the formula
[tex]18.0=\dfrac{9\times10^{9}\times q}{1.20}[/tex]
[tex]q=\dfrac{18.0\times1.20}{9\times10^{9}}[/tex]
[tex]q=2.4\times10^{-9}\ C[/tex]
We need to calculate the electric potential at the center of the sphere
Using formula of potential
[tex]V=\dfrac{kq}{r}[/tex]
Put the value into the formula
[tex]V=\dfrac{9\times10^{9}\times2.4\times10^{-9}}{0.600}[/tex]
[tex]V=36\ V[/tex]
Hence, The electric potential at the center of the sphere is 36 V.
The electric potential at the center of the sphere is 36 V.
The given parameters;
radius of the, R = 0.6 mdistance of the charge, r = 1.2 m potential difference, V = 18 VThe magnitude of the charge is calculated as follows;
[tex]V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\q = \frac{18 \times 1.2}{9\times 10^9} \\\\q = 2.4 \times 10^{-9} \ C[/tex]
The electric potential at the center of the sphere;
[tex]V = \frac{kq}{r} \\\\V = \frac{9\times 10^9 \times 2.4 \times 10^{-9}}{0.6} \\\\V = 36 \ V[/tex]
Thus, the electric potential at the center of the sphere is 36 V.
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The total mass of all the planets is much less than the mass of the Sun. (T/F)
Answer:
True
Explanation:
The total mass of all the planets is much less than the mass of the Sun is a True statement.
Sun contributes to most of the mass of the solar system. Sun contributes to 99.8% mass of the solar system. Only 0.2% of the mass of the solar system is given by the planets. Hence, the above statement is absolutely correct.
A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint between the two particles.
The electric potential at the midpoint between a proton and an electron is zero.
Explanation:The electric potential at the midpoint between a proton and an electron can be calculated using the formula:
V = k * (q1 / r1 + q2 / r2)
where V is the electric potential, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r1 and r2 are the distances between the particles and the midpoint. In this case, the charges of the proton and the electron are equal in magnitude but opposite in sign, so q1 = -q2. The distances from the particles to the midpoint are equal because they are fixed in space, so r1 = r2. Plugging in the values, we get:
V = k * (-q / r1 + q / r1) = 0
Therefore, the electric potential at the midpoint between the two particles is zero.
Does the speedometer of a car measure speed or velocity? Explain.
The speedometer of a car measures speed, which is a scalar quantity indicating how fast the car is moving without regard to direction. Velocity, on the other hand, includes both speed and direction, which the speedometer does not display. The odometer measures total distance traveled, not displacement, and dividing distance by time gives average speed, not velocity.
Explanation:The speedometer of a car measures speed, not velocity. Speed is a scalar quantity, which means it only describes how fast an object is moving regardless of its direction. On the other hand, velocity is a vector quantity that describes both the speed and the direction of an object's movement. For example, if a car is moving at 60 miles per hour (mph), the speedometer shows this speed, but it does not indicate whether the car is traveling north, south, east, or west – that would be necessary information to determine the car's velocity.
A car's odometer, in contrast, measures the total distance traveled by the car. This distance is a scalar quantity as well, which means it does not account for the direction of travel, only the cumulative distance covered. When you divide the total distance traveled, as shown on the odometer, by the total time taken for the trip, you are calculating the average speed of the car, not the magnitude of average velocity. These two quantities – average speed and the magnitude of average velocity – are the same when the car moves in a straight line without changing its direction.
A wave on a string has a wavelength of 0.90 m at a frequency of 600 Hz. If a new wave at a frequency of 300 Hz is established in this same string under the same tension, what is the new wavelength? Group of answer choices
Answer:
λ₂ = 1.8 m
Explanation:
given,
wavelength of the string 1 = 0.90 m
frequency of the string 1 = 600 Hz
wavelength of string 2 = ?
frequency of the string 2 = 300 Hz
we now,
[tex]f\ \alpha\ \dfrac{1}{\lambda}[/tex]
now,
[tex]\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}[/tex]
[tex]\dfrac{600}{300}=\dfrac{\lambda_2}{0.9}[/tex]
λ₂ = 2 x 0.9
λ₂ = 1.8 m
Hence, the wavelength of the second string is equal to λ₂ = 1.8 m
Compared with ultraviolet radiation, infrared radiation has greater (a) wavelength; (b) amplitude; (c) frequency; (d) energy.
Answer:
(a) WaveLength
Explanation:
Compared with ultraviolet radiation, infrared radiation has greater wavelength. Option A.
UV vs IRCompared with ultraviolet radiation, infrared radiation has greater wavelength.
Wavelength refers to the distance between two consecutive peaks or troughs of a wave. In the electromagnetic spectrum, infrared radiation has longer wavelengths than ultraviolet radiation.
Ultraviolet radiation has shorter wavelengths and higher energy, while infrared radiation has longer wavelengths and lower energy.
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Two college students are sliding down a hill on excellent sleds so you can ignore friction. One has a mass of 85 kg and one has a mass of 75 kg. Which will reach the bottom of the hill first? a. they will both reach at the same time. b. 85 kg person c. 75 kg person
Answer:
a. They both reach at the same time.
Explanation:
On a frictionless incline, the only force that moves the person downwards is the x-component of the persons weight. (x-direction is the direction along the incline.)
[tex]F = mg\sin(\theta)[/tex]
Here, θ is the angle of the incline above horizontal.
This force is equal to 'ma' according to Newton's Second Law.
Comparing the weights of the two persons gives
[tex]F_1 = 85g\sin(\theta) = 85a_1\\F_2 = 75g\sin(\theta) = 75a_1\\a_1 = g\sin(\theta)\\a_2 = g\sin(\theta)[/tex]
Since the accelerations of both persons are the same, they reach the bottom at the same time.
The crucial point here is that the acceleration on a frictionless incline is independent from the mass of the object. If there were friction on the surface, then the person with smaller mass would reach the bottom first.
The rocket-driven sled Sonic Wind No. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s (500 mi/h) in 0.900 s. (a) Compute the acceleration in m/s2, assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body (g)? (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled de-creased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40 g . Are these figures consistent?
Answer:
a) The acceleration of the rocket is 249 m/s².
b) The acceleration of the rocket is 25 times the acceleration of a free-falling body (25 g),
c) The distance traveled in 0.900 s was 101 m.
d) The figures are not consistent. The acceleration of the rocket was 20 g.
Explanation:
Hi there!
a) To calculate the acceleration of the rocket let's use the equation of velocity of the rocket:
v = v0 + a · t
Where:
v = velocity of the rocket.
v0 = initial velocity.
a = acceleration.
t = time.
We know that at t = 0.900 s, v = 224 m/s. The initial velocity, v0, is zero because the rocket starts from rest.
v = v0 + a · t
Solving for a:
(v - v0) / t = a
224 m/s / 0.900 s = a
a = 249 m/s²
The acceleration of the rocket is 249 m/s²
b) The acceleration of gravity is ≅ 10 m/s². The ratio of the acceleration of the rocket to the acceleration of gravity will be:
249 m/s² / 10 m/s² = 25
So, the acceleration of the rocket is 25 times the acceleration of gravity or 25 g.
c) The equation of traveled distance is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the rocket at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since x0 and v0 are equal to zero, then, the equation of position gets reduced to:
x = 1/2 · a · t²
x = 1/2 · 249 m/s² · (0.900 s)²
x = 101 m
The distance traveled in 0.900 s was 101 m.
d) Now, using the equation of velocity, let's calculate the acceleration. We know that at 1.40 s the velocity of the rocket is zero and that the initial velocity is 283 m/s.
v = v0 + a · t
0 m/s = 283 m/s + a · 1.40 s
-283 m/s / 1.40 s = a
a = -202 m/s²
The figures are not consistent because 40 g is equal to an acceleration of 400 m/s² and the magnitude of the acceleration of the rocket was ≅20 g.
A snorkeler with a lung capacity of 4.3 L inhales a lungful of air at the surface, where the pressure is 1.0 atm. The snorkeler then descends to a depth of 49 m , where the pressure increases to 5.9 atm. What is the volume of the snorkeler's lungs at this depth? (Assume constant temperature.)
To find the volume of the snorkeler's lungs at a depth of 49m, use Boyle's Law by calculating the volume at the new pressure using the initial volume and pressure.
The volume of the snorkeler's lungs at a depth of 49 m can be calculated using Boyle's Law. Boyle's Law states that pressure and volume are inversely proportional when temperature is constant. To find the volume at depth, you can use P1V1 = P2V2, where P1V1 is the initial condition and P2V2 is the final condition.
Using the given data:
P1 = 1.0 atm, V1 = 4.3 L (lung capacity at the surface)P2 = 5.9 atm (pressure at 49 m depth), V2 = unknown (volume at 49 m depth)By rearranging the formula:
V2 = (P1 * V1) / P2 = (1.0 atm * 4.3 L) / 5.9 atm = 0.72 L
Therefore, at a depth of 49 m, the volume of the snorkeler's lungs would be approximately 0.72 liters.
Two electrons are separated by a distance of 1.00 nm and held fixed in place. A third electron, initially very far away, moves toward the other two electrons and stops at the point exactly midway between them. Calculate the speed of the third electron when it was very far away from the other electrons.
Answer:
The speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s
Explanation:
qV = 0.5Mv²
where;
V is the potential difference, measured in Volts
q is the charge of the electron in Coulomb's = 1.6 × 10⁻¹⁹ C
Mass is the mass of the electron in kg = 9 × 10⁻³¹ kg
v is the velocity of the electron in m/s
Applying coulomb's law, we determine the Potential difference V
V = kq/r
V = (8.99X10⁹ * 1.6 × 10⁻¹⁹)/(1X10⁻⁹)
V = 14.384 X 10¹⁹ V
The speed of the electron can be determined as follows;
v² = (2qV)/M
v = √(2qV)/M)
v = √(2*1.6 × 10⁻¹⁹* 14.384 X 10¹⁹)/(9 × 10⁻³¹)
v = √(5.1143 X 10³¹) = 7 X 10¹⁵m/s
Therefore, the speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s
A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 13°, and that the angle of depression to the bottom of the tower is 4°. How far is the person from the monument? (Round your answer to three decimal places.)
Answer:
665 ft
Explanation:
Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.
The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is
[tex]dtan13^0 = 0.231d[/tex]
Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees
[tex]dtan4^0 = 0.07d [/tex]
Since the 2 sides length above make up the 200 foot monument, their total length is
0.231d + 0.07d = 200
0.301 d = 200
d = 200 / 0.301 = 665 ft
To find the distance to the 200-foot tall monument from a specified point, we use tangent trigonometric functions with the given angles of elevation and depression. The total horizontal distance is calculated by separately finding the distances to the top and the bottom of the monument and combining them.
Explanation:The question involves solving a problem using trigonometry to find the distance to a monument given the angles of elevation and depression from a certain point. Since the height of the monument is given as 200 feet, and we have the angles of elevation (13°) to the top of the monument and the angle of depression (4°) to the bottom, we can use trigonometric functions to calculate the distances to the top and the bottom of the monument separately and then sum them to find the total distance from the observer to the monument.
To find the horizontal distances (D) from the observer to the top and bottom of the monument, we can use the tangent function (tan) from trigonometry. The tangent of the angle of elevation (13°) is equal to the height of the monument divided by the distance to the top (Dtop), and the tangent of the angle of depression (4°) is equal to the height from the window to the base of the monument divided by the distance to the bottom (Dbottom).
Assuming the window is at the same height as the base of the monument (which is not given explicitly, but implied by the angle of depression), we get two equations:
tan(13°) = 200 / Dtoptan(4°) = 0 / DbottomWe can then solve for Dtop and Dbottom using these equations and add the two distances to find the total horizontal distance to the monument.
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An oscillator with angular frequency of 1.00 s-1has initial displacement of 1.00 m and initial velocity of 1.72 m/s. What is the amplitude of oscillation?
Final answer:
The amplitude of the oscillator is 1.989 m.
Explanation:
The amplitude of an oscillator can be determined using the initial displacement and initial velocity of the system. In this case, the initial displacement is given as 1.00 m and the initial velocity as 1.72 m/s. The amplitude, also known as the maximum displacement, is equal to the square root of the sum of the squares of the initial displacement and initial velocity.
Using the formula:
X = √(x₀² + v₀²)
Where X is the amplitude, x₀ is the initial displacement, and v₀ is the initial velocity.
Substituting the given values into the formula:
X = √(1.00² + 1.72²)
= √(1 + 2.9584)
= √3.9584
= 1.989 m
The amplitude of the oscillator is 1.989 m.
You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.46 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 2.2 m and then starts to slow down.What is the maximum speed of the elevator?
Answer:
Final velocity of the elevator will be 4.453 m/sec
Explanation:
Let mass is m
Acceleration due to gravity is g m/sec^2
Distance s = 2.2 m
As the elevator is moving upward so net force on elevator
[tex]F=mg+ma[/tex]
So according to question
[tex]1.46mg=mg+ma[/tex]
0.46 mg = ma
a = 0.46 g
a = 0.46×9.8 = 4.508 [tex]m/sec^2[/tex]
Initial velocity of elevator is 0 m/sec
From third equation of motion
[tex]v_f^2=v_i^2+2as[/tex]
[tex]v_f^2=0^2+2\times 4.508\times 2.2[/tex]
[tex]v_f=4.453m/sec[/tex]
So final velocity of the elevator will be 4.453 m/sec
Which of the following statements are true?
a. Electric field lines and equipotential surfaces are always mutually perpendicular.
b. When all charges are at rest, the surface of a conductor is always an equipotential surface.
c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point.
d. The potential energy of a test charge increases as it moves along an equipotential surface.
e. The potential energy of a test charge decreases as it moves along an equipotential surface.
Answer:
a,b and c are true.
Explanation:
Following are true statements
a. Electric field lines and Equipotential surfaces are always mutually perpendicular is a true statement.
b. When all charges are at rest, the surface of a conductor is always an equipotential surface.
c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point.
Following are False statements
d. The potential energy of a test charge increases as it moves along an equipotential surface.
e. The potential energy of a test charge decreases as it moves along an equipotential surface.
Reason: A t any point in an equipotential surface, the potential is same throughout. There is no increase or decrease in potential energy as the test charge moves in an equipotential environment.
Statements a, b, and c are correct: Electric field lines and equipotential surfaces are always mutually perpendicular; when all charges are at rest, the surface of a conductor is always an equipotential surface; and an equipotential surface is a three-dimensional surface on which the electric potential is the same at every point. Statements d and e are not correct because the potential energy of a test charge does not change as it moves along an equipotential surface.
Explanation:The following statements are true about electrical fields and potential:
a. Electric field lines and equipotential surfaces are always mutually perpendicular. This statement is correct. An electric field line shows the direction of the force a positive test charge would experience. An equipotential line or surface is one where the potential is the same at any point on the line or surface. As such, they will always be perpendicular to each other.b. When all charges are at rest, the surface of a conductor is always an equipotential surface. This statement is correct. In static conditions, the surface of a conductor is at a uniform potential because charges flow until they reach an equilibrium.c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point. This statement is right. That's why we named it equipotential (equal potential).d. The potential energy of a test charge increases as it moves along an equipotential surface. This statement is not correct. Since it's an equipotential surface, the potential energy stays the same.e. The potential energy of a test charge decreases as it moves along an equipotential surface. This statement is also not correct for the same reason stated above.Learn more about Electric Field and Potential here:https://brainly.com/question/31256352
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How do the magnitudes of the inertial (the density times acceleration term), pressure, and viscous terms in the Navier-Stokes equation compare?
Answer:
The general equation of movement in fluids is obtained from the application, at fluid volumes, of the principle of conservation of the amount of linear movement. This principle establishes that the variation over time of the amount of linear movement of a fluid volume is equal to that resulting from all forces (of volume and surface) acting on it. Expressed in This equation is called the Navier-Stokes equation.
The equation is shown in the attached file
Explanation:
The derivative of velocity with respect to time determines the change in the velocity of a particle of the fluid as it moves in space. It also includes convective acceleration, expressed by a nonlinear term that comes from convective inertia forces). With this equation, Stokes studied the motion of an infinite incompressible viscous fluid at rest at infinity, and in which a solid sphere of radius r makes a rectilinear and uniform translational motion of velocity v. It assumes that there are no external forces and that the movement of the fluid relative to a reference system on the sphere is stationary. Stokes' approach consists in neglecting the nonlinear term (associated with inertial forces due to convective acceleration).
Calculate the acceleration of a 1400-kg car that stops from 39 km/h "on a dime" (on a distance of 1.7 cm).
Answer:
[tex]a=-3449.67\frac{m}{s^2}[/tex]
Explanation:
The car is under an uniforly accelerated motion. So, we use the kinematic equations. We calculate the acceleration from the following equation:
[tex]v_f^2=v_0^2+2ax[/tex]
We convert the initial speed to [tex]\frac{m}{s}[/tex]
[tex]39\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=10.83\frac{m}{s}[/tex]
The car stops, so its final speed is zero. Solving for a:
[tex]a=\frac{v_0^2}{2x}\\a=-\frac{(10.83\frac{m}{s})^2}{2(1.7*10^{-2}m)}\\a=-3449.67\frac{m}{s^2}[/tex]
To calculate the acceleration of the car, use the formula a = (vf - vi) / t. Convert the initial velocity and distance to the appropriate units before substituting them into the formula.
To calculate the acceleration of the car, we can use the formula:
a = (vf - vi) / t
Where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken. In this case, the initial velocity is 39 km/h, which is converted to m/s by vi = 39 km/h * (1000 m/1 km) * (1 h/3600 s). The final velocity is 0 m/s since the car stops. The time taken can be found by t = d / vi, where d is the distance and vi is the initial velocity. The distance is given as 1.7 cm, which is converted to m by d = 1.7 cm * (1 m/100 cm). Substituting these values into the formula, we get the acceleration of the car.
So the acceleration of the car is approximately -8.45 m/s^2.
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A person is placed in a large, hollow, metallic sphere that is insulated from ground. (a) If a large charge is placed on the sphere, will the person be harmed upon touching the inside of the sphere? Yes No Correct: Your answer is correct. (b) Explain what will happen if the person also has an initial charge whose sign is opposite that of the charge on the sphere.
Answer:
Explanation:
It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.
If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.
If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.
If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere
A person inside a charged, hollow, metallic sphere, known as a Faraday cage, would not be harmed upon touching the interior, regardless of their own charge, due to the neutralization of the electric field inside the conductor.
A person placed inside a large, hollow, metallic sphere that is insulated from the ground and charged will not be harmed upon touching the inside of the sphere. The phenomenon that explains this is known as the Faraday cage effect. According to this principle, an external static electric field will cause the charges within a conductor to rearrange themselves in such a way that the electric field inside the conductor cancels out.
Now, if the person inside the sphere also has a charge of the opposite sign to that of the sphere, an interesting interaction occurs. However, due to the conductor's nature, the electric field inside the metallic sphere remains null. The charges within the conductor would still redistribute to neutralize the electric field within the conductor. Therefore, if a person inside touched the inner surface of the sphere, they would not be directly harmed by the electric field, as it is neutralized within the conductor. The charges from the person would likely redistribute on the sphere's inner surface to maintain electrostatic equilibrium without causing harm.
Charge of uniform surface density (0.20 nC/m2 ) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z = 2.0 m
Answer:
E= 11.3 N/C.
Explanation:
The electric filed at any point on the z-axis is given by the formula
[tex]E= \frac{\sigma}{2\epsilon_0}[/tex]
here, sigma is the charge density and ε_o is the permitivity of free space.
therefore,
[tex]E= \frac{0.2\times10^{-9}}{2\times8.85\times10^{-12}}[/tex]
solving it we get
E= 11.3 N/C.
Hence, the required Electric field is E= 11.3 N/C.