Answer:
Explanation:
Given
Voltage of battery [tex]V=45\ V[/tex]
electric field(E) is given by
[tex]E=|-\frac{dV}{dx}|[/tex]
i.e. Change in Potential over a distance x
(a)When Plate are separated by 3 cm apart
[tex]E=|\frac{45}{3\times 10^{-2}}|[/tex]
[tex]E=1500\ V/m[/tex]
(b)When Plates are 5 cm apart
[tex]E=|-\frac{45}{5\times 10^{-2}}|[/tex]
[tex]E=900\ V/m[/tex]
Force on electron is given by
[tex]F=charge\times Electric\ Field[/tex]
[tex]F=q\times E[/tex]
For case a
[tex]F=1.6\times 10^{-19}\times 1500[/tex]
[tex]F=24\times 10^{-17}\ N[/tex]
(b)[tex]F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N[/tex]
A ball is thrown vertically upward with a speed of 19.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started?
Answer:
Explanation:
initial speed, u = 19 m/s
(a) Let it rises upto height h.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity and it is zero.
0 = 19 x 19 - 2 x 9.8 x h
h = 18.4 m
(c) Let the ball takes time t to reach to the maximum height.
use first equation of motion
v = u - gt
0 = 19 - 9.8 x t
t = 1.94 s
(c) The time taken by the ball to reach to the ground = 2 x time to reach to maximum height
T = 2 x t = 2 x 1.94 = 3.88 s
(d) When the ball reaches the ground, let the velocity is v.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity
v² = 0 + 2 x 9.8 x 18.4
v = 19 m/s
(a) The ball reaches a height of approximately 18.68 meters. (b) It takes approximately 1.94 seconds to reach its highest point (c) The time taken to hit the ground is 1.94 seconds. When the ball returns to the level from which it started, its velocity is -19.0 m/s.
Explanation:(a) To find the height that the ball reaches, we can use the kinematic equation:
Δy = v2 / (2g)
where Δy is the change in height, v is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have:
Δy = (19.0 m/s)2 / (2 * 9.8 m/s2)
Calculating, we find that the ball rises to a height of approximately 18.68 meters.
(b) The time it takes for the ball to reach its highest point can be calculated using the equation:
t = v / g
where t is the time, v is the initial velocity, and g is the acceleration due to gravity. Substituting the given values, we get:
t = (19.0 m/s) / (9.8 m/s2)
Simplifying, we find that it takes approximately 1.94 seconds for the ball to reach its highest point.
(c) Since the time it takes for the ball to reach its highest point is the same as the time it takes for it to fall back down, the time it takes for the ball to hit the ground after reaching its highest point is also 1.94 seconds.
(d) When the ball returns to the level from which it started, its velocity is equal in magnitude but opposite in direction to its initial velocity. Therefore, the velocity when it returns is -19.0 m/s.
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What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?
Answer:
0.044 V
Explanation:
E = Electric field = [tex]5.5\times 10^6\ V/m[/tex]
d = Thickness of membrane = 8 nm
When the electric field strength is multiplied by the membrane thickness we get the voltage
Voltage across a gap is given by
[tex]V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V[/tex]
The voltage across the membrane is 0.044 V
Final answer:
The voltage across an 8.00 nm-thick membrane with an electric field strength of 5.50 MV/m is calculated using the formula V = Ed, resulting in a voltage of 44.0 mV.
Explanation:
The voltage across a membrane can be determined by using the relationship between electric field strength (E), voltage (V), and the distance (d) the electric field spans. The electric field is uniform and the formula to use is V = Ed. In this case, the electric field (E) is given as 5.50 MV/m or 5.50 x 10⁶V/m, and the thickness of the membrane (d) is 8.00 nm or 8.00 x 10⁻⁹ m.
To find the voltage across the membrane, we simply multiply the electric field by the thickness of the membrane:
V = (5.50 x 10⁶ V/m) x (8.00 x 10⁻⁹ m)
Therefore:
V = 44 x 10⁻³ V
V = 44.0 mV
So, the voltage across an 8.00 nm-thick membrane with the given electric field strength is 44.0 mV.
A mass is attached to a spring with spring constant k = 20 N m . The spring is stretched to 10 cm past its resting position. How much work (in J) does the spring do when the object is released and the mass travels back to its initial position?
The spring does 0.1 Joules of work when the object is released and travels back to its initial position.
We have,
The work done by the spring can be calculated using the formula for the potential energy stored in a spring:
Potential Energy (PE) = 0.5 * k * x²,
where k is the spring constant and x is the displacement from the equilibrium position.
Given:
Spring constant (k) = 20 N/m,
Displacement (x) = 0.10 m (10 cm).
Calculate the potential energy when the spring is stretched to 10 cm:
PE = 0.5 * k * x²
= 0.5 * 20 N/m * (0.10 m)²
= 0.5 * 20 * 0.01 J
= 0.1 J.
The spring stores 0.1 Joules of potential energy when stretched to 10 cm.
When the mass is released and returns to its initial position, this potential energy is converted back into kinetic energy as the mass accelerates.
Therefore,
The spring does 0.1 Joules of work when the object is released and travels back to its initial position.
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Final answer:
The spring does 0.1 Joules of work when the mass travels back to its initial position.
Explanation:
To find the work done by a spring when an object is released and travels back to its initial position, we can use the formula:
Work = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant is 20 N/m and the displacement is 10 cm (which is 0.1 m). Plugging these values into the formula, we get:
Work = (1/2) * 20 N/m * (0.1 m)^2 = 0.1 J
So, the spring does 0.1 Joules of work when the mass travels back to its initial position.
3 An empty hot tub has a mass of 320 kg. When filled, the tub holds 600 gallons of water (rho = 62.4 lbm/ft3). The local acceleration due to gravity is 32 ft/s2. Determine the total weight of the hot tub and water in pounds-force (lbf)
Answer:
Total weight of the hot tub and water is 5676.6 pounds-force
Explanation:
rho = 62.4lbm/ft^3 × 1ft^3/7.481gal = 8.34lbm/gal
Mass of water = rho × volume = 8.34lbm/gal × 600 gallons = 5004lbm = 5004×0.45359kg = 2269.8kg
Total mass of hot tub and water = 320kg + 2269.8kg = 2589.8kg
Local acceleration due to gravity = 32ft/s^2 = 32ft/s^2 × 1m/3.2808ft = 9.75m/s^2
Total weight of hot tub and water = 2589.8kg × 9.75m/s^2 = 25250.55N = 25250.55/4.4482 lbf = 5676.6 lbf
The distance from the earth to the sun is about 1.50×1011 m . Find the total power radiated by the sun.
Answer:
Power, [tex]P=3.93\times 10^{26}\ W[/tex]
Explanation:
Given that,
The distance from the earth to the sun is about, [tex]d=1.5\times 10^{11}\ m[/tex]
let us assume that the average intensity of solar radiation at the upper atmosphere of earth,. [tex]I=1390\ W/m^2[/tex]
We need to find the total power radiated by the sun. The intensity is defined as the total power divided by the area of a sphere of radius equal to the average distance between the earth and the sun. It is given by :
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
P is total power
[tex]P=4\pi r^2\times I[/tex]
[tex]P=3.93\times 10^{26}\ W[/tex]
So, the total power radiated by the sun is [tex]P=3.93\times 10^{26}\ W[/tex]. Hence, this is the required solution.
Final answer:
The total power radiated by the Sun is calculated using the area of the Sun and the power radiated per square meter at its surface. The Sun's total power output is found to be 3.82×1026 W. This value is important for understanding the energy Earth receives from the Sun, known as the solar constant (1360 W/m²).
Explanation:
The distance from the Earth to the Sun is about 1.50×1011 meters. The total power radiated by the Sun can be determined by using the following physics principle: The Sun, behaving as a perfect black body with an emissivity of exactly 1, radiates power uniformly across its surface area. The power radiated per square meter on the Sun's surface is found to be 6.3×107 W/m². The sun's radius is approximately 7.00×108 meters. Using the formula Power = Area × Intensity, we calculate the Sun's total power output.
The area of a sphere is given by 4πR2, where R is the radius of the sphere. Therefore, the total power output of the Sun is 4πR2σT4, which calculates to 3.82×1026 W. This immense amount of power is what we refer to as the solar luminosity.
At the distance of the Earth, this power is spread over a spherical area with a radius equal to the Earth-Sun distance. We can find the power per square meter at this distance, known as the solar constant, which is approximately 1360 W/m².
A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
a) What is the magnitude of the force of the wall on ladder?b) What is the magnitude of the normal force of the ground on ladder?c) What is the minimum coefficient of friction so the ladder does not slip?d) What is the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder?
A) Force of the wall on the ladder: 186.3 N
B) Normal force of the ground on the ladder: 725.2 N
C) Minimum value of the coefficient of friction: 0.257
D) Minimum absolute value of the coefficient of friction: 0.332
Explanation:
a)
The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:
[tex]W=mg[/tex]: weight of the ladder, with m = 20 kg (mass) and [tex]g=9.8 m/s^2[/tex] (acceleration of gravity)
[tex]W_M=Mg[/tex]: weight of the person, with M = 54 kg (mass)
[tex]N_1[/tex]: normal reaction exerted by the wall on the ladder
[tex]N_2[/tex]: normal reaction exerted by the floor on the ladder
[tex]F_f = \mu N_2[/tex]: force of friction between the floor and the ladder, with [tex]\mu[/tex] (coefficient of friction)
Also we have:
L = 4.1 m (length of the ladder)
d = 3.0 m (distance of the man from point A)
Taking the equilibrium of moments about point A:
[tex]W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}[/tex]
where
[tex]Wsin 21^{\circ}[/tex] is the component of the weight of the ladder perpendicular to the ladder
[tex]W_M sin 21^{\circ}[/tex] is the component of the weight of the man perpendicular to the ladder
[tex]N_1 sin 69^{\circ}[/tex] is the component of the normal force perpendicular to the ladder
And solving for [tex]N_1[/tex], we find the force exerted by the wall on the ladder:
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N[/tex]
B)
Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of [tex]N_2[/tex].
We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.
Therefore, we have:
[tex]\sum F_y = 0\\N_2 - W - W_M =0[/tex]
And substituting and solving for N2, we find:
[tex]N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N[/tex]
C)
Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.
The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.
Therefore, we can write:
[tex]\sum F_x = 0\\F_f - N_1 = 0[/tex]
And re-writing the equation,
[tex]\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257[/tex]
So, the minimum value of the coefficient of friction is 0.257.
D)
Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.
From part C), we saw that the coefficient of friction can be written as
[tex]\mu = \frac{N_1}{N_2}[/tex]
This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}[/tex]
We see that this quantity is maximum when d is maximum, so when
d = L
Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)[/tex]
And substituting, we get
[tex]N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N[/tex]
And therefore, the minimum coefficient of friction in order for the ladder not to slip is
[tex]\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332[/tex]
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A vehicle of mass 1,500 kg is traveling at a speed of 50 km/hr. What is the kinetic energy stored in its mass? Calculate the energy that can be recovered by slowing the vehicle to a speed of 10 km/hr.
Answer:
Explanation:
Given
mass of vehicle [tex]m=1500\ kg[/tex]
Speed of vehicle [tex]u=50\ km/hr\approx 13.89\ m/s[/tex]
Kinetic Energy Possessed by mass
[tex]K_i=\frac{1}{2}mu^2[/tex]
[tex]K_i=\frac{1}{2}\times 1500\times (13.89)^2[/tex]
[tex]K_i=144.69\ kJ[/tex]
when vehicle is slowed down to speed of [tex]v=10\ km/hr\approx 2.78\ m/s[/tex]
Kinetic Energy at this speed
[tex]K_f=\frac{1}{2}mv^2[/tex]
[tex]K_f=\frac{1}{2}\times 1500\times (2.78)^2[/tex]
[tex]K_f=5.78\ kJ[/tex]
Energy Recovered [tex]=K_i-K_f[/tex]
Energy Recovered[tex]=144.69-5.78=138.9\ kJ[/tex]
How strong is the attractive force between a glass rod with a 0.700 μC charge and a silk cloth with a –0.600 μC charge, which are 12.0 cm apart, using the approximation that they act like point charges?
Answer:
[tex]F=0.26N[/tex]
Explanation:
Assuming that thet act like point charges, the attractive force is given by Coulomb's law:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
Where k is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the magnitudes of the point charges and d is the distance of separation between them. Thus, we replace the given values and get how strong is the attractive force between them:
[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(0.7*10^{-6}C)(-0.6*10^{-6}C)}{(12*10^{-2}m)^2}\\F=0.26N[/tex]
A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , where x is in cm and t is in seconds. What is the frequency of the oscillation?
Answer:
1/2 Hz
Explanation:
A simple harmonic motion has an equation in the form of
[tex]x(t) = Acos(\omega t - \phi)[/tex]
where A is the amplitude, [tex]\omega = 2\pi f[/tex] is the angular frequency and [tex]\phi[/tex] is the initial phase.
Since our body has an equation of x = 5cos(π t + π/3) we can equate [tex]\omega = \pi[/tex] and solve for frequency f
[tex]2\pi f = \pi[/tex]
f = 1/2 Hz
0.5Hz
Explanation:The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;
x = A cos (ωt ± ∅) --------------------------(i)
where;
A = amplitude of the wave
ω = angular velocity of the wave
∅ = phase constant of the wave
From the question;
x = 5cos(π t + π/3) -----------------------------(ii)
Comparing equations (i) and (ii), the following deductions among others can be made;
A = 5cm
ω = π
But the angular velocity (ω) of the wave is related to its frequency (f) as follows;
ω = 2 π f --------------------(iii)
Substitute the value of ω = π into equation (iii) as follows;
π = 2 π f
Divide through by π;
1 = 2f
Solve for f;
f = 1/2
f = 0.5
Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz
Given two vectors A⃗ =4.00i^+7.00j^ and B⃗ =5.00i^−2.00j^ , find the vector product A⃗ ×B⃗ (expressed in unit vectors).
Answer:
[tex]-43\hat{k}[/tex]
Explanation:
given,
[tex]\vec{A} = 4 \hat{i} + 7 \hat{j}[/tex]
[tex]\vec{B} = 5 \hat{i} - 2 \hat{j}[/tex]
vector product [tex] \vec{A} \times \vec{B} = ?[/tex]
[tex]\vec{A} \times \vec{B}[/tex] = [tex]\begin{bmatrix}i & j & k\\ 4 & 7 &0 \\ 5 & -2 & 0\end{bmatrix}[/tex]
now, expanding the vector
[tex]\vec{A} \times \vec{B}= \hat{k}(-2\times 4 - 7\times 5)[/tex]
[tex]\vec{A} \times \vec{B}= -43\hat{k}[/tex]
the vector product is equal to [tex]-43\hat{k}[/tex]
How far apart are two conducting plates that have an electric field strength of 4.5 × 103V/m between them, if their potential difference is 12.5 kV?
Answer:
Explanation:
Given
Electric Field Strength [tex]E=4.5\times 10^{3}\ V/m[/tex]
Potential Difference between Plates is given by [tex]V=12.5\ kV[/tex]
In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates
The potential difference is given by
[tex]\Delta V=Ed[/tex]
where E=Electric Field strength
d=distance between Plates
[tex]d=\frac{\Delta V}{E}[/tex]
[tex]d=\frac{12.5\times 10^3}{4.5\times 10^{3}}[/tex]
[tex]d=2.78\ m[/tex]
The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is sufficiently loud to cause permanent hearing damage if the operator doesn't wear ear protection. What are (a) the sound intensity and (b) the sound intensity level for a person watching from 50 m away?
Answer:
(a) [tex]I_{1}=3.2*10^{-3}W/m^{2}[/tex]
(b) [tex]\beta =95dB[/tex]
Explanation:
Given data
Distance r₁=50 m
Distance r₂=2 m
Intensity I₂=2.0 W/m²
To find
(a) The Sound Intensity I₁
(b) The Sound Intensity level β
Solution
For (a) the Sound Intensity I₁
[tex]\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}[/tex]
For (b) the Sound Intensity level β
The Sound Intensity level β is calculated as follow
[tex]\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB[/tex]
At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph ?
Answer:
T= 89.25 K
Explanation:
Given that
V= 590 mph
We know that
1 mph = 0.44 m/s
That is why ,V= 263.75 m/s
We know that speed of the gas molecule is given as
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
R= 8.314 J/mol.k
M= 32 g/mol = 0.032 kg/mol
T=Temperature in Kelvin unit
[tex]V^2=\dfrac{3RT}{M}[/tex]
[tex]T=\dfrac{V^2\times M}{3R}[/tex]
[tex]T=\dfrac{263.76^2\times 0.032}{3\times 8.314}\ K[/tex]
T= 89.25 K
Therefore the average temperature ,T = 89.25 K
Answer:
temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph = 89.24 K
Explanation:
Average speed of oxygen molecule is given by
[tex]v= \sqrt{\frac{3RT}{M} }[/tex]
[tex]590\times0.44704 = 263.75[/tex] m/s
R= 8.314 J/mol K = universal gas constant
M= molecular weight of oxygen = 32 g/mol =0.032 Kg/mol
now plugging these values to find T we get
[tex]263.75=\sqrt{\frac{3(8.314)(T)}{0.032}}[/tex]
solving the above equation we get
T= 89.24 K
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizontal?
Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:
[tex]W_{work}=2.67*10^{-3}J[/tex]
Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
[tex]W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J[/tex]
There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2.0 × 107 m/s in a direction opposite to the field. Its speed increases to 4.0 × 107 m/s over a distance of 1.2 cm. What is the magnitude of the electric field?
Answer:
Explanation:
Given
speed of Electron [tex]u=2\times 10^7\ m/s[/tex]
final speed of Electron [tex]v=4\times 10^7\ m/s[/tex]
distance traveled [tex]d=1.2\ cm[/tex]
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=Final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}[/tex]
[tex]a=5\times 10^{16}\ m/s^2[/tex]
acceleration is given by [tex]a=\frac{qE}{m}[/tex]
where q=charge of electron
m=mass of electron
E=electric Field strength
[tex]5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}[/tex]
[tex]E=248.3\ kN/C[/tex]
Two point charges, +2.20 μC and -8.00 μC, are separated by 2.60 m. What is the electric potential midway between them? Number Units
Answer:
Electric potential, [tex]V=-4.01\times 10^4\ volts[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=2.2\ \mu C[/tex]
Point charge 2, [tex]q_2=-8\ \mu C[/tex]
Distance between charges, d = 2.6 m
We need to find the electric potential midway between them. The electric potential is given by :
[tex]V=\dfrac{kq}{r}[/tex]
In this case, r = 1.3 m (midway between charges)
[tex]V=\dfrac{kq_1}{r}-\dfrac{kq_2}{r}[/tex]
[tex]V=\dfrac{k}{r}(q_1-q_2)[/tex]
[tex]V=\dfrac{9\times 10^9}{1.3}(2.2\times 10^{-6}-8\times 10^{-6})[/tex]
[tex]V=-40153.84\ volts[/tex]
or
[tex]V=-4.01\times 10^4\ volts[/tex]
So, the electric potential midway between the charges is [tex]V=-4.01\times 10^4\ volts[/tex]. Hence, this is the required solution.
Final answer:
The electric potential midway between two point charges of +2.20 μC and -8.00 μC, separated by 2.60 m, is calculated separately for each charge using the formula V = kq/r and summed up. The total electric potential at the midpoint is -4.00 × 10⁴ V.
Explanation:
To find the electric potential midway between two point charges, we need to consider the contribution from each charge separately and then sum them up.
The electric potential V due to a single point charge q at a distance r is given by the formula:
V = kq/r
where k is the Coulomb's constant (k ≈ 8.99 × 109 Nm²/C²).
In this case, we have two charges, +2.20 μC and -8.00 μC, and they are separated by 2.60 m. So the distance from the midpoint to each charge is 1.30 m (half of 2.60 m).
Calculating the potential due to the +2.20 μC charge:
V₁ = (8.99 × 109)(+2.20 × 10⁻⁶) / 1.30 = 1.53 × 10⁴ V
And for the -8.00 μC charge:
V₂ = (8.99 × 10⁹)(-8.00 × 10⁻⁶) / 1.30 = -5.53 × 10⁴ V
The total electric potential at the midpoint is the sum of V₁ and V₂:
Vtotal = V₁ + V₂ = 1.53 × 10⁴ V - 5.53 × 10⁴ V = -4.00 × 10⁴V
The electric potential midway between the two charges is -4.00 × 10⁴V.
How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr?
The reaction produces -4.95 kJ of heat when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr.
The equation of the reaction is;
CO(g) + H2(g) -------> CH2O(g)
The heat of reaction is obtained from;
Enthalpy of products - Enthalpy of reactants = (-116kJ/mol) - (-110.5 kJ/mol)
= -5.5 kJ/mol
Number of moles of CO is obtained from;
PV = nRT
P = 112 kPa or 1.1 atm
T = 85°C + 273 = 358 K
n = ?
R = 0.082 atmLK-1mol-1
V = 15.0 L
n = PV/RT
= 1.1 atm × 15.0 L/ 0.082 atmLK-1mol-1 × 358 K
= 0.56 moles
Number of moles of H2
n = PV/RT
P= 744 torr or 0.98 atm
V = 14.4 L
T = 75°C + 273 = 348 K
n = 0.98 atm × 14.4 L/0.082 atmLK-1mol-1 × 348 K
n = 0.49 moles
We can see that H2 is the limiting reactant here hence 0.49 moles of formaldehyde is produced.
If 1 mole of formaldehyde produces -5.5 kJ of heat
0.49 moles of formaldehyde produces -5.5 kJ × 0.49 moles / 1 mole
= -4.95 kJ of heat
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How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 30.0°C greater than when they were laid? Their original length is 30.0 m.
Answer:
1.2 cm
Explanation:
Thermal Expansion
It's the tendency that materials have to change its size and/or shape under changes of temperature. It can be in one (linear), two (surface) or three (volume) dimensions.
The formula to compute the expansion of a material under a change of temperature from [tex]T_o[/tex] to [tex]T_f[/tex] is given by.
[tex]\Delta L=L_o.\alpha .(T_f-T_o)[/tex]
Where Lo is the initial length and [tex]\alpha[/tex] is the linear temperature expansion coefficient, which value is specific for each material. The data provided in the problem is as follows:
[tex]L_o=30\ m,\ T_f-T_o=30^oC,\ \alpha=13\times 10^{-6}\ ^oC^{-1}[/tex]
Computing the expansion we have
[tex]\Delta L=30\times 13\times 10^{-6}(30)=0.0117=1.17\ cm[/tex]
The expansion gap should be approximately 1.2 cm
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 9.1 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
In order for the net effect at point A to be a downward force, the tension in cable AC (T) should be equal to the tension in cable AB (9.1 kN). The magnitude of the resulting downward force (R) would be the sum of the tensions in both cables, thus 2 * 9.1 kN = 18.2 kN.
Explanation:To understand this scenario, it is essential to apply the principles of equilibrium and vector sum in Physics. The tension in the wires can be considered as forces experienced by point A. According to the question, the net effect of these forces should be a downward force, implying that they should negate the opposite upward force.
To find the tension T in cable AC, it's logical to assume that the force due to this tension needs to be equal and opposite to the force exerted by the tension in wire AB, which is 9.1 kN. Therefore, T should also be 9.1 kN for the net effect at point A to be a downward force.
The magnitude R of the downward force can be determined by considering the combined effect of tensions in cables AB and AC. Since point A is in equilibrium, R will be the result of the total upward forces. Hence, R is equal to 2 times the tension in any one cable (as they are equal), which gives us R = 2 * 9.1 kN = 18.2 kN.
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An exploration submarine should be able to descend 1200 m down in the ocean. If the ocean density is 1020 kg/m3, what is the maximum pressure on the submarine hull?
Answer:
11995200 N/m²
Explanation:
Pressure: The is the ratio of force to the surface area in contact. The S.I unit of pressure is N/m².
Generally pressure in fluid can be expressed as
P = ρgh......................... Equation 1
Where P = maximum pressure on the submarine hull, ρ = Density of ocean, h = depth of ocean, g = acceleration due to gravity.
Given: h = 1200 m, ρ = 1020 kg/m³
Constant: g = 9.8 m/s²
Substitute into equation 1
P = 1200(1020)(9.8)
P = 11995200 N/m²
Hence the maximum pressure on the submarine hull = 11995200 N/m²
Final answer:
The maximum pressure on a submarine hull that descends 1200 m in the ocean, with ocean density of 1020 kg/m³, is calculated using the formula P = [tex]P_{o}[/tex] + ρgh. With an atmospheric pressure of 101325 Pa, the final pressure would be 12116485 Pa at that depth.
Explanation:
To calculate the maximum pressure on a submarine hull that can descend 1200 m in the ocean, we use the formula for the pressure exerted by a fluid at a certain depth. The formula is P = [tex]P_{o}[/tex] + ρgh, where P is the pressure at depth, [tex]P_{o}[/tex] is the atmospheric pressure on the surface, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Atmospheric pressure [tex]P_{o}[/tex] is approximately 101325 Pa (or 1 atm). The density of sea water is given as 1020 kg/m³ and the depth is 1200 m. Taking g as 9.8 m/s², the standard acceleration due to gravity, we can substitute the values into the formula:
P = 101325 Pa + (1020 kg/m³)×(9.8 m/s²)×(1200 m)
P = 101325 Pa + 12003360 Pa
P = 12116485 Pa
Therefore, the maximum pressure on the submarine hull at a depth of 1200 m would be 12116485 Pascal (Pa).
A lightbulb with an intrinsic resistance of 270 \OmegaΩ is hooked up to a 12-volt battery. How much power is output by the lightbulb? Give your answer out to the thousandths place in units of watts (W).
Answer:
P = 0.533 W
Explanation:
given,
Resistance of the bulb, R = 270 Ω
Potential of the battery, V= 12 V
Power output of the bulb = ?
we know,
P = I² R
also, V = IR
[tex]P = \dfrac{V^2}{R}[/tex]
[tex]P =\dfrac{12^2}{270}[/tex]
[tex]P =\dfrac{144}{270}[/tex]
P = 0.533 W
Hence, the Power delivered by the bulb is equal to 0.533 W.
The destination airport has one runway, 08-26, and the wind is calm. The normal approach in calm wind is a left hand pattern to runway 08. There is no other traffic at the airport. A thunderstorm about 6 miles west is beginning to develop to its mature stage, and rain is starting to reach the ground. The pilot decides to a. Fly normal approach to runway 8b. Fly a right hand approach to runway 8c. Fly the approach to runway 26d. Fly to the west for a fun ride
Answer:
C
Explanation:
Correct answer: C. Fly the approach to runway 26. The winds from the storm could suddenly gust up and you don’t want to be in a short final or even in the flare when a tailwind from a storm gusts up. No other traffic and calm winds currently means that you can land on any runway you want. Go for the safe bet and land on 26.
In this scenario, the pilot should fly a left hand approach to runway 08.
Explanation:In this scenario, with a calm wind, the pilot should fly a left hand approach to runway 08.
Since the wind is calm, there is no need for the pilot to adjust the approach. Flying in a clockwise pattern to runway 08 is the normal procedure.
Choosing any other option would not be appropriate or necessary.
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How much stronger is the gravitational pull of the Sun on Earth, at 1 AU, than it is on Saturn at 10 AU?
A power supply maintains a potential difference of 52.3 V 52.3 V across a 1570 Ω 1570 Ω resistor. What is the current in the resistor?
Answer:
0.033 A
Explanation:
Current: This can be defined as the rate of flow of electric charge in a circuit.
The S.I unit of current is Ampere (A)
From Ohm's law.
V = IR ............................ Equation 1
Where V = Potential difference, I = current, R = resistance.
Making I the subject of the equation,
I = V/R................... Equation 2
Given: V = 52.3 V, R = 1570 Ω
Substitute into equation 2
I = 52.3/1570
I = 0.033 A.
Hence the current in the resistor = 0.033 A
A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibrationa. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelengthe. The length of the string is equal to four times the wavelength
Answer:
d. The length of the string is equal to one-half of a wavelength
Explanation:
A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration
a. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelength
e. The length of the string is equal to four times the wavelength
A stretched string of length L fixed at both ends is vibrating in its third harmonic H
How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration
d. The length of the string is equal to one-half of a wavelength
There are two points during vibration , the node and the antinode
the node is the point where the amplitude is zero.
from the third harmonics, there are two nodes. The first node is half of the wavelength which is the closest to the fixed point.
for third harmonics=3/2lamda
A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s?
Answer:
Frequency of the vibration due to crevice is 1500 Hz.
Explanation:
Frequency is defined as rate of vibration per second. Its S.I. unit is s⁻¹ or Hz.
According to the problem, we have to find number of crevice car makes in 1 second.
Speed of car, u = 30 m/s
Distance covered by car in 1 second, d = 30 m
For every 0.02 m, one crevice occurred by the tire of car.
Number of crevice occurred in 1 second by the car = [tex]\frac{30}{0.02}[/tex]
= 1500
Since, each crevice makes a single vibration. Thus, the frequency of these vibrations is 1500 Hz.
The frequency of the vibrations generated by the tire treads in this case, given that each crevice is 2.00 cm apart and the car moves at 30.0 m/s, is 1500 Hz.
Explanation:In this scenario, it is important to understand that the frequency of the vibrations is determined by the speed of the tire and the tread pattern. Here, the crevice, which causes the vibration, occurs every 2.00 cm (or 0.02 m). This means for every meter the car moves, 0.02 m into 1 m gives a total of 50 vibrations. Therefore, at a speed of 30.0 m/s, we will have 50 vibrations/m multiplied by 30 m/s, giving a frequency of 1500 Hz or vibrations per second.
Frequency in this question refers to the number of times the vibration from the crevices occur in a unit of time (in this case, per second). This concept is crucial in understanding wave motions and vibration patterns, and is often applied in physics-related disciplines.
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How do astronomers set about looking for extrasolar planets?
Explanation:
Implicit techniques for the discovery of extra-solar planets are used by astronomers. Evidence from the radial velocity of a change in the rotation of the star indicates the presence of a planet. If, with reference to Earth, the inclination of the planet's orbit is viewed as an edge-on, the detection of light originating from the star throughout its planetary transit would then be a verification.
By what factor does the energy of a 1-nm X-ray photon exceed that of a 10-MHz radio photon? How many times more energy has a 1-nm gamma ray than a 10-MHz radio photon?
To solve this problem we will apply the concepts related to the relationship between energy and frequency, from the latter we will obtain similar expressions that relate to the wavelength to find the two energy states between the given values. Finally we will make the comparative radius between the two. The relation between energy and frequency is given as,
[tex]E = hf[/tex]
Here,
E = Energy
h = Planck's constant
The relation between the speed of the electromagnetic waves (c), frequency (f) and wavelength ([tex]\lambda[/tex] ) is,
[tex]c = f\lambda[/tex]
Rearrange the above equation for frequency f as follows
[tex]f = \frac{c}{\lambda}[/tex]
Substitute,
[tex]E = h\frac{c}{\lambda}[/tex]
The wavelength x-ray or gamma ray photon is
[tex]\lambda = 1.0nm (\frac{1nm}{10^{9}nm})[/tex]
[tex]\lambda = 10^{-9} m[/tex]
Therefore the energy would be,
[tex]E_1 = \frac{hc}{\lambda}[/tex]
[tex]E_1 = \frac{(6.63*!0^{-34}J\cdo s)(3*10^{8}m/s)}{10^{-9}m}[/tex]
[tex]E_1 = 19.89*10^{-17} J[/tex]
The frequency is given as,
[tex]f = 10MHz (\frac{10^6z}{1.0MHz})[/tex]
[tex]f = 10^7Hz[/tex]
Now the second energy would be
[tex]E_2 = hf[/tex]
[tex]E_2 = (6.63*10^{-27}J\cdot s)(10^7Hz)[/tex]
[tex]E_2 = 6.63*10^{-27}J[/tex]
Therefore the ratio between them is
[tex]\frac{E_1}{E_2} = \frac{19.89*10^{-17}J}{6.63*10^{-27}J}[/tex]
[tex]\frac{E_1}{E_2} = 3*10^{20}[/tex]
Therefore the energy of 1nm x ray or gamma ray photon is [tex]3*10^{20}[/tex] times more than energy of 10MHz radio photon
Which of the following statements about insulating materials is correct?a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.b.The electric field from a charged object is able to penetrate through an insulating material.c.Insulating materials exhibit a linear relationship between voltage and current.d.Silver and copper are good insulators.
Answer:
a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.
Explanation:
A conductor has low resistance, while an insulator has much higher resistance. Devices called resistors control amounts of resistance into an electrical circuits. Electric charges inside an insulator are bound to the individual atoms or molecules, not being able to move inside the material.
When you turn up the resistance, the electric current flowing through the circuit is reduced
Ohm's law:
V = I * R
R = V/I
An Ohmic conductor would have a linear relationship between the current and the voltage. With non-Ohmic conductors, the relationship is not linear. Most metals are good conductors example silver, copper etc.
A 12,000-lb bus collides with a 2800-lb car. The velocity of the bus before the collision is vB = 18i (ft/s) and the velocity of the car is vC =33j (ft/s). The two vehicles become entangled and remain together after the collision. The coefficient of kinetic friction between the vehicles’ tires and the road is μk = 0.6. Find (a) velocity of combined center of mass immediately after the collision, (b) final position (assume brakes are on and you have skidding, not rolling). (c) If the collision lasted 0.5 seconds, what was the impulsive force of the bus on the car?
Answer:
(a) 20.84 ft/s
(b) 11.24 ft
(c) -68160 N
Explanation:
Parameters given:
Mass of Bus, Mb = 12000 lb
Mass of car, Mc = 2800 lb
Initial speed of bus before collision, u(b) = 18 ft/s
Initial speed of car before collision, u(c) = 33 ft/s
Coefficient of friction, μk = 0.6
(a)Combined velocity after collision.
Since the bus and car are entangled and move together after the collision, they have the same velocity after collision.
Using the law of conservation of momentum, we have:
Mb*u(b) + Mc*u(c) = Mb*v(b) + Mc*v(c)
Where v(b) = velocity of the bus,L after collision,
v(c) = velocity of the car after collision.
Since v(b) = v(c) = v,
Mb*u(b) + Mc*u(c) = (Mb + Mc)*v
=> (12000 * 18) + (2800 * 33) = (12000 + 2800) * v
=> 308400 = 14800 * v
=> v = 20.84 ft/s
Combined velocity after collision is 20.84 ft/s
(b) The force acting on the bus and car after collision is given as:
F = m*a
Where F = force,
m = combined mass of bus and car,
a = acceleration of the bus and car after collision.
We know that the only force acting on the combined mass of the bus and car is the Frictional force, Fr and it is given as:
Fr = μk*m*g
Where g = acceleration due to gravity
Hence,
Fr = ma
=> - μk*m*g = m*a
The negative sign signifies that the fictional force is acting in the opposite direction to the lotion.
=> a = - μk*g
a = - 0.6 * 32.2
a = - 19.32 ft/s^2
Using one of the equations of linear motion, we can find the distance moved by the car and bus after collision:
v*v = u*u + 2*a*s
Where s = distance moved
The final velocity, v, of the car and bus is 0 because they come to rest and the initial velocity, u, is 20.84 ft/s, the velocity of the car and bus after collision.
Hence,
0 = (20.84*20.84) + (-2*19.32*s)
=> s = -434.3056/-38.64
s = 11.24 ft
(c) Impulsive force is the force that two bodies which are colliding exert on one another. It is given mathematically as
I. F. = (momentum change)/time
Momentum change of the bus is:
Momentum change = final momentum - initial momentum
Momentum change = Mb*v(b) - Mb*u(b)
Momentum change = (12000*20.84) - (12000*18)
Momentum change = 250080 - 216000
Momentum change = - 34080 kgft/s
=> I. F. = - 34080/0.5
I. F. = - 68160 N
The Impulsive force of the bus on the car is - 68160N
The negative sign means the force is acting opposite to the motion of the car.