Answer:
Double the charge to 2Q
Explanation:
The magnitude of a electric field caused by a charge Q at a distance R can be find by the following expression:
[tex]E = K\frac{Q}{R^2}[/tex]
Where K is the Coulomb constant.
As the relationship between the charge and the electric fiel is proportional, by simply doubling the charge, you would double the magnitude of the electric field.
Notice that the distance affects the magnitude of the electric field by the inverse square. So if you half the distance, the magnitude of the field will quadruple.
To double the electric field's magnitude at point P from a point charge Q at distance R, double the charge to 2Q while keeping the distance R unchanged. Option B is the correct option.
To double the magnitude of the electric field at point P due to a point charge Q a distance R away from P, you can follow Coulomb's law. Electric field (E) is directly proportional to the charge (Q) and inversely proportional to the square of the distance (R) between the point charge and the observation point.
If you double the charge to 2Q while keeping the distance R the same, the new electric field magnitude at point P will be double the original E. This is because E ∝ Q. Other options like reducing the distance to R/2 or R/4, or doubling the distance to 2R, would not result in a doubling of the field magnitude but rather lead to different field strengths according to the inverse square law, and altering both the charge and distance simultaneously is not necessary to achieve this specific goal. Option B is the correct option.
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The appropriate question is :
The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2 OR
A) reduce the distance to R/4.
B) double the charge to 2Q.
C)reduce the distance to R/2.
D) double the distance to 2R.
If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
Answer:
Frequency of the wave will be 31.84 Hz
Explanation:
We have given the equation [tex]y=0.02sin(30x-200t)[/tex]-----eqn 1
The standard equation of sine wave is given by [tex]y=Asin(kx-\omega t)[/tex]----eqn 2
On comparing eqn 1 and eqn 2
[tex]\omega =200[/tex]
Angular frequency [tex]\omega[/tex] is given by [tex]\omega =2\pi f[/tex]
So [tex]200=2\times 3.14\times f[/tex]
[tex]f=31.84Hz[/tex]
5. Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make angles of 52 and 40 with the horizontal. Find the tension in each wire and the magnitude of each tension. Hint: The lengths of the wires have nothing to do with the tension in each wire. slader
The tension in the rope must equal the weight of the supported mass. For the given situation, the tension can be calculated by multiplying the mass by the acceleration due to gravity.
Explanation:Tension in the rope must equal the weight of the supported mass, as we can prove using Newton's second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus, T - w = 0.
Thus, for a 5.00-kg mass (neglecting the mass of the rope), we can find that T = mg = (5.00 kg) (9.80 m/s²) = 49.0 N.
Thus, Tension in the rope equals the weight of the supported mass (5.00 kg) due to Newton's second law, making T = mg = 49.0 N.
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A computer hard drive has a disc that rotates at 10,000 rpm. The reader head is positioned 0.0005 in. above the disc’s surface. Estimate the shear force on the reader head due to the air between the disc and head.
Answer:
shearing force is [tex]3.40\times 10^{-4} lb[/tex]
Explanation:
we know that
force can be determined [tex]F = \tau \times A[/tex]
Area can be determine as
[tex]A = \frac{\pi}{4} d^2 = \frac{\pi}{4} [\frac{0.2}{12}]^2 = 2.18\times 10^{-4} ft^2[/tex]
linear velocity can be determines as
[tex]\tau = \mu_{air} \frac{U}{b}[/tex]
dynamic viscosity of air [tex]\mu_{air} = 3.74\times 10^{-7} lb-s/ft^2[/tex]
veolcity of disc
[tex]U =\omega R[/tex]
[tex]U = \frac{2\pi N}{60} \times R = \frac{2\pi 10,000}{60} \times \frac{2}{12}[/tex]
U = 174.5 ft/s
so
[tex]\tau = 3.74\times 10^{-7} \times \frac{174.5}{\frac{0.0005}{12}}[/tex]
[tex]\tau = 1.56 lb/ft^2[/tex]
[tex]F = 1.56\times 2.18\times 10^{-4} = 3.40\times 10^{-4} lb[/tex]
shearing force is [tex]3.40\times 10^{-4} lb[/tex]
A can contains 375 mL of soda. How much is left after 308 mL is removed?
Answer:
Volume left, v = 67 mL
Explanation:
Given that,
Volume of soda contained in a can, V = 375 mL
We need to find the volume of soda left after 308 mL of soda is removed, V' = 308 mL
Let v is the left volume of soda. It can be calculated using simple calculations as :
v = V - V'
v = 375 mL - 308 mL
v = 67 mL
So, the left volume in the can is 67 mL. Hence, this is the required solution.
A car is driven east for a distance of 47 km, then north for 28 km, and then in a direction 35° east of north for 27 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
Answer:
(a) 82 Km
(b) 32°
Explanation:
First you should draw the vectors in the cartesian plane (please see the picture below).
As the car is driven east for a distance of 47 Km, your first vector should be drawn from the origin (0,0) and on the x axis
Then the car is driven north for a distance of 28 Km, so your second vector should be drawn from the origin and on the y axis.
And the car finally goes east of north for 27Km, so the third vector should be drawn from the origin east of north forming an 35° angle with x axis.
Then you should find the components of the vector in x and y:
For Vector 1 ([tex]V_{1}[/tex])
[tex]V_{1x}=47Km[/tex]
[tex]V_{1y}=0[/tex]
For Vector 2 ([tex]V_{2}[/tex])
[tex]V_{2x}=0[/tex]
[tex]V_{2y}=28Km[/tex]
For Vector 3 ([tex]V_{3}[/tex])
[tex]V_{3x}=27cos37^{o}[/tex]
[tex]V_{3x}=21.56[/tex]
[tex]V_{3y}=27sin37^{o}[/tex]
[tex]V_{3y}=16.25[/tex]
To find the magnitud of the car´s total displacement, (R) you should add up all the x and y components.
For the x component:
[tex]R_{x}=V_{1x}+V_{2x}+V_{3x}[/tex]
[tex]R_{x}=47+0+21.56[/tex]
[tex]R_{x}=68.56Km[/tex]
For the x component:
[tex]R_{y}=V_{1y}+V_{2y}+V_{3y}[/tex]
[tex]R_{y}=0+28+16.25[/tex]
[tex]R_{y}=44.25Km[/tex]
Now please see the second picture that is showing the components x and y as the sides of a right triangle, and we are going to use the Pythagorean theorem to find the resultant, R.
[tex]R=\sqrt{R_{x}^{2}+R_{y}^{2}[/tex]
[tex]R=\sqrt{68.56^{2}+44.25^{2}}[/tex]
[tex]R=82Km[/tex]
And to find the angle of the car´s total displacement (α), we use the same right triangle with the relationship between its legs.
[tex]tan(\alpha)=\frac{44.25}{68.56}[/tex]
[tex]tan(\alpha)=0.64[/tex]
[tex]\alpha=tan^{-1}(0.64)[/tex]
[tex]\alpha =32^{o}[/tex]
Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm per year. (a) What distance does it move in 71 seconds at this speed?
m
(b) What is its speed in miles per million years?
mi/My
Answer:
1.35×10⁻⁷ m
37.278 mi/My
Explanation:
Speed of the tectonic plate= 6 cm/yr
Converting to seconds
[tex]6=\frac{6}{365.25\times 24\times 60\times 60}[/tex]
So in one second it will move
[tex]\frac{6}{365.25\times 24\times 60\times 60}[/tex]
In 71 seconds
[tex]71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm[/tex]
The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m
Convert to mi/My
1 cm = 6.213×10⁻⁶ mi
1 M = 10⁶ years
[tex]6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My[/tex]
Speed of the tectonic plate is 37.278 mi/My
The height of a door is measured by four people and their measured values are 217.6 cm, 217.2 cm, 216.8 cm, and 217.9 cm. (a) What is the average value of these four measurements? (Enter your answer to the nearest tenths place.)
(b) What is the standard deviation for the four measurements?
Answer:
a) To the nearest tenth the average value of the measures is 217.4cm
b) The standard deviation for the four measurements is 0.415
Explanation:
The average is the result of adding all the values and dividing by the number of measures, in this case 4 then
[tex]Average = \frac{217.6 + 217.2 + 216.8 + 217.9}{4} \\[/tex]
[tex]Average = \frac{869.5}{4} \\[/tex]
[tex]Average = 217.375 [/tex]
a) To the nearest tenth the average value of the measures is 217.4cm
The standard deviation equals to the square root of the variance, and the variance is the adition of all the square of the difference between the measure and the average for each value divided by the number of measurements
So first, we must calculate the variance:
[tex]Variance = \frac{(217.6-217.4)^2+(217.2-217.4)^2+(216.8-217.4)^2+(217.9-217.4)^2}{4}[/tex].
[tex]Variance = \frac{(0.2)^2+(-0.2)^2+(-0,6)^2+(0.5)^2}{4}[/tex].
[tex]Variance = \frac{0.04+0.04+0.36+0.25}{4}[/tex].
[tex]Variance = \frac{0.69}{4}[/tex].
[tex]Variance = 0.1725[/tex].
This represent the difference between the average and the measurements.
Now calculate the standard deviation
Standard deviation = [tex]\sqrt{Variance}[/tex]
Standard deviation = [tex]\sqrt{0.1725}[/tex]
Standard deviation = 0.415
b) The standard deviation for the four measurements is 0.415
This measure represent a standard way to know what is normal in this sample. so the differences between the average should be of ±0.415
The average value of the four door height measurements is 217.4 cm, and the standard deviation of these measurements is approximately 0.4 cm.
Explanation:To find the average value of the four measurements: 217.6 cm, 217.2 cm, 216.8 cm, and 217.9 cm, you add them up and then divide by the number of measurements, which is four:
Average = (217.6 + 217.2 + 216.8 + 217.9) / 4 = 217.375
Rounded to the nearest tenths place, the average is 217.4 cm.
For the standard deviation, we first calculate the variance. First, find the difference of each measurement from the mean, square that difference, and then find the average of those squared differences. Finally, take the square root of that average to find the standard deviation:
Substituting the values and performing the calculations gives us a standard deviation of approximately 0.4 cm.
A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 18 km in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm^3?
Answer:
[tex]Density=6.51*10^{14}g/cm^{3[/tex]
Explanation:
Sun mass:
Ms=1.989 × 10^30 kg
Neutron star has the same mass.
Radius Neutron:
R=9Km (because diameter is 18Km)
R=9*10^3m
Density neutron star:
[tex]D=Mn/Vol=Ms/(4/3*\pi*R^3)=1.989*10^{30} /(4/3*\pi*(9*10^{3})^{3})=6.51*10^{17}kg/m^{3}[/tex]
[tex]D=6.51*10^{17}kg/m^{3}*(1000g/1Kg)*(1m^{3}/1000000cm^{3})=6.51*10^{14}g/cm^{3[/tex]
The density of a neutron star is about 10¹⁴ grams / cm³.
What is neutron star?Any member of the class of extremely dense, compact stars known as neutron stars is assumed to be predominantly made of neutrons. The average diameter of a neutron star is 18 km (12 miles). Their masses range from 1.18 to 1.97 times those of the Sun, with the majority being 1.35 times the Sun. They have exceptionally high mean densities, almost 10¹⁴ times that of water.
This is similar to the density of an atomic nucleus, and one may think of a neutron star as a massive nucleus. Where the pressure is greatest, at the centre of the star, it is not known for sure what is there; hypotheses include hyperons, kaons, and pions. Most of the particles in the intermediate layers are neutrons, which are likely in a "superfluid" condition.
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If a train is travelling 200km/hour eastward for 1800 seconds how far does it travel?
Answer:
Distance, d = 99990 meters
Explanation:
It is given that,
Speed of the train, v = 200 km/h = 55.55 m/s
Time taken, t = 1800 s
Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]d=v\times t[/tex]
[tex]d=55.55\times 1800[/tex]
d = 99990 m
So, the distance covered by the train is 99990 meters. Hence, this is the required solution.
A compound microscope operated in near point adjustment comprises an objective lens of focal length 5.0 cm and an eyepiece lens of focal length 12.0 cm. If the near-point distance is 25.0 cm, and the overall magnification of the microscope is required to be 12.0 x, how far apart must the lenses be placed?
Answer:
28.8 cm
Explanation:
Magnification in a microscope is:
M = Mo * Me
Where
Mo: magnification of the objective,
Me: Magnification of the eyepiece.
The magnification of the objective if:
Me = npd/fe
Where:
npd: near point distance
fe: focal length of the eyepiece
The magnification of the objective:
Mo = d/fo
Where
d: the distance between lenses
fo: focal length of the objective
Then
M = npd/fe * d/fo
d = M * fe * fo / npd
d = 12 * 5 * 12 / 25 = 28.8 cm
In quantum mechanics, the fundamental constant called Planck's constant, h, has dimensions of [ML^2T^-1 ]. Construct a quantity with the dimensions of length using h, a mass m, and c, the speed of light.
Answer:
h/(m*c)
Explanation:
Hi!
Lets denote the units of X as [X]
Since the dimentions of h are:
[tex][h] = \frac{ML^{2}}{T}[/tex]
If we divide [h] by the units of mass, we get:
[tex]\frac{[h]}{[m]} = \frac{L^{2}}{T}}[/tex]
Also we know that:
[tex][c] = \frac{L}{T}[/tex]
So:
[tex][\frac{h}{mc}] = \frac{L^{2}}{T}}*\frac{T}{L}=L[/tex]
Therefore
h/(mc) has dimentiosn of length
To construct a quantity with the dimensions of length using Planck's constant (h), mass (m), and the speed of light (c), the formula L = h/(mc) can be used. This length scale is relevant in the realms of quantum mechanics and high-energy physics.
Explanation:The student has asked how to construct a quantity with the dimensions of length using Planck's constant (h), a mass (m), and the speed of light (c). To achieve this, we can use the formula for the Planck length given by:
Lp = √hG/c³
Where G is Newton's gravitational constant. However, since we need to construct a length using just h, m, and c without G, we can rearrange the Planck length equation to:
L = h/(mc)
This gives a length L which is dependent on the mass m in addition to Planck's constant h and the speed of light c. This length scale is significant in quantum mechanics and high-energy physics, where extremely small distances are relevant.
A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 70 m above the ground. After an additional 4.70 s, it is 1.15 km above the ground. Part A
Calculate the magnitude of the average velocity of the rocket for the 4.70 s part of its flight.
Express your answer in meters per second. Part B
Calculate the magnitude of the average velocity of the rocket the first 5.85 s of its flight.
Express your answer in meters per second.
Answer:
a) [tex]v=230 m/s[/tex]
b) [tex]v=196.5 m/s[/tex]
Explanation:
a) The formula for average velocity is
[tex]v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }[/tex]
For the first Δt=4.7s
[tex]v=\frac{(1150-70)m}{(4.7)s} =230 m/s[/tex]
b) For the secont Δt=5.85s we know that the displacement is 1150m. So, the average velocity is:
[tex]v=\frac{(1150)m}{(5.85)s}=196.5m/s[/tex]
Which of the following must always be the same: a. Time - a second for example b. Distance - a meter for example The speed of Light - 3E8 m/s d. Weight - how much does an apple weight.
Answer:
The speed of light remains the same.
Explanation:
(a) Time : second, hour minutes are example of time.
(b) Distance : meter, kilometers etc are some example of distance.
(c) The speed of light : It always remains constant. It is equal to [tex]3\times 10^8\ m/s[/tex].
(d) The weight of an object is given by the product of mass of an object and the acceleration due to gravity. As the value of g is not same everywhere, its weight varies.
So, the speed of light is always remains the same. Hence, the correct option is (c).
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.
Answer:
The answer is 9 m.
Explanation:
Using the kinematic equation for an object in free fall:
[tex]y = y_o - v_o-\frac{1}{2}gt^{2}[/tex]
In this case:
[tex]v_o = \textrm{Initial velocity} = 1.5[m/s]\\t = \textrm{air time} = 1.2 [s][/tex][tex]y_o = 0[/tex]
[tex]g = \textrm{gravity} = 9.8 [m/s^{2} ][/tex]
Plugging those values into the previous equation:
[tex]y = 0 - 1.5*1.2-\frac{1}{2}*9.8*1.2^{2} \\y = -8.85 [m] \approx -9 [m][/tex]
The negative sign is because the reference taken. If I see everything from the rescuer point of view.
A vertical container with base area measuring 14 cm by 16 cm is being filled with identical pieces of candy, each with a volume of 50.0 mm^3 and a mass of 0.0200 g. Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of 0.21 cm/s, at what rate does the mass of the candies in the container increase?
Rate of change of mass is given by,
[tex]\frac{dm}{dt}=18.816\,g/s[/tex]
Explanation:In the question,
We have the Base Area of the vertical container = 14 cm x 16 cm
Now,
Let us take the height of the container = h
Rate of change of height with time, dh/dt = 0.21 cm/s = 2.1 mm/s
So,
Volume of the container = Base Area x Height
So,
V = 14 x 16 x h
V = 140 x 160 x h (because, 1 cm = 10 mm)
V = (22400)h
Now,
Volume of one of the candy = 50 mm³
Mass of the candy = 0.0200 g
So,
Density of the candy = Mass/ Volume
So,
[tex]Density=\frac{0.0200}{50}\\Density=0.0004[/tex]
Now,
V = (22400)h
On differentiating with respect to time, t, we get,
[tex]\frac{dV}{dt}=\frac{22400h}{dt}\\\frac{d}{dt}(\frac{mass}{density})=22400.\frac{dh}{dt}\\\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\[/tex]
Therefore, on putting the value of density in the equation and also the value of rate of change of height with time, we get,
[tex]\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\\frac{1}{0.0004}.\frac{dm}{dt}=22400(2.1)\\\frac{dm}{dt}=18.816\,g/s[/tex]
Therefore, the rate of change of mass with respect to time is given by,
[tex]\frac{dm}{dt}=18.816\,g/s[/tex]
What is the acceleration of an electron that has moved between -1.5V and 3.0V? Assume is began at rest.
Answer:
Insufficient data.
Explanation:
Hi!
The data you have is only potential diference, so you know that the variation in potential energy of the electron when moving from -1.5V to 3.0V, is 4.5 V.
But you cannot know the acceleration. For that you need to know the electric field, so you can calculate force.
The density of a nuclear matter is about 10^18kg/m^3. given that 1mL is equal in volume to cm^3, what is the density of nuclear matter in megawatts per micrometer that is Mg/uL?
Answer:
The density of nuclear matter is [tex]10^{6}\ Mg/\mu L[/tex]
Explanation:
Given that,
Density [tex]\rho= 10^{18}\ kg/m^3[/tex]
Using unit conversation,
[tex]\Rightarrow \dfrac{10^{18}\ kg}{1\ m^3}\times\dfrac{1 Mg}{10^{6}g}\times\dfrac{1000\ g}{1\ kg}\times\dfrac{1\ m^3}{10^6\ cm^3}\times\dfrac{1\ cm^3}{1\ mL}\times\dfrac{1000\ mL}{1\ L}\times\dfrac{10^{-6}\ L}{1\ \muL}[/tex]
[tex]\Rightarrow \dfrac{10^{18}\times10^{3}\times10^{3}\times10^{-6}}{10^{6}\times10^{6}\times10^{6}}[/tex]
The density of nuclear matter is
[tex]\rho=10^{6}\ Mg/\mu L[/tex]
Hence, The density of nuclear matter is [tex]10^{6}\ Mg/\mu L[/tex]
The electric field at 4 cm from the center of a long copper rod of radius 2 cm has a magnitude of 4 N/C and is directed outward from the axis of the rod. (a) How much charge per unit length (in C/m) exists on the copper rod?
(b) What would be the electric flux (in N · m^2/C) through a cube of side 3 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
To find the charge per unit length on the copper rod, use the formula E = k * λ / r. Plug in the given values to find the charge per unit length. The electric flux through the cube can be calculated by multiplying the electric field by the area of one face of the cube.
Explanation:(a) To find the charge per unit length on the copper rod, we can use the formula for an electric field (E = k * λ / r), where E is the electric field, k is the electrostatic constant, λ is the charge per unit length, and r is the distance from the center of the rod. Rearranging the formula, we can solve for λ: (λ = E * r / k). Plugging in the given values, we have: λ = (4 N/C) * (0.04 m) / (9 x 10^9 Nm^2/C^2) = 1.78 x 10^-10 C/m.
(b) The electric flux through a surface is given by the formula (Φ = E * A), where Φ is the electric flux, E is the electric field, and A is the area of the surface. In this case, the electric flux through the cube can be calculated by multiplying the electric field (4 N/C) by the area of one face of the cube (3 cm)^2 = (0.03 m)^2 = 9 x 10^-4 m^2. Therefore, the electric flux is Φ = (4 N/C) * (9 x 10^-4 m^2) = 3.6 x 10^-3 Nm^2/C.
(a) The charge per unit length on the copper rod is [tex]\( {8.90 \times 10^{-10} \, \text{C/m}} \).[/tex]
(b) The electric flux through the cube is [tex]\( {0.0036 \, \text{N} \cdot \text{m}^2/\text{C}} \).[/tex]
(a): Charge per Unit Length on the Copper Rod
The electric field ( E ) at a distance ( r ) from the axis of a long charged rod is given by:
[tex]\[E = \frac{2k\lambda}{r}\][/tex]
From the given data:
[tex]\[4 = \frac{2 \cdot 8.99 \times 10^9 \cdot \lambda}{0.04}\][/tex]
Solving for [tex]\( \lambda \):[/tex]
[tex]\[\lambda = \frac{4 \cdot 0.04}{2 \cdot 8.99 \times 10^9}\][/tex]
[tex]\[\lambda = \frac{0.16}{1.798 \times 10^9}\][/tex]
[tex]\[\lambda \approx 8.90 \times 10^{-10} \, \text{C/m}\][/tex]
(b): Electric Flux through a Cube
To find the electric flux [tex]\( \Phi_E \)[/tex] through the cube, we use Gauss's law, which states:
[tex]\[\Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}\][/tex]
Since the rod is long and the cube is oriented such that the rod passes through opposite sides perpendicularly, the electric flux [tex]\( \Phi_E \)[/tex] through the cube is:
[tex]\[\Phi_E = E \cdot A\][/tex]
Calculate ( A ):
[tex]\[A = a^2 = (0.03)^2 = 0.0009 \, \text{m}^2\][/tex]
Now, calculate [tex]\( \Phi_E \):[/tex]
[tex]\[\Phi_E = 4 \times 0.0009\][/tex]
[tex]\[\Phi_E = 0.0036 \, \text{N} \cdot \text{m}^2/\text{C}\][/tex]
An architect is redesigning a rectangular room on the blueprints of a house. She decides to double the width of the room, increase the length by 50 percent, and increase the height by 20 percent. By what factor has the volume of the room increased?
Answer:3.6
Explanation:
Given
Architect decide to double the width
length increases by 50 percent
height increases by 20 percent
Let L,B,H be length,height and width of rectangular room
L'=1.5L
B'=2B
H'=1.2H
therefore new volume V'=L'\times B'\times H'
[tex]V'=1.5L\times 2B\times 1.2H[/tex]
V'=3.6LBH
new volume is 3.6 times the original
What is the potential energy at the origin due to an electric field of 5 x 10^6 N/C located at x=43cm,y=28cm?
Answer:
potential energy at origin is [tex]2.57*10^{6} volt[/tex]
Explanation:
given data:
electric field E = 5*10^{6} N/C
at x = 43 cm, y = 28 cm
distance btween E and origin
[tex]\Delta r = \sqrt{43^2 +28^2}[/tex]
[tex]\Delta r = 51.313 cm[/tex]
potential energy per unit charge [tex]\Delta V = - Edr[/tex]
[tex]\Delta V = 5*10^6*51.313*10^{-2} J/C[/tex]
[tex]\Delta V = 2.57*10^{6} volt[/tex]
potential energy at origin is 2.57*10^{6} volt
For a monatomic ideal gas, temperature is proportional to : the square of the average atomic velocity.
the average atomic velocity.
the atomic mean free path.
the number of atoms.
Answer:
the square of the average atomic velocity.
Explanation:
From the formulas for kinetic energy and temperature for a monoatomic gas, which has three translational degrees of freedom, the relationship between root mean square velocity and temperature is as follows:
[tex]v_{rms}=\sqrt{\frac{3RT}{M}}[/tex] (1)
Where [tex]v_{rms}[/tex] is the root mean square velocity, M is the molar mass of the gas, R is the universal constant of the ideal gases and T is the temperature.
The root mean square velocity is a measure of the velocity of the particles in a gas. It is defined as the square root of the mean square velocity of the gas molecules:
[tex]v_{rms}=\sqrt{<v^2>}[/tex] (2)
substituting 2 in 1, we find the relationship between mean square speed and temperature:
[tex]\sqrt{<v^2>}=\sqrt{\frac{3RT}{M}}\\T=\frac{M<v^2>}{3R}\\\\T\sim <v^2>[/tex]
Final answer:
The temperature of a monatomic ideal gas is directly proportional to the average atomic velocity.
Explanation:
The temperature of a monatomic ideal gas is directly proportional to the average atomic velocity. As the temperature increases, the average atomic velocity also increases. This relationship is a result of the fact that temperature is a measure of the kinetic energy of the gas particles, and the average velocity is related to the kinetic energy. Therefore, temperature and average atomic velocity are directly proportional in a monatomic ideal gas.
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much time will elapse before it returns to its starting point?
Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × [tex]10^{7}[/tex] m/s
uniform electric field E = 1.18 × [tex]10^{4}[/tex] N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × [tex]10^{-31}[/tex] kg ×a = 1.18 × [tex]10^{4}[/tex] × 1.602 × [tex]10^{-19}[/tex]
a = 20.75 × [tex]10^{14}[/tex] m/s²
so acceleration is 20.75 × [tex]10^{14}[/tex] m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × [tex]10^{7}[/tex] = 0 + 20.75 × [tex]10^{14}[/tex] (t)
t = 11.80 × [tex]10^{-9}[/tex] s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 × [tex]10^{-9}[/tex]
time is 23.6 × [tex]10^{-9}[/tex] s
time will elapse before it return to its staring point is 23.6 ns
A concave mirror produces a real image that is three times as large as the object. If the object is 20 cm in front of the mirror, what is the image distance? What is the focal length of this mirror?
The distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.
What is focal length of the lens?The focal length of the lens is length of the distance between the middle of the lens to the focal point.
It can be find out using the following formula as,
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.
Here, the concave mirror produces a real image that is three times as large as the object. The object is 20 cm in front of the mirror, and the concave mirror produces a real image that is three times as large as the object.
Hence, the value of magnification of the mirror is 3.
The object distance is 20 cm thus the image distance, using the magnification formula, can be given as,
[tex]m=\dfrac{v}{u}\\3=\dfrac{v}{20}\\v=60\rm cm[/tex]
Put the values in the lens formula as,
[tex]\dfrac{1}{60}+\dfrac{1}{20}=\dfrac{1}{f}\\f=30 \rm \; cm[/tex]
Hence, the distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.
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Final answer:
Using the mirror equation and magnification formula, the image distance is found to be -60 cm and the focal length of the concave mirror is calculated to be 30 cm.
Explanation:
The question involves determining the image distance and the focal length of a concave mirror that produces a real image three times larger than the object. The object is placed 20 cm in front of the mirror. To find these values, we can use the mirror equation which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.
Given that the magnification (m) is -3 (negative sign since the image is real and inverted), and magnification is also equal to -di/do, we can express di as -3do. We are told that do is 20 cm, so di is -3(20 cm) = -60 cm (negative because the image is on the same side as the object).
Now, we can plug these values into the mirror equation to find the focal length (f):
1/f = 1/do + 1/di1/f = 1/20 + 1/(-60)1/f = 1/20 - 1/601/f = (3-1)/601/f = 2/60f = 60/2f = 30 cmTherefore, the image distance is -60 cm and the focal length of the concave mirror is 30 cm.
A squirrel runs along an overhead telephone wire that stretches from the top of one pole to the next. The creature is initially at position xi = 3.65 m, as measured from the center of the wire segment. It then undergoes a displacement of Δx = -6.81 m. What is the squirrel\'s final position xf? xf = _____ m
Answer:
xf = - 3.16 m
Explanation:
the squirrel was initially in the position xi = 3.65 m, then it had a displacement of Δx = -6.81 m.
The negative sign indicates that it moved in the opposite direction, so we must subtract this displacement Δx = -6.81 m to the initial position xi = 3.65 m, to find its final position.
3.65 m - 6.81 m = - 3.16 m
Two point charges totaling 8.90 μC exert a repulsive force of 0.120 N on one another when separated by 0.460 m. What is the charge on each? Assume that the two charges are negative. What is the charge on each if the force is attractive?
Explanation:
Let [tex]q_1[/tex] is the first charge and [tex]q_2[/tex] is the second charge.
Force between them, F = 0.12 N
Distance between charges, d = 0.46 m
(a) Force acting between two point charges is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]q_1q_2=\dfrac{Fd^2}{k}[/tex]
[tex]q_1q_2=\dfrac{0.12\times (0.46)^2}{9\times 10^9}[/tex]
[tex]q_1q_2=2.82\times 10^{-12}[/tex]..............(1)
Also,
[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(2) (both charges are negative)
On solving equation (1) and (2) :
[tex]q_1=-8.571\ C[/tex]
and
[tex]q_2=-0.329\ C[/tex]
(b) If the force is attractive, F = -0.12 N
[tex]q_1q_2=\dfrac{Fd^2}{k}[/tex]
[tex]q_1q_2=\dfrac{-0.12\times (0.46)^2}{9\times 10^9}[/tex]
[tex]q_1q_2=-2.82\times 10^{-12}[/tex]..............(3)
[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(4)
Solving equation (3) and (4) we get :
[tex]q_1=-0.306\ C[/tex]
[tex]q_2=9.206\ C[/tex]
Hence, this is the required solution.
Using Coulomb's law, set up a system of equations with q1 and q2 as unknowns and solve to find the values of the individual charges for the case of repulsive force. Repeat the process for attractive forces, keeping in mind that charges will have the opposite sign.
Explanation:This problem can be solved using the formula for Coulomb's law, which states that the force between two charges is equal to the product of the charges divided by the distance squared, times the Coulomb constant: F = k*q1*q2/r², where F is the force, k is the Coulomb constant (8.99 * 10^9 N.m²/C²), q1 and q2 are the charges, and r is the distance between them.
For repulsive force, both charges have the same sign. But in this particular problem, we are given that the total charge is 8.90 μC, so let's take q1 and q2 as unknowns. Now q1 + q2 = 8.90 μC and using the above formula we get another equation. Now you have two equations to solve the unknown charges. Same procedure applies for the attractive force, but know that charges are of opposite sign for attractive force.
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A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 23 km in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm^3? Express your answer in grams per cubic centimeter.
Answer:
3.122×10¹⁴ g/cm³
Explanation:
Diameter of neutron star = 23 km = 2300000 cm
Radius of neutron star = 2300000/2 = 1150000 cm = r
Mass of neutron star = 1.989 × 10³⁰ kg = 1.989 × 10³³ g = m
Volume of neutron star
[tex]v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 1150000^3[/tex]
Density = Mass / Volume
[tex]\rho=\frac{m}{v}\\\Rightarrow \rho=\frac{1.989\times 10^{33}}{\frac{4}{3}\pi 1150000^3}\\\Rightarrow \rho=3.122\times 10^{14}\ g/cm^3[/tex]
∴ Density of neutron star is 3.122×10¹⁴ g/cm³
Final answer:
The typical density of a neutron star is about 10¹⁴g/cm³. This is calculated by dividing the mass of the star in grams by its volume in cubic centimeters, and taking into account that a neutron star's mass is typically 1.4 solar masses and its diameter is about 20 kilometers.
Explanation:
To calculate the density of a neutron star, we consider it as a sphere with a typical mass of 1.4 solar masses and a diameter of about 20 kilometers. The formula for density (d) is mass (m) divided by volume (V), and the volume of a sphere is given by the formula V = ⅔πr3, where r is the radius of the sphere.
First, we convert the solar mass to kilograms (1 solar mass = 1.99 × 1030 kg) and 1.4 solar masses to kg gives us 2.786 × 1030 kg. Next, we convert the diameter to radius in centimeters (10,000 cm), then calculate the volume. Now we can find the density:
Mass: 2.786 × 1030 kg
Volume: ⅔π(105 cm)³= 4.18879 × 1015 cm³
Density = Mass/Volume
Density = 2.786 × 1030 kg / 4.18879 × 1015 cm³
The density, in grams per cubic centimeter (g/cm³), is calculated by converting the mass from kilograms to grams. This gives us a typical neutron star density of about 1014 g/cm³, which is exceedingly high compared to materials we experience on Earth.
Calculate the de Broglie wavelength of an electron and a one-ton car, both moving with speed of 100 km/hour. Based on your calculation could you predict which will behave like a "quantum particle" and why. Please explain each step in words and detail.
Answer :
(a). The wavelength of electron is 26.22 μm.
(b).The wavelength of car is [tex]2.38\times10^{-38}\ m[/tex]
Explanation :
Given that,
Speed = 100 km/hr
Mass of car = 1 ton
(a). We need to calculate the wavelength of electron
Using formula of wavelength
[tex]\lambda_{e}=\dfrac{h}{p}[/tex]
[tex]\lambda_{e}=\dfrac{h}{mv}[/tex]
Put the value into the formula
[tex]\lambda_{e}=\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times100\times\dfrac{5}{18}}[/tex]
[tex]\lambda=0.00002622[/tex]
[tex]\lambda=26.22\times10^{-6}\ m[/tex]
[tex]\lambda=26.22\ \mu m[/tex]
(II). We need to calculate the wavelength of car
Using formula of wavelength again
[tex]\lambda_{e}=\dfrac{6.63\times10^{-34}}{1000\times100\times\dfrac{5}{18}}[/tex]
[tex]\lambda=2.38\times10^{-38}\ m[/tex]
The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.
Therefore, electron is quantum particle and car is classical.
Hence, (a). The wavelength of electron is 26.22 μm.
(b).The wavelength of car is [tex]2.38\times10^{-38}\ m[/tex].
A small block with mass 0.0475 kg slides in a vertical circle of radius 0.425 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.95 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.670 N. How much work was done on the block by friction during the motion of the block from point A to point B?
Answer:0.10283 J
Explanation:
Given
mass of block(m)=0.0475 kg
radius of track (r)=0.425 m
when the Block is at Bottom Normal has a magnitude of 3.95 N
Force acting on block at bottom
[tex]N-mg=\frac{mu^2}{r}[/tex]
[tex]N=mg+\frac{mu^2}{r}[/tex]
[tex]3.95=0.0475\timess 9.81+\frac{0.0475\times u^2}{0.425}[/tex]
[tex]u^2=31.172[/tex]
[tex]u=\sqrt{31.172}[/tex]
u=5.583 m/s
At top point
[tex]N+mg=\frac{mv^2}{r}[/tex]
[tex]0.670+0.0475\timess 9.81=\frac{mv^2}{r}[/tex]
[tex]v^2=10.1639[/tex]
v=3.188 m/s
Using Energy conservation
[tex]\frac{mu^2}{2}=\frac{mv^2}{2}+mg\left ( 2r\right )+W_f[/tex]
[tex]W_f=0.4989-0.39607[/tex]
[tex]W_f=0.10283 J[/tex]
i.e. 0.10283 J of energy is wasted while moving up.
The work done on the block by friction during the motion of the block from point A to point B is 0.102 J.
Velocity of the block at top circle
The velocity of the block at the top of the vertical circle is calculated as follows;
[tex]W + mg = \frac{mv^2}{r} \\\\0.67 + 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\1.136 = 0.112 v^2\\\\v^2 = 10.14\\\\v = 3.18 \ m/s[/tex]
Velocity of the block at bottom circleThe velocity of the block at the bottom of the vertical circle is calculated as follows;
[tex]W - mg = \frac{mv^2}{r} \\\\3.95 - 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\3.485 = 0.112 v^2\\\\v^2 = 31.12\\\\v = 5.58 \ m/s[/tex]
Work done by frictionThe work done by friction is the change in the kinetic energy of the block.
[tex]W _f =P.E_f - \Delta K.E \\\\W_f = mgh - \frac{1}{2} m(v_f^2 - v_i^2)\\\\Wf = 0.0475\times 9.8(2 \times 0.425) - \frac{1}{2} \times 0.0475(5.58^2 - 3.18^2)\\\\W_f = 0.396 - 0.498\\\\W_f = -0.102 \ J[/tex]
Thus, the work done on the block by friction during the motion of the block from point A to point B is 0.102 J.
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Suppose that a constant force is applied to an object with a mass of 12kg, it’s creates an acceleration of 5m/s^2. The acceleration of another object produced by the same force is 4m/s^2, what is the mass of this object?
Answer:
Mass of second object will be 15 kg
Explanation:
We have given mass of first object = 12 kg
Acceleration [tex]a=5m/sec^2[/tex]
According to second law of motion we know that force F = MA
So force [tex]F=12\times 5=60N[/tex]
As the same force is applied to the second object of acceleration [tex]a=4m/Sec^2[/tex]
So force = ma
[tex]60=m\times 4[/tex]
m = 15 kg
So mass of second object will be 15 kg
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an amplitude of 0.157 mm and a frequency of 92.2 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x,t) = ym sin(kx + ωt), what are (a) ym, (b) k, and (c) ω, and (d) the correct choice of sign in front of ω?
Answer:
Part a)
[tex]y_m = 0.157 mm[/tex]
part b)
[tex]k = 101.8 rad/m[/tex]
Part c)
[tex]\omega = 579.3 rad/s[/tex]
Part d)
here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive
Explanation:
As we know that the speed of wave in string is given by
[tex]v = \sqrt{\frac{T}{m/L}}[/tex]
so we have
[tex]T = 17.5 N[/tex]
[tex]m/L = 5.4 g/cm = 0.54 kg/m[/tex]
now we have
[tex]v = \sqrt{\frac{17.5}{0.54}}[/tex]
[tex]v = 5.69 m/s[/tex]
now we have
Part a)
[tex]y_m [/tex] = amplitude of wave
[tex]y_m = 0.157 mm[/tex]
part b)
[tex]k = \frac{\omega}{v}[/tex]
here we know that
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(92.2) = 579.3 rad/s[/tex]
so we have
[tex]k = \frac{579.3}{5.69}[/tex]
[tex]k = 101.8 rad/m[/tex]
Part c)
[tex]\omega = 579.3 rad/s[/tex]
Part d)
here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive