Answers :
1. All points of a conductor are at the same potential. - True
2. Charges prefer to be uniformly distributed throughout the volume of a conductor. - False
3 The electric field inside the conducting material is always zero. -True
4.Just outside the surface of a conductor, the electric field is always zero. - False
a) True
b) False
c) True
d) False
What is a conductor ?
A conductor is a substance or material that allows electricity to flow through it.
a) All points of a conductor are at the same potential is True as charge distribution on the surface of the conductor is uniform
b) Charges prefer to be uniformly distributed throughout the volume of a conductor is False because all the charge comes on the surface and get distributed uniformly on the surface of the conductor and their is no charge inside the conductor
c) The electric field inside the conducting material is always zero is True
Since , all the charge is on the surface of the conductor so , there will not be any charges inside the conductor , this is why there will not be electric field .
d)Just outside the surface of a conductor, the electric field is always zero is False as due to charge on the surface there will be electric field outside the surface of conductor .
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Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks a stone with a horizontal velocity of 3.50 m/s, and it lands in the water a horizontal distance of 5.40 m from where Billy-Joe is standing what is the height of the bridge? b) If the stone had been kicked harder, how would this affect the time it would take to fall
(a) The height is [tex]h = 11.66\ m[/tex]. (b) If the speed is greater, the time required will be longer.
The height can be computed from the second equation of motion. However, for height to be computed, time is required. Therefore, the required time can be computed from distance and speed.
Given:
Horizontal velocity, [tex]u = 3.5 m/s\\[/tex]
Horizontal distance, [tex]x = 5.4 m/s\\[/tex]
(a)
The time is computed as:
[tex]t = x/u\\t = 5.4/3/5\\t = 1.5428/ s[/tex]
The height is given as:
[tex]h = 1/2gt^2\\h = 1/2 \times9.8\times1.5428^2\\h = 11.66\ m[/tex]
Hence, the height is [tex]h = 11.66\ m[/tex].
(b)
The relation between time and velocity is given as:
[tex]t= x/u\\t \alpha1/u[/tex]
The time and speed are inversely proportional. Therefore, if the velocity is larger then the time will be shorter.
Hence, if the speed is greater, the time required will be longer.
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A car initially contains just its driver. The combined mass of car and driver is 869 kg. When the gas pedal is pushed down all the way, the car reaches its maximum acceleration of 5.6 m/s2.If the driver then picks up three of his friends, the combined mass is increased by 300 kg. What is the maximum acceleration possible in units m/s2 now if the force provided by the engine remains the same as before? Round to one decimal place.
Answer:
4.2 m/s²
Explanation:
m = combined mass of car and driver = 869 kg
a = maximum acceleration reached by the car with driver in it = 5.6 m/s²
Force provided by the engine is given as
[tex]F_{eng}[/tex] = ma
[tex]F_{eng}[/tex] = (869) (5.6)
[tex]F_{eng}[/tex] = 4866.4 N
M = Combined mass of car, driver and his three friends = 869 + 300 = 1169 kg
a' = maximum acceleration reached with friends in the car = ?
Force provided by the engine remains same and is then given as
[tex]F_{eng}[/tex] = M a'
4866.4 = (1169) a'
a' = 4.2 m/s²
13. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compared to what it was at the lake surface? The area of the eardrum is 0.60 cm².
The increase in force exerted on a diver's eardrum as they descend to a depth of 3.2m in a freshwater lake is due to the increased underwater pressure from the weight of the water above them. By calculating the pressure increase, and then multiplying this by the area of the eardrum, we find the increase in force is approximately 1.88 N.
Explanation:The increase in pressure experienced by a diver as they go deeper underwater is due to the weight of the water above them; this weight results from the water's density, which is approximately 775 times greater than air. Hence, the force exerted increases with the increasing depth. To calculate the pressure increase, we use the formula: Pressure = fluid density * gravity * depth, and then the force on the eardrum is obtained by: Force = Pressure * Area.
In this case, the fluid density of freshwater is 1000 kg/m³, gravity is 9.81 m/s², the depth is 3.2 m, and the eardrum area is 0.60 cm² (or 0.00006 m² when converted to square meters).
So, the Pressure increase due to depth = 1000kg/m³ * 9.81m/s² * 3.2m = 31428 Pascal (or Pa); therefore, increase in force = 31428 Pa * 0.00006m² = 1.88 N, approximately. So, the increase in the force on the eardrum of the diver when she swims down to a depth of 3.2 m in the freshwater lake is around 1.88 Newtons.
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Final answer:
The increase in force pushing in on the eardrum at a depth of 3.2 m compared to the surface is 1.8816 N.
Explanation:
The student has asked about the increase in force on a diver's eardrum when diving to a depth of 3.2 m in a freshwater lake. The increase in force is due to the increase in water pressure with depth. To find the increase in the force on the eardrum, we need to calculate the water pressure at the depth of 3.2 m and multiply it by the area of the eardrum.
Water pressure increases by approximately 9.8 kPa per meter of depth (this is due to the weight of the water above). Therefore, at a depth of 3.2 m, the pressure is 3.2 m * 9.8 kPa/m = 31.36 kPa. The area of the eardrum is given as 0.60 cm², which is 0.60 * 10^-4 m² in SI units.
The increase in force F = pressure * area = 31.36 kPa * 0.60 * 10^-4 m². To get the force in newtons, we convert kPa to Pa by multiplying by 1,000, giving F = 31.36 * 1,000 Pa * 0.60 * 10^-4 m² = 1.8816 N.
The increase in force pushing in on the eardrum at a depth of 3.2 m compared to the surface is 1.8816 N.
The pressure at constant velocity, flowing through a pipe measuring 8 inches in diameter with a flow rate of 2,000 gpmS 1.1 psi 32.2 psi 34.8 psi 28.9 psi
Answer:
The pressure at constant velocity is 1.1 psi.
(1). is correct option.
Explanation:
Given that,
Diameter = 8 inch = 20.32 cm = 0.2032 m
Flow rate =2000 g/m
We need to calculate the velocity
[tex]v = \dfrac{Flow\ rate}{Area}[/tex]
[tex]2000\ g/m=\dfrac{0.00378\times2000}{60}[/tex]
[tex]2000\ g/m=0.1262 m^3/s[/tex]
[tex]v=\dfrac{0.1262}{\pi (0.1016)^2}[/tex]
[tex]v=3.89\ m/s[/tex]
We need to calculate the pressure
Using formula of pressure
[tex]P = \dfrac{1}{2}\rho v^2[/tex]
[tex]P=\dfrac{1}{2}\times1000\times(3.89)^2[/tex]
[tex]P=7566\ Pa[/tex]
[tex]P=1.09\ psi=1.1\ psi[/tex]
Hence, The pressure at constant velocity is 1.1 psi.
A unit vector is used in the symbolic representation of the electric field. What physical units does the unit vector have? meters. The unit vector has no units. coulombs. The units depend on how it is used.
Answer:
Explanation:
A unit vector is a vector whose magnitude is always unity that means 1 . It gives the direction of that quantity which is represented by this vector.
If the electric field is represented by a unit vector so it has same unit as electric field that means Newton per coulomb.
The unit of unit vector depend on the quantity for which it is used.
Final answer:
A unit vector is dimensionless and has no physical units, serving solely to represent direction in a vector space, such as the direction of an electric field.
Explanation:
In physics, particularly when discussing electric fields, the concept of vectors is crucial for understanding the direction and magnitude of forces. A unit vector is a standard tool used to simply indicate direction in space. Unlike other physical quantities, the unit vector is unique because it is dimensionless; that is, it has no units. This absence of units allows it to universally represent direction, irrespective of the nature of the physical quantity being described, such as an electric field.
When representing an electric field, which is a vector field, the direction in which a positive test charge would be pushed is shown by the unit vector. Vectors, as in the case of an electric field vector, are drawn as arrows with their length proportional to the magnitude and their orientation showing the direction. The unit vector associated with these fields provides direction but does not influence the physical units of the field itself, which are newtons per coulomb (N/C) for electric fields.
A wave on a string is reflected from a fixed end. The reflected wave 1. Is in phase with the original wave at the end. 2. Has a larger speed than the original wave. 3. Is 180◦ out of phase with the original wave at the end. 4. Cannot be transverse. 5. Has a larger amplitude than the original wave.
Answer:
3. Is 180◦ out of phase with the original wave at the end.
Explanation:
Here when wave is reflected by the rigid boundary then due to the rigidly bounded particles at the end or boundary they have tendency not to move and remains fixed at their position.
Due to this fixed position we can say when wave reach at that end the particles will not move and they apply equal and opposite force at the particles of string
Due to this the reflected wave is transferred back into the string in opposite phase with respect to the initial wave
so here correct answer will be
3. Is 180◦ out of phase with the original wave at the end.
As a runner crosses the finish line of a race, she starts decelerating from a velocity of 9 m/s at a rate of 2 m/s^2. Take the runner's velocity as she crosses the finish line to be in the positive direction. What is the runner’s displacement, in meters, during the first 5 seconds after crossing the finish line?
What is her velocity, in meters per second, 5 seconds after crossing the finish line?
Answer:
- 1 m/s, 20 m
Explanation:
u = 9 m/s, a = - 2 m/s^2, t = 5 sec
Let s be the displacement and v be the velocity after 5 seconds
Use first equation of motion.
v = u + a t
v = 9 - 2 x 5 = 9 - 10 = - 1 m/s
Use second equation of motion
s = u t + 1/2 a t^2
s = 9 x 5 - 1/2 x 2 x 5 x 5
s = 45 - 25 = 20 m
Final answer:
The runner's displacement in the first 5 seconds after crossing the finish line is 22.5 meters. Her velocity 5 seconds after crossing the finish line is 1 m/s in the positive direction. These calculations are based on the formulas for displacement and velocity under constant acceleration.
Explanation:
As a runner crosses the finish line, she decelerates from a velocity of 9 m/s at a rate of 2 m/s2. To find her displacement during the first 5 seconds after crossing the finish line, we use the formula for displacement under constant acceleration, δx = v0t + ½at2, where v0 = 9 m/s, a = -2 m/s2 (deceleration means acceleration is in the opposite direction to velocity), and t = 5 s. Substituting these values gives us a displacement of 22.5 meters. To find her velocity 5 seconds after crossing the finish line, we use the formula v = v0 + at, which yields a final velocity of 1 m/s in the positive direction.
This result makes sense as the runner is slowing down but not yet stopped 5 seconds after crossing the finish line. The calculated displacement of 22.5 meters is the total distance covered during these 5 seconds of deceleration.
Assume that typographical errors committed by a typesetter occur completely randomly. Suppose that a book of 600 pages contains 600 such errors. Using the Poisson distribution, calculate the probability (a) that page 1 contains no errors (b) that page 1 contains at least three errors
Answer: (a) 0.3679
(b) 0.0803
Explanation:
Given : A book of 600 pages contains 600 such errors.
Then , the average number of errors per page = [tex]\dfrac{600}{600}=1[/tex]
[tex]\text{i.e. }\lambda=1[/tex]
The Poisson distribution function is given by :-
[tex]P(x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
Then , the probability that that page 1 contains no errors ( Put [tex]x=0[/tex] and [tex]\lambda=1[/tex]) :-
[tex]P(x=0)=\dfrac{e^{-1}1^0}{0!}=0.3678794411\approx0.3679[/tex]
Now, the probability that page 1 contains at least three errors :-
[tex]P(x\geq3)=1-(P(0)+P(1)+P(2))\\\\=1-(\dfrac{e^{-1}1^0}{0!}+\dfrac{e^{-1}1^1}{1!}+\dfrac{e^{-1}1^2}{2!})=0.0803013970714\approx0.0803[/tex]
A girl with a mass of 40 kg is swinging from a rope with a length of 3.3 m. What is the frequency of her swinging?
Answer:
Frequency, f = 0.274 Hz
Explanation:
It is given that,
Mass of the girl, m = 40 kg
Length of the rope, L = 3.3 m
We need to find the frequency of her swinging. It is an example of simple harmonic motion whose time period is given by :
[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{3.3\ m}{9.8\ m/s^2}}[/tex]
T = 3.64 seconds
The frequency, [tex]f=\dfrac{1}{T}[/tex]
[tex]f=\dfrac{1}{3.64\ s}[/tex]
f = 0.274 Hz
So, the frequency of her swinging is 0.274 Hz. Hence, this is the required solution.
Given that the internal energy of water at 28 bar pressure is 988 kJ kg–1 and that the specific volume of water at this pressure is 0.121 × 10–2 m3 kg–1, calculate the specific enthalpy of water at 55 bar pressure.
Answer:
1184 kJ/kg
Explanation:
Given:
water pressure P= 28 bar
internal energy U= 988 kJ/kg
specific volume of water v= 0.121×10^-2 m^3/kg
Now from steam table at 28 bar pressure we can write
[tex]U= U_{f}= 987.6 kJ/Kg[/tex]
[tex]v_{f}=v=1.210\times 10^{-3}m^{^{3}}/kg[/tex]
therefore at saturated liquid we have specific enthalpy at 55 bar pressure.
that the specific enthalpy h = h at 50 bar +(55-50)/(60-50)*( h at 50 bar - h at 60 bar)
[tex]h= 1154.5 + \frac{5}{10}\times(1213-1154)[/tex]
h= 1184 kJ/kg
Light that passes through a series of three polarizing filters emerges from the third filter horizontally polarized with an intensity of 280 W/m2. 1) If the polarization angle between the filters increases by 25∘ from one filter to the next, find the intensity of the incident beam of light, assuming it is initially unpolarized. (Express your answer to two significant figures.
Answer:
8.3×10² W/m²
Explanation:
I₀ = Intensity of unpolarized light
θ = angle between the filters = 25°
I = Intensity of the polarized light after passing through three filters = 280 W/m²
[tex]I=\frac{I_0}{2}cos^2{25}cos^2{25}\\\Rightarrrow I_0=\frac{2I}{cos^4{25}}\\\Rightarrow I_0=\frac{2\times 280}{cos^4{25}}\\\Rightarrow I_0=830.01\ W/m^2[/tex]
∴ Intensity of unpolarized light is 8.3×10² W/m²
To find the intensity of the incident beam of light passing through three polarizing filters, we can use the formula for the intensity of polarized light after passing through a filter. The incident light is initially unpolarized, so we need to calculate the intensity after passing through each filter. Given the intensity after passing through the third filter and the angle between the filters, we can find the intensity of the incident beam.
Explanation:To find the intensity of the incident beam of light, we need to use the formula for the intensity of polarized light after passing through a polarizing filter: I = Io * cos^2(theta). Since the incident light is initially unpolarized, the intensity of the incident beam is twice the intensity of the light after passing through the first filter. Let's call this intensity I1. The intensity after passing through the second filter is I2 = I1 * cos^2(theta), and the intensity after passing through the third filter is I3 = I2 * cos^2(theta). Given that I3 = 280 W/m^2 and theta increases by 25 degrees from one filter to the next, we can calculate the intensity of the incident beam, Io, using these formulas.
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The line of action of the force \vec{F}=5\vec{i}-10\vec{j} F =5 i −10 j N passes through the point with coordinates (in meters) (2, 2). What is the moment of this force about the coordinate origin?
Answer:
[tex]\tau = - 30 \hat k[/tex]
Explanation:
Position vector of the point of application of point of application of force is given as
[tex]\vec r = 2\hat i + 2\hat j[/tex]
now we have have force
[tex]\vec F = 5 \hat i - 10\hat j[/tex]
now the moment of force is given as
[tex]\tau = \vec r \times \vec F[/tex]
[tex]\tau = (2\hat i + 2\hat j) \times (5\hat i - 10\hat j)[/tex]
[tex]\tau = -20\hat k - 10 \hat k[/tex]
[tex]\tau = - 30 \hat k[/tex]
The moment of the force about the coordinate origin is 5 Nm.
Explanation:To find the moment of a force about the coordinate origin, we need to calculate the cross product of the force vector and the position vector from the origin to the point where the force is applied. The position vector is given by: r = 2i + 2j. The cross product is obtained by taking the determinant of a matrix that includes the unit vectors for i, j, and k. The moment is then the magnitude of the cross product.
So, the cross product is: (i, j, k) ⨯ (5i, -10j, 0k) = (0, 0, 5)
Since the moment is the magnitude of the cross product, the moment of the force about the coordinate origin is 5 Nm.
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Organ pipe A, with both ends open, has a fundamental frequency of 310 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are (a) pipe A and (b) pipe B? (Take the speed of sound to be 343 m/s.)
Answer:
Part a)
55.3 cm
Part b)
41.5 cm
Explanation:
Pipe A is open at both ends so the fundamental frequency of this pipe is given as
[tex]f_o = \frac{V}{2L}[/tex]
here we know that
V = 343 m/s
[tex]f_o = 310 Hz[/tex]
now we have
[tex]310 = \frac{343}{2L}[/tex]
[tex]L = 55.3 cm[/tex]
Now we also know that second harmonic of pipe A and third harmonic of pipe B has same frequency
so we will have
[tex]\frac{2V}{2L_a} = \frac{3V}{4L_b}[/tex]
[tex]L_b = \frac{3}{4}L_a[/tex]
[tex]L_b = \frac{3}{4}(55.3) = 41.5 cm[/tex]
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/sec^2) find the maximum height
Final answer:
To find the maximum height, we need to use the initial velocity and the angle of the baseball's trajectory. By breaking down the initial velocity into its horizontal and vertical components, we can then use the equation h = v_y^2 / (2g) to find the maximum height.
Explanation:
To find the maximum height, we first need to break down the initial velocity into its horizontal and vertical components. The initial velocity of the baseball is given as 100 ft/s at an angle of 45° with respect to the ground.
The horizontal component of velocity can be found using the equation: vx = v * cos(θ), where v is the initial velocity and θ is the angle.
The vertical component of velocity can be found using the equation: vy = v * sin(θ).
Once we have the vertical component of velocity, we can use the equation h = vy2 / (2g) to find the maximum height.
Substituting the given values, we have:
h = (100 * sin(45°))2 / (2 * 32)
Calculating this will give us the maximum height of the baseball above its initial position.
A 76-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 840 N. What is the acceleration of the elevator? 1.25 Correct: Your answer is correct. m/s2 upward
Answer:
1.25 m/s^2
Explanation:
m = 76 kg, R = 840 N
Let a be the acceleration of teh elevator in upward direction.
By use of Newton's second law
R - mg = m a
R = m ( g + a)
840 = 76 ( 9.8 + a)
a = 1.25 m/s^2
The acceleration of the elevator is [tex]\( {1.25 \, \text{m/s}^2} \)[/tex] upward.
Given:
- Mass of the man (m): 76 kg
- Weight of the man (W): mg where g is the acceleration due to gravity [tex](\( 9.8 \, \text{m/s}^2 \))[/tex]
- Normal force (read by the scale): 840 N (this includes the weight of the man plus any additional force due to acceleration)
The forces acting on the man in the elevator are:
- Downward force (weight of the man): ( mg )
- Upward normal force (read by the scale): [tex]\( 840 \, \text{N} \)[/tex]
The net force acting on the man is:
[tex]\[ F_{\text{net}} = \text{Normal force} - \text{Weight} \][/tex]
Since the elevator is accelerating upward with acceleration a, we have:
[tex]\[ F_{\text{net}} = m(g + a) \][/tex]
Given that the normal force read by the scale is 840 N, we equate it to [tex]\( m(g + a) \)[/tex]:
[tex]\[ 840 = 76 \times (9.8 + a) \][/tex]
Now, solve for a:
[tex]\[ 840 = 76 \times 9.8 + 76a \]\[ 840 = 744.8 + 76a \]\[ 76a = 840 - 744.8 \]\[ 76a = 95.2 \]\[ a = \frac{95.2}{76} \]\[ a \approx 1.25 \, \text{m/s}^2 \][/tex]
A 0.380 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 28.0 N/m force constant that is at its equilibrium length. A 0.0600 kg wad of Play-Doh is thrown horizontally at the block with a speed of 2.60 m/s and sticks to it. Determine the amount in centimeters by which the Play-Doh-block system compresses the spring.
Initial momentum of the Play-doh: 0.0600 x 2.60 = 0.156 kg/m/s
Total mass of the block and play-doh: 0.38 + 0.0600 = 0.44 kg.
Final momentum is mass x velocity = 0.44v
V = Initial momentum / mass
V = 0.156 / 0.44 = 0.3545 m/s
Work done by spring is equal to the Kinetic enrgy.
Work Done by spring = 1/2 *28.0 * distance^2 = 14 * d^2
KE = 1/2 * 0.44* 0.3545^2
set to equal each other:
14 * d^2 = 0.22 *0.12567
Solve for d:
d = √(0.22*0.12567)/14
d = 0.44 meters = 4.4cm
Answer:
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A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration 0.155 rad/s2. After making 2870 revolutions, its angular speed is 127 rad/s. (a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
Answer:
a) Initial angular velocity of the turbine = 102.66 rad/s
b) Time elapsed while the turbine is speeding up = 157 s
Explanation:
a) Considering angular motion of turbine:-
Initial angular velocity, u = ?
Acceleration , a = 0.155 rad/s²
Final angular velocity, v = 127 rad/s
Angular displacement, s = 2π x 2870 = 18032.74 rad
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
127² = u² + 2 x 0.155 x 18032.74
u = 102.66 rad/s
Initial angular velocity of the turbine = 102.66 rad/s
b) We have equation of motion v = u + at
Initial angular velocity, u = 102.66 rad/s
Acceleration , a = 0.155 rad/s²
Final angular velocity, v = 127 rad/s
Substituting
v = u + at
127 = 102.66 + 0.155 x t = 157 s
Time elapsed while the turbine is speeding up = 157 s
A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the other at 6.8 m/s, 35° north of east. What is the original speed of the mess kit?
Answer:
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Explanation:
Let north represent positive y axis and east represent positive x axis.
Here momentum is conserved.
Let the initial velocity be v.
Initial momentum = 4.4 x v = 4.4v
Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s
Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s
Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s
We have
Initial momentum = Final momentum
4.4v = 12.364 i + 15.092 j
v =2.81 i + 3.43 j
Magnitude
[tex]v=\sqrt{2.81^2+3.43^2}=4.43m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0[/tex]
50.67° north of east.
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
A copper (shear modulus 4.2 x 1010 N/m2) cube, 0.242 m on a side, is subjected to two shearing forces, each of magnitude F = 3.07 x 10 6 N (see the drawing). Find the angle (in degrees), which is one measure of how the shape of the block has been altered by shear deformation.
The shear deformation in a copper block due to applied forces can be calculated using the formula Δx = (F * L₀) / (S * A), where F is the force applied, L₀ is the initial length of the block, S is the shear modulus of copper, and A is the area of the surface on which the force is applied. The angle can then be calculated by taking the inverse tangent of the ratio of deformation (Δx) to the original length (L₀) of the block.
Explanation:The subject of the question is related to the physics concept of shear strain and shear stress. The shear strain caused on a material, in this case copper, is defined by the sideways deformation of the material due to the applied force, and it is given by the formula Δx = (F * L₀) / (S * A). Here, F represents the force applied which is 3.07 X 10 6 N, L₀ is the height of the block which is 0.242 m. The shear modulus S of copper is given as 4.2 X 10¹⁰ N/m², and A is the area of the surface on which the force is applied, which can be calculated as 0.242 m² since a cube has all its sides equal. By substituting these values in the equation, we can calculate the value of the deformation Δx, which will give us an idea of how the shape of the block has been altered. The angle of shear can be obtained by taking the inverse tangent of the ratio of deformation to the original height of the cube.
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a wave travels in a string at 58 m/s. a second string of 10% greater linear density has the same tension applied as in the first string. what will be the resulting wave speed in the second string
Answer:
The speed of wave in the second string is 55.3 m/s.
Explanation:
Given that,
Speed of wave in first string= 58 m/s
We need to calculate the wave speed
Using formula of speed for first string
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]...(I)
For second string
[tex]v_{2}=\sqrt{\dfrac{T}{\mu_{2}}}[/tex]...(II)
Divided equation (II) by equation (I)
[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\dfrac{T}{\mu_{2}}}{\dfrac{T}{\mu_{1}}}}[/tex]
Here, Tension is same in both string
So,
[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{\mu_{2}}}[/tex]
The linear density of the second string
[tex]\mu_{2}=\mu_{1}+\dfrac{10}{100}\mu_{1}[/tex]
[tex]\mu_{2}=\dfrac{110}{100}\mu_{1}[/tex]
[tex]\mu_{2}=1.1\mu_{1}[/tex]
Now, Put the value of linear density of second string
[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{1.1\mu_{1}}}[/tex]
[tex]v_{2}=v_{1}\times\sqrt{\dfrac{1}{1.1}}[/tex]
[tex]v_{2}=58\times\sqrt{\dfrac{1}{1.1}}[/tex]
[tex]v_{2}=55.3\ m/s[/tex]
Hence, The speed of wave in the second string is 55.3 m/s.
Final answer:
The resulting wave speed in the second string with a 10% greater linear density and the same tension will be the same as the wave speed in the first string.
Explanation:
The wave speed in a string can be determined by the tension and linear mass density of the string. In this case, the wave speed in the first string is 58 m/s. To find the resulting wave speed in the second string with a 10% greater linear density and the same tension, we can use the formula for wave speed:
v = √(T/μ)
Let's assume the linear mass density of the first string is μ and the linear mass density of the second string is 1.1μ. Since the tension is the same in both strings, we have:
v1 = v2
√(T/μ) = √(T/(1.1μ))
Cross multiplying and simplifying, we get:
Tμ = 1.1T
μ = 1.1
So, the resulting wave speed in the second string will be 58 m/s. Therefore, the wave speed remains the same.
A star has a mass of 1.48 x 1030 kg and is moving in a circular orbit about the center of its galaxy. The radius of the orbit is 1.9 x 104 light-years (1 light-year = 9.5 x 1015 m), and the angular speed of the star is 1.8 x 10-15 rad/s. (a) Determine the tangential speed of the star. (b) What is the magnitude of the net force that acts on the star to keep it moving around the center of the galaxy?
Answer:
a)
3.25 x 10⁵ m/s
b)
8.7 x 10²⁰ N
Explanation:
(a)
w = angular speed of the star = 1.8 x 10⁻¹⁵ rad/s
r = radius of the orbit = 1.9 x 10⁴ ly = 1.9 x 10⁴ (9.5 x 10¹⁵) m = 18.05 x 10¹⁹ m
tangential speed of the star is given as
v = r w
v = (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)
v = 32.5 x 10⁴ m/s
v = 3.25 x 10⁵ m/s
b)
m = mass of the star = 1.48 x 10³⁰ kg
Net force on the star to keep it moving is given as
F = m r w²
F = (1.48 x 10³⁰) (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)²
F = 8.7 x 10²⁰ N
What is the maximum speed at which a car could safely negotiate a curve on a highway, if the radious of the curve is 250 m and the coefficient of friction is 0.4? Give the answer in meter/sec.
Answer:
31.3 m/s
Explanation:
The relation between the coefficient of friction and the velocity, radius of curve path is given by
μ = v^2 / r g
v^2 = μ r g
v^2 = 0.4 x 250 x 9.8 = 980
v = 31.3 m/s
An object moves with simple harmonic motion. If the amplitude and the period are both doubled, the object's maximum speed is 1. quartered 2. halved 3. quadrupled. 4. doubled 5. unchanged
Answer:
The object's maximum speed remains unchanged.
Explanation:
The speed of a particle in SHM is given by :
[tex]v(t)=A\omega\ sin(\omega t)[/tex]
Maximum speed is, [tex]v_{max}=A\omega[/tex]
If A' = 2A and T' = 2T
[tex]v'_{max}=(2A)\dfrac{2\pi}{2T}[/tex]
[tex]v'_{max}=(A)\dfrac{2\pi}{T}[/tex]
[tex]v'_{max}=v_{max}[/tex]
So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.
If the amplitude and period of an object's simple harmonic motion are both doubled, the object's maximum speed will be halved.
Explanation:When an object moves with simple harmonic motion, doubling the amplitude and period will affect various properties of the motion. In this case, if the amplitude and the period are both doubled, the object's maximum speed will be halved. This means that the object will reach its maximum velocity at a slower rate compared to its original motion.
To understand why this happens, it's important to know that in simple harmonic motion, the maximum speed occurs when the object passes through equilibrium. Doubling the period means that the object will take twice as long to complete one full cycle, which results in a decrease in its maximum speed by a factor of 2.
A curve of radius 40 m is banked so that a 1100 kg car traveling at 40 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. The acceleration of gravity is 9.81 m/s 2 . θ Find the minimum speed at which a car can travel around this curve without skidding if the coefficient of static friction between the road and the tires is 0.2. Answer in units of m/s.
Final answer:
The minimum speed at which a car can travel around a curve without skidding is approximately 19.81 m/s.
Explanation:
To find the minimum speed at which a car can travel around a curve without skidding, we need to consider the forces acting on the car. The centripetal force required to keep the car moving in a circular path is given by the equation[tex]Fc = (mv^2)/r[/tex], where m is the mass of the car, v is the velocity, and r is the radius of the curve.
In this case, the centripetal force is provided by the friction force between the tires and the icy road. When the road is icy, the coefficient of static friction is approximately zero, so the car will not be able to rely on friction to round the curve. Instead, the car will rely on the component of the car's weight perpendicular to the road surface.
The perpendicular component of the weight is given by the equation Wp = mg * cos(θ), where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of banking of the curve.
Setting the centripetal force equal to the perpendicular component of the weight, we have[tex](mv^2)/r = mg * cos(\theta).[/tex]Rearranging the equation, we find v = sqrt(rg * cos(θ)). Substituting the given values of r = 40 m and θ = 0 (since there is no angle of banking), we can calculate the minimum speed as v = sqrt(40 * 9.81 * cos(0)) = sqrt(392.4) ≈ 19.81 m/s.
A series circuit consists of a 0.55-H inductor with internal resistance of 8.0 Ω connected in series with a 4.0-Ω resistor, an open switch, and an ideal 12-V battery. (a) When the switch is closed, what is the initial current through the 4.0-Ω resistor? (b) What is the current through the 4.0-Ω resistor a very long time after the switch is closed?
Answer:
Part a)
i = 0
Part b)
i = 1 A
Explanation:
Part a)
As per Lenz law we know that inductor in series circuit opposes the sudden change in current
And if the flux in the circuit will change then it will induce back EMF to induce opposite current in it
Now when we close the switch at t = 0
then initially it will induce the opposite EMF in such a way that net EMF of the circuit will be ZERO
so current = 0
Part b)
After long time the induced EMF in the circuit will be zero as the flux will become constant
so here we can say
[tex]EMF = i(R_1 + R_2)[/tex]
[tex]12 = i (8 + 4)[/tex]
[tex]i = 1 A[/tex]
The initial current through the resistor in a series circuit with an inductor and resistor can be calculated using Ohm's Law. After a very long time, the current through the resistor becomes zero.
Explanation:In a series circuit with an inductor and a resistor, the initial current through the resistor can be calculated using Ohm's Law. The total resistance in the circuit is the sum of the resistance of the inductor and the resistor. Using the formula I = V / R, where V is the voltage from the battery and R is the total resistance, we can calculate the initial current.
After a very long time, the inductor reaches a state of equilibrium with no change in current. So the current through the resistor a very long time after the switch is closed would be zero.
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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.
Answer:
The coordinates of the point is (0,0.55).
Explanation:
Given that,
First charge [tex]q_{1}=9\times10^{-6}\ C[/tex] at origin
Second charge [tex]q_{2}=6\times10^{-6}\ C[/tex]
Second charge at point P = (0,1)
We assume that,
The net electric field between the charges is zero at mid point.
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}[/tex]
[tex]\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}[/tex]
[tex]\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1[/tex]
[tex]\dfrac{1}{d}=1.82[/tex]
[tex]d=\dfrac{1}{1.82}[/tex]
[tex]d=0.55\ m[/tex]
Hence, The coordinates of the point is (0,0.55).
It is a physical field occupied by a charged particle on another particle in its surrounding. The coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).
What is an electric field?It is a physical field occupied by a charged particle on another particle in its surrounding.
The following data are given as
q₁ is the first charge at the origin
q₂ is the Second charge
point of the Second charge is P = (0,1)
As we know the net electric field between the charges is zero at the midpoint.
The relation of the electric field with the distance between the charged particle is given by the formula
[tex]\rm E=\frac{Kq}{d^2} \\\\0=\frac{K\times9\times10^{-6}}{d^2} +\frac{K\times6\times10^{-6}}{(1-d)^2} \\\\\frac{1-d}{d} =\sqrt{\frac{6}{9} } \\\\\frac{1}{d} =\sqrt{\frac{{6} }{9} } +1\\\\\rm\frac{1}{d} =1.82\\\\\rm d=0.55m[/tex]
Hence the coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).
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A tank of water has a length 10.0 m, width 5.00 m, and depth 2.50 m. What the absolute pressure at the bottom of the tank?
Answer:
1.28 x 10^5 Pa
Explanation:
The absolute pressure at the bottom of the tank of water is given by:
[tex]p= p_0 + \rho g h[/tex]
where
[tex]p_0 = 1.03 \cdot 10^5 Pa[/tex] is the atmospheric pressure
[tex]\rho = 1000 kg/m^3[/tex] is the water density
g = 9.8 m/s^2 is the acceleration of gravity
h = 2.50 m is the heigth of the column of water
Substituting into the formula, we find
[tex]p=1.03\cdot 10^5 +(1000)(9.8)(2.50)=1.28\cdot 10^5 Pa[/tex]
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance r1 (r1 < R) from the center of the sphere, the electric field has magnitude E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R, the magnitude of the electric field at the same distance r1 from the center would be equal to:
Answer:
[tex]E' = \frac{E}{8}[/tex]
Explanation:
As we know that that electric field inside the solid non conducting sphere is given as
[tex]\int E.dA = \frac{q_{en}}{\epsilon_0}[/tex]
[tex]\int E.dA = \frac{\frac{Q}{R^3}r_1^3}{\epsilon_0}[/tex]
[tex]E(4\pi r_1^2) = \frac{Qr_1^3}{R^3 \epsilon_0}[/tex]
so electric field is given as
[tex]E = \frac{Qr_1}{4\pi \epsilon_0 R^3}[/tex]
now if another sphere has same charge but twice of radius then the electric field at same position is given as
[tex]E' = \frac{Qr_1}{4\pi \epsilon_0 (2R)^3}[/tex]
so here we have
[tex]E' = \frac{E}{8}[/tex]
Final answer:
The electric field at distance r1 from the center of a uniformly charged sphere with charge Q and radius 2R would be one-eighth of the field when the charge is within a sphere with radius R.
Explanation:
To find the electric field at a distance r1 from the center of a uniformly charged sphere, we can use Gauss's law. Inside a uniformly charged sphere, the electric field is proportional to the distance from the center because only the charge enclosed by a Gaussian surface contributes to the field at a point. The field inside the sphere can be described using the formula E = (Qr) / (4πε₀R³), where Q is the total charge, r is the distance from the center, R is the radius of the sphere, and ε0 is the permittivity of free space.
When the same total charge Q is distributed uniformly throughout a sphere of radius 2R, for a point at a distance r1 inside the sphere, which is still less than 2R, the electric field magnitude would be calculated with the new radius. Since the enclosed charge within the distance r1 would be the same and the volume of the larger sphere is greater, the electric field at r1 would be one-eighth of its original value, due to the volume of the sphere increasing by eight times when the radius is doubled (since the volume is proportional to the cube of the radius). Therefore, the new electric field magnitude at distance r1 would be E/8.
For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 ×106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in2 ) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?
Answer:
a) P = 44850 N
b) [tex]\delta l =0.254\ mm[/tex]
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:
[tex]\sigma=\frac{\textup{Load}}{\textup{Area}}[/tex]
on substituting the values, we get
[tex]345\times10^6=\frac{\textup{Load}}{0.00013}[/tex]
or
Load, P = 44850 N
Hence the maximum load that can be applied is 44850 N = 44.85 KN
b)The deformation ([tex]\delta l[/tex]) due to an axial load is given as:
[tex]\delta l =\frac{PL}{AE}[/tex]
on substituting the values, we get
[tex]\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}[/tex]
or
[tex]\delta l =0.254\ mm[/tex]
The maximum load that may be applied to the specimen without causing plastic deformation is 44850 N. The maximum length to which the specimen can be stretched without causing plastic deformation is 76.254 mm.
Explanation:To find the maximum load that may be applied to the specimen without causing plastic deformation, we need to calculate the stress.
Stress = Force / Area
Where the force can be calculated by multiplying the stress with the cross-sectional area of the specimen.
Hence, maximum load = Stress × Area = 345 MPa × 130 mm² = 44850 N
To find the maximum length to which the specimen can be stretched without causing plastic deformation, we need to calculate the strain.
Strain = Change in length / Original length
Maximum length = Original length + (Strain × Original length) = 76 mm + (345 MPa / 103 GPa × 76 mm) = 76 mm + 0.254 mm = 76.254 mm
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Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by E = (4i+3j)N/C. What is the potential difference VA - VB?
To calculate the potential difference VA - VB, first determine the displacement from point A to B. This results in a vector, which you dot product with the electric field vector to find the work done. This work done is the potential difference.
Explanation:To compute the potential difference VA - VB, we first need to determine the displacement from point A to point B. This is determined by subtracting the coordinates of point A from point B, so (5 - 2)i + (7 - 3)j = 3i + 4j (meters). The next step involves calculating the work done by the electric field in moving a unit positive charge from A to B, which is obtained by taking the dot product of the displacement and the electric field vectors. Hence, the work done is (3i + 4j) . (4i + 3j) = 24 Joules/Coulomb. Now, since potential difference is defined as the work done per unit charge in moving a positive charge from one point to another, the potential difference VA - VB in this case would just be equal to this work done. So VA - VB = 24 Volts.
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