Answers
Answer: The length L of the wire used to construct the coil is 191.4m
Explanation: Please see the attachments below
Related Questions
Let's return to our friend the sphere, with surface charge density σ(θ, f) = σ0sinθcos2f . Find the net polarization of this sphere, assuming the charge density is bound charge. You can assume the sphere is not rotating (f is azimuthal angle). What is the electric potential a large distance from the sphere?
Answers
Answer:
Explanation:
solution solved below
Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B is at rest. A period of slipping follows and disk B makes 2 revolutions before reaching its final angular velocity. Assuming that the angular acceleration of each disk is constant and inversely proportional to the cube of its radius, determine (a) the angular acceleration of each disk, (b) the time during which the disks slip.
Answers
Answer:
a) αA = 4.35 rad/s²
αB = 1.84 rad/s²
b) t = 3.7 rad/s²
Explanation:
Given:
wA₀ = 240 rpm = 8π rad/s
wA₁ = 8π -αA*t₁
The angle in B is:
[tex]\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}[/tex]
[tex]\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad[/tex]
[tex]w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}[/tex]
The velocity at the contact point is equal to:
[tex]v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}[/tex]
[tex]v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}[/tex]
Matching both expressions:
[tex]1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s[/tex]
b) The time during which the disks slip is:
[tex]t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s[/tex]
a) The angular acceleration of each disk is
[tex]\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)[/tex]
[tex]\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)[/tex]
The angular acceleration of each disk and the time during which the disks slip cannot be determined with the given information.
Explanation:(a) To determine the angular acceleration of each disk, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time. For disk A, the initial angular velocity is 240 rpm counterclockwise and the final angular velocity is 0 rpm. The time during which disk A stops is not given, so we cannot determine its angular acceleration at this time. For disk B, the initial angular velocity is 0 rpm and the final angular velocity is also 0 rpm. Since disk B is at rest initially and then starts moving, it undergoes an angular acceleration that brings it to rest. However, the specific time during which disk B slips is not given, so we cannot determine its angular acceleration during slipping.
(b) The time during which the disks slip is not provided in the question, so we cannot determine the exact time. However, we know that disk B makes 2 revolutions before reaching its final angular velocity, so during this period, disk B is slipping.
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An air-plane has an effective wing surface area of 17.0 m² that is generating the lift force. In level flight the air speed over the top of the wings is 66.0 m/s, while the air speed beneath the wings is 40.0 m/s. What is the weight of the plane?(The density of air is 1.29 kg/m³)
Answers
Answer:
Explanation:
Given that,
Surface area A= 17m²
The speed at the top v" = 66m/s
Speed beneath is v' =40 m/s
The density of air p =1.29kg/m³
Weight of plane?
Assuming that,
the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.
Using Bernoulli equation
P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''
Where
P' is pressure at the bottom in N/m²
P" is pressure at the top in N/m²
v' is velocity at the bottom in m/s
v" is velocity at the top in m/s
Then, Bernoulli equation becomes
P'+ ½pv'² = P'' + ½pv''²
Rearranging
P' — P'' = ½pv"² —½pv'²
P'—P" = ½p ( v"² —v'²)
P'—P" = ½ × 1.29 × (66²-40²)
P'—P" = 1777.62 N/m²
Lift force can be found from
Pressure = force/Area
Force = ∆P ×A
Force = (P' —P")×A
Since we already have (P'—P")
Then, F=W = (P' —P")×A
W = 1777.62 × 17
W = 30,219.54 N
The weight of the plane is 30.22 KN
Answer:
Weight of plane ; W = 30219.54 N
Explanation:
For us to determine the lift force of the system, let's multiply the pressure difference with the effective wing surface area given that the area is obtained by Bernoulli equation. Thus,
P_b + (1/2)ρ(v_b)² + ρg(y_b) = P_t + (1/2)ρ(v_t)² + ρg(y_t)
Now, since the flight is level, the height is constant.
Thus, (y_b) = (y_t)
So, we now have;
P_b + (1/2)ρ(v_b)² = P_t + (1/2)ρ(v_t)²
Rearranging, we have ;
P_b - P_t = (1/2)ρ(v_t)² - (1/2)ρ(v_b)²
P_b - P_t = (1/2)ρ[(v_t)² - (v_b)²]
Now, weight is given by the formula;
W = (P_b - P_t) •A
Thus,
W = (1/2)ρ[(v_t)² - (v_b)²] •A
From the question,
Density; ρ = 1.29 kg/m³
Velocity over top of wings; v_t = 66 m/s
Velocity beneath the wings; v_b = 40 m/s
Surface Area; A = 17 m²
Thus;
W = (1/2)1.29[(66)² - (40)²] •17
W = (1/2)•1.29•17[2756]
W = 30219.54 N
People who do very detailed work close up, such as jewelers, often can see objects clearly at a much closer distance than the normal 25.0 cm . What is the power of the eyes of a woman who can see an object clearly at a distance of only 7.25 cm ? Assume a distance from the eye lens to the retina of 2.00 cm .
Answers
The power of the eye for a person seeing an object which is distanced a distance 7.25 cm is 63.75 diptore
Explanation:
To find the power of the eye
we have the formula,
P=1/f= 1/d0 +1/di
Where,
do denotes the distance between eyes length and the object
di denotes the distance between eyes length and the image
Given data
do=7.25 cm di=2.00 cm
substitute in the formula
P=1/f= 1/d0 +1/di
P= 1/0.0725 +1/0.02=13.79
P=63.79 D
The power of the eye for a person seeing an object which is distanced a distance 7.25 cm is 63.75 diptore
Consider a thin film surrounded with two other transparent materials. Index of refraction of the top material is 1.2; index of refraction of a film 1.3 and of the bottom material 1.4 For normal incidence this film will look bright for light which wavelength satisfies the following condition a. Thickness of a film is equal to of wavelength of light in bottom material b. Thickness of a film is equal ½ of wavelength in the film c. Thickness of a film is equal ¼ of wavelength in the film. d. Thickness of a film is equal to ½ of wavelength of light in upper material e. Thickness of a film is equal ¼ of wavelength of light in upper material f. Thickness of a film is equal 4 of wavelength in bottom material
Answers
Answer:
Answer: The thickness of film is equal to 1/2 of wavelength in the film
Explanation:
Consider a thin film surrounded with two other transparent materials. Index of refraction of the top material is 1.2; index of refraction of a film 1.3 and of the bottom material 1.4 For normal incidence this film will look bright for light which wavelength satisfies the following condition a. Thickness of a film is equal to of wavelength of light in bottom material b. Thickness of a film is equal ½ of wavelength in the film c. Thickness of a film is equal ¼ of wavelength in the film. d. Thickness of a film is equal to ½ of wavelength of light in upper material e. Thickness of a film is equal ¼ of wavelength of light in upper material f. Thickness of a film is equal 4 of wavelength in bottom material
For constructive interference
Zt = ∧/n
t=∧/n
2
thickness = 1/2 x wavelength in the film
The thickness of film is equal to 1/2 of wavelength in the film
The condition for a thin film to look bright is when its thickness is ½ of the wavelength in the film itself, due to constructive interference resulting from the phase shift upon reflection.
The question deals with thin film interference, a phenomenon where light waves reflected off the top and bottom surfaces of a thin film interfere with each other. For a thin film to appear bright for light of a certain wavelength, the condition for constructive interference must be satisfied. This often involves the film's thickness being a multiple or a fraction of the wavelength of light in the film itself or in the surrounding media.
In the context of thin films surrounded by materials with different indices of refraction, the correct condition for the film to look bright is when its thickness is equal to ½ of the wavelength in the film itself. This accounts for the phase shift that occurs upon reflection, leading to constructive interference when the optical path difference is a multiple of the wavelength.
It's important to note that the index of refraction of the material influences the effective wavelength of light within it. Therefore, the wavelength in the material can be determined by dividing the wavelength in vacuum by the material's index of refraction.
For a certain optical medium the speed of light varies from a low value of 1.80 × 10 8 m/s for violet light to a high value of 1.92 × 10 8 m/s for red light. Calculate the range of the index of refraction n of the material for visible light.
Answers
Answer:
1.56 - 1.67
Explanation:
Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.
Mathematically, it is given as:
n = c/v
Where c is the speed of light in a vacuum and v is the speed of light in the medium.
Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.
When the speed is 1.8 * 10^8 m/s, the refractive index is:
n = (3 * 10^8) / (1.8 * 10^8)
n = 1.67
When the speed is 1.92 * 10^8 m/s, the refractive index is:
n = (3 * 10^8) / (1.92 * 10^8)
n = 1.56
Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.
Final answer:
The range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.
Explanation:
The index of refraction, n, of a material can be calculated using the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.
For this specific optical medium, the speed of light varies from 1.80 × 10^8 m/s for violet light to 1.92 × 10^8 m/s for red light. To calculate the range of the index of refraction, we need to determine the ratio of the speed of light in a vacuum to the speed of light in the material for both violet and red light.
The range of the index of refraction, n, can be calculated as:
For violet light: n = c/v = (3.00 × 10^8 m/s) / (1.80 × 10^8 m/s) ≈ 1.67
For red light: n = c/v = (3.00 × 10^8 m/s) / (1.92 × 10^8 m/s) ≈ 1.56
Therefore, the range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.
A 60.0-kg man jumps 1.70 m down onto a concrete walkway. His downward motion stops in 0.025 seconds. If he forgets to bend his knees, what force (N) is transmitted to his leg bones?
Answers
Answer:
The magnitude of the force transmitted to his leg bones is 1413.6 N
Explanation:
Recall that force is defined as the change in linear momentum per unit time:
[tex]F=\frac{P_f-P_i}{\Delta t}[/tex]
We can use this formula to find the force transmitted to his legs. We know that the final momentum ([tex]P_f[/tex]) is 0 since the person is not moving on the floor, but we need to find what the person's momentum was an instant before he touches the ground. Since we know he person's mass, all we need for the initial momentum is his velocity.
For such we use conservation of energy in free fall, knowing that he jumped from 1.7 meters:
[tex]Potential \,\,Energy \,\,at \,\,the \,\,top \,\,of \,\,the\,\, jump = U_i=m * g * h = 60*9.8*1.7 \,J\\Kinetic \,\,Energy \,\,when \,\,touching \,\,ground =KE_f= \frac{1}{2} m*v^2=\frac{60\,kg}{2} v^2\\\\KE_f=U_i\\\frac{60\,kg}{2} v^2=60*9.8*1.7 \,J\\v^2=2*9.8*1.7 \,\frac{m^2}{s^2} \\v=0.589\,\frac{m}{s}[/tex]
Now with this velocity, we know the [tex]P_i[/tex] (initial momentum) just before impact.
[tex]P_i=60 \,kg * 0.589 \frac{m}{s} =35.34 \,kg\,\frac{m}{s}[/tex]
And since the impact lasted 0.025 seconds, we can find the force using the first formula we recalled:
[tex]F=\frac{P_f-P_i}{\Delta t}=\frac{0-35.34_i}{\0.025}: N= -1413.6\,N[/tex]
so the magnitude of the force is 1413.6 N
An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass 1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.
Answers
Answer:
A) The new amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please find the attached files for the solution
In fission processes, which of the following statements is true? The total number of protons and the total number of neutrons both remain the same. The total number of protons and the total number of mass nuclei both remain the same. Only the total number of protons remains the same. Only the total number of mass nuclei remains the same. Only the total number of neutrons remains the same.
Answers
Answer: A. The total number of protons and the total number of neutrons both remain the same.
Explanation:
Nuclear fission is the process in which a large nucleus splits into two smaller nuclei with the release of energy. In other words, fission is the process in which a nucleus is divided into two or more fragments, and neutrons and energy are released.
In fission processes, the total number of protons and neutrons both remain the same because nucleons are neither created nor destroyed, they just rearrange into new nuclei.
Explanation:In fission processes, it is true that the overall number of protons and the total number of neutrons remains the same after the event. This is because nucleons, which include both protons and neutrons, are neither created nor destroyed during fission; they simply rearrange themselves into new nuclei. The mass number, which is the sum of protons and neutrons, is preserved, and though there can be a conversion between protons and neutrons (e.g., via beta decay), the total count of nucleons stays constant. However, the total mass of the product nuclei is less than the mass of the reactants due to the release of nuclear energy as a result of the conversion of mass to energy according to Einstein's equation E=mc2.
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A dielectric cube of side a, centered at the origin, carries a "frozen-in" polarization P = kr, where k is a constant. Find all the bound charges and check that they add up to zero.
Answers
The total volume of bound charge is zero.
Explanation:
We have to the volume and surface bounded charge densities.
ρb = - Δ . p = - Δ .k ([tex]x^{X}[/tex] +[tex]y^{Y}[/tex] +[tex]x^{Y}[/tex])
= - 3k
On the top of the cube the surface charge density is
σb = p . z
= [tex]\frac{ka}{2}[/tex]
By symmetry this holds for all the other sides. The total bounded charge should be zero
Qtot = (-3k)a³ + 6 . [tex]\frac{ka}{2}[/tex] . a² = 0
σb = -3K σb = [tex]\frac{ka}{2}[/tex]
Qtot = 0
Hence, the total volume of bound charge is zero.
Martha is measuring the barometric pressure in her back yard every thirty minutes. She records her data in the table below.
Answers
Answer:a low pressure system is moving into the area
Explanation:
This is the only reasonable answer
Answer: a low pressure system is moving into the area
Explanation: It says it on study island
Consider a comet with an elliptic orbit whose aphelion and perihelion distances are rA = 5.00109 km and rP = 8.00107 km. e. Find the speed of the comet at aphelion and at perihelion
Answers
Answer:
Explanation:
To find this speed find the attached document below;
The speed of a comet can be calculated at aphelion and perihelion by using the vis-viva equation, which involves the gravitational constant, the mass of the Sun, the distance of the comet from the Sun at specific points, and the semi-major axis of the comet's orbit.
Explanation:Calculating Comet Speed at Aphelion and PerihelionTo find the speed of a comet at aphelion (ra) and perihelion (rp), we will use the conservation of energy and angular momentum principles for orbital motion. The specific mechanical energy (the sum of kinetic and potential energy) in an elliptical orbit around a much larger body, like the sun, is constant at any point along the orbit. This allows us to create an equation relating the speed of the comet at its closest and furthest points from the Sun.
Using the vis-viva equation:
v = √GM(rac{2}{r} - rac{1}{a})
We can calculate the speed at perihelion (vp) and aphelion (va) using the given distances, where a is the semi-major axis (the average of aphelion and perihelion distances) and r is the comet's distance to the Sun at a specific point:
Calculate the semi-major axis a = (½)(ra + rp).Substitute a, ra, and rp into the vis-viva equation to find va and vp.For the given values of rA = 5.00 x 109 km and rP = 8.00 x 107 km, we calculate a, and then use these values to find the comet's speeds at its closest and furthest points from the Sun.
A 38000-Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 195 kN. The time required is ____ min ___ s.
Answers
Answer:
216.59 s.
Explanation:
Using,
F = ma................. Equation 1
Where F = force exerted by the tugboat, m = mass of the ocean liner, a = acceleration of the ocean liner
make a the subject of the equation
a = F/m.................. Equation 2
Given: F = 195 kN = 195000 N, m = 38000 Mg = 38000000 kg.
Substitute into equation 2
a = 195000/38000000
a = 5.13×10⁻³ m/s²
Also using,
Assuming the liner is decelerating
a = (u-v)/t............ Equation 3
Where v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (u-v)/a............. Equation 4
Given: u = 4 km/h = 4(1000/3600) = 1.111 m/s, v = 0 m/s, a = 0.00513 m/s²
Substitute into equation 4
t = (1.111-0)/0.00513
t = 216.59 s.
A car goes from 100 m/s to a full stop for in 9.5 seconds. What is the acceleration?
Answers
Answer:Calculate displacement of an object that is not acceleration, given initial position and velocity.
Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.
Four men racing up a river in their kayaks.
Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr).
We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.
Notation: t, x, v, a
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is
Δ
t
=
t
f
−
t
0
, taking
t
0
=
0
means that
Δ
t
=
t
f
, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is,
x
0
is the initial position and
v
0
is the initial velocity. We put no subscripts on the final values. That is,
t
is the final time,
x
is the final position, and
v
is the final velocity. This gives a simpler expression for elapsed time—now,
Δ
t
=
t
. It also simplifies the expression for displacement, which is now
Δ
x
=
x
−
x
0
. Also, it simplifies the expression for change in velocity, which is now
Δ
v
=
v
−
v
0
. To summarize, using the simplified notation, with the initial time taken to be zero,
Explanation:
12. A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.
How much work does the woman do in lifting the child?
Answers
Answer:
515.025J
A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.
How much work does the woman do in lifting the child?
Explanation:
Given;
Mass m = 35kg
Distance of lifting l = 1.5m
Acceleration due to gravity g = 9.81 m/s^2
Workdone = mgl = 35×1.5 × 9.81 = 515.025 J
The change in potential energy is equal to the work that must be done, to lift the child a certain distance. Therefore, the work done by the woman in lifting the child is:
[tex]W=\Delta U\\W=mgh_f-mgh_0\\W=mg(h_f-h_0)\\W=35kg*1.5m(1.5m-0m)\\W=515.03J[/tex]
A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. Initially the magnetic field in the region is pointed out of the page and has a magnitude of 5.5 T, but it is decreasing at a rate of 24.5 G/s. Due to the changing magnetic field, an electric field will be induced in this space which causes the acceleration of charges in the region. What is the direction of acceleration of a proton placed in at 1.5 cm from the center?
Answers
Answer:
The acceleration is [tex]a = 3.45*10^{3} m/s^2[/tex]
Explanation:
From the question we are told that
The radius is [tex]d = 6.5 cm = \frac{6.5}{100} = 0.065 m[/tex]
The magnitude of the magnetic field is [tex]B = 5.5 T[/tex]
The rate at which it decreases is [tex]\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s[/tex]
The distance from the center of field is [tex]r = 1.5 cm = \frac{1.5}{100} = 0.015m[/tex]
According to Faraday's law
[tex]\epsilon = - \frac{d \o}{dt}[/tex]
and [tex]\epsilon = \int\limits {E} \, dl[/tex]
Where the magnetic flux [tex]\o = B* A[/tex]
E is the electric field
dl is a unit length
So
[tex]\int\limits {E} \, dl = - \frac{d}{dt} (B*A)[/tex]
[tex]{E} l = - \frac{d}{dt} (B*A)[/tex]
Now [tex]l[/tex] is the circumference of the circular loop formed by the magnetic field and it mathematically represented as [tex]l = 2\pi r[/tex]
A is the area of the circular loop formed by the magnetic field and it mathematically represented as [tex]A= \pi r^2[/tex]
So
[tex]{E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}[/tex]
[tex]E = \frac{r}{2} [ - \frac{db}{dt} ][/tex]
Substituting values
[tex]E = \frac{0.015}{2} (24*10^{-4})[/tex]
[tex]E = 3.6*10^{-5} V/m[/tex]
The negative signify the negative which is counterclockwise
The force acting on the proton is mathematically represented as
[tex]F_p = ma[/tex]
Also [tex]F_p = q E[/tex]
So
[tex]ma = qE[/tex]
Where m is the mass of the the proton which has a value of [tex]m = 1.67 *10^{-27} kg[/tex]
[tex]q = 1.602 *10^{-19} C[/tex]
So
[tex]a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}[/tex]
[tex]a = 3.45*10^{3} m/s^2[/tex]
dam is used to block the passage of a river and to generate electricity. Approximately 58.4 x 103 kg of water falls each second through a height of 20.1 m. If one half of the gravitational potential energy of the water were converted to electrical energy, how much power (in MW) would be generated
Answers
Answer:
8.049 MW
Explanation:
The expression for gravitational potential energy is given as
Ep = mgh............. Equation 1
Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.
Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²
Substitute into equation 1
Ep = 58.4×10³(20.1)(9.81)
Ep = 1.6098×10⁷ J.
If one half the gravitational potential energy of the water were converted to electrical energy
Electrical energy = Ep/2
Electrical energy = (1.6098×10⁷)/2
Electrical energy = 8.049×10⁶ J
In one seconds,
The power generated = 8.049×10⁶ W
Power generated = 8.049 MW
Which front formed widespread clouds rain or snow
Answers
Answer: A cold front occurs when cold, denser air replaces the rising, less dense air mass. The reason this front brings in the rain is that as the rising warm air cools (as it rises to the cooler upper atmosphere) the moisture in it condenses into clouds that precipitate down as rain or snow.
Answer:
cold front
Explanation:
An object with a mass of 3 kg has a force of 9 newtons applied to it. What is the resulting acceleration of the object?
Answers
Answer: 3 m/s²
Explanation:
m = 3 kg
F = 9 N
a = ?
F = m × a
9 = 3 × a
[tex]\frac{9}{3}[/tex] = a
a = 3 m/s²
An iron wire and a copper wire of the same length have the same potential difference applied to them. What must be the ratio of their radii if the currents in the two wires are to be the same? (Ratio of the radius of iron wire to that of copper wire). For the resistivities of iron and copper use: rhoiron = 1.0 times 10-7 capital omega m; rhocopper = 1.7 times 10-8 capital omega m;
Answers
Answer: The ratio of their radii (Ratio of the radius of iron wire to that of copper wire) is 2.43
Explanation: Please see the attachments below
Final answer:
The ratio of the radius of an iron wire to that of a copper wire that have the same potential difference and current is approximately 1.9.
Explanation:
To find the ratio of the radii of an iron wire and a copper wire with the same potential difference and current, we can use Ohm's law and the formulas for resistance and resistivity.
The resistance is given by R = rho * (L/A), where rho is the resistivity, L is the length, and A is the cross-sectional area of the wire. Since the potential difference and current are the same for both wires, we can equate the resistances and cross-sectional areas, resulting in rho_iron * (L/A_iron) = rho_copper * (L/A_copper). Rearranging the equation, we get (A_iron/A_copper) = (rho_iron/rho_copper).
Since the radii are related to the areas of the wires by the equation A = pi * r^2, we can substitute A_iron = pi * (r_iron)^2 and A_copper = pi * (r_copper)^2 into the equation. This gives us (pi * (r_iron)^2)/(pi * (r_copper)^2) = (rho_iron/rho_copper). By canceling out the pi terms, we find (r_iron/r_copper)^2 = (rho_iron/rho_copper). Simplifying further, we get r_iron/r_copper = sqrt(rho_iron/rho_copper).
Substituting the given resistivities into the equation, we have r_iron/r_copper = sqrt((1.0 * 10^-7) / (1.7 * 10^-8)). Evaluating the expression, we find r_iron/r_copper ≈ 1.9. Therefore, the ratio of the radius of the iron wire to that of the copper wire should be approximately 1.9.
A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill? where X = 16.8.
Answers
Answer:h=19.4 m
Explanation:
Given
mass of automobile [tex]m=750\ kg[/tex]
Initial height of automobile [tex]h_o=5\ m[/tex]
Velocity at this instant [tex]v=16.8\ m/s[/tex]
If the car stops somewhere at a height [tex]h[/tex]
Thus conserving total energy we get
[tex]K_i+U_i=K_f+U_f[/tex]
[tex]\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh[/tex]
[tex]\frac{v^2}{2g}+h_o=h[/tex]
[tex]h=5+\frac{16.8^}{2\times 9.8}[/tex]
[tex]h=5+14.4[/tex]
[tex]h=19.4\ m[/tex]
A piston is compressed from a volume of 8.47 L to 2.62 L against a constant pressure of 1.93 atm. In the process, there is a heat gain by the system of 360. J. 371.2 Incorrect: Your answer is incorrect. J
Answers
Final answer:
The question relates to thermodynamics in physics, focusing on gas expansion or compression in a piston and its related work, heat transfer, and temperature change.
Explanation:
The question involves the concept of thermodynamics, which is a branch of physics dealing with heat, work, and energy transfer. When gas in a piston expands or compresses, it can perform work on its surroundings, and there may also be a transfer of heat between the system and its surroundings. The specifics of the temperature change, work done, and energy transfer depend on factors such as initial and final volume, pressure, and the heat capacity of the system or the environment it's in contact with.
A projectile proton with a speed of 1100 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 45° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?
Answers
Answer:
So, the target proton's speed is 777.82 m/s
And, the projectile proton's speed is 777.82 m/s
Explanation:
as per the system, it conserves the linear momentum,
so along x axis :
Mp V1 (i) = Mp V1 (f) cos θ1 + Mp V2 (f) cos θ2
along y axis :
0 = -Mp V1 (f) sin θ1 + Mp V2 (f) sin θ2
let us assume before collision it was moving on positive x axis, hence target angle will be θ2 = 45° from x axis
V2(f) = V1 (i) sin θ1 / ( cosθ2 sin θ1 + cos θ1 sin θ2)
= 1100 * sin 45 / ( cos 45 sin 45 + cos 45 sin 45 )
= 1100 * 0.7071 /( 0.7071 * 0.7071 +0 .7071 * 0.7071 )
= 777.82 /( 0.5 + 0.5)
= 777.82 m/s
(b)
the speed of projectile , V1 (f) = sinθ2 * V2(f) / sinθ1
= sin 45 * 777.82 / sin 45
= 777.82 m/s
So, the target proton's speed is 777.82 m/s
And, the projectile proton's speed is 777.82 m/s
Find the terminal speed of the rod if it has mass m = 2.8 grams , length l = 21 cm , and resistance R = 0.0011 Ω . It is falling in a uniform horizontal field B = 0.052 T . Neglect the resistance of the U-shaped conductor.
Answers
Answer:
0.253 m/s
Explanation:
As the conductor falls down, the magnetic flux throug the coil formed by the conductor and the the rest of the circuit changes, therefore an electromotive force is induced in the rod; its magnitude is given by
[tex]E=BvL[/tex]
where
B = 0.052 T is the strength of the magnetic field
v is the speed at which the rod is falling
L = 21 cm = 0.21 m is the length of the rod
Due to this electromotive force, a current is also induced in the rod and the circuit, and this current is given by
[tex]I=\frac{E}{R}[/tex]
where
[tex]R=0.0011 \Omega[/tex] is the resistance of the rod
So the current is
[tex]I=\frac{BvL}{R}[/tex] (1)
At the same time, we know that a current-carrying wire in a magnetic field experiences a force, which is given by
[tex]F_B = IBL[/tex] (2)
where in this case:
I is the induced current given by eq(1)
B is the strength of the magnetic field
L is the lenght of the rod
Inserting eq(1) into (2), we find that the magnetic force on the rod is:
[tex]F_B=\frac{BvL}{R}\cdot BL = \frac{B^2 L^2 v}{R}[/tex]
However, there is another force acting on the rod: the force of gravity, given by
[tex]F_g=mg[/tex]
where
[tex]m=2.8 g = 0.0028 kg[/tex] is the mass of the rod
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
The falling rod will reach its terminal speed when its net acceleration becomes zero; this occurs when the net force on it is zero, so when the magnetic force is balanced by the force of gravity, so when
[tex]F_B = F_g[/tex]
So
[tex]\frac{B^2 L^2v}{R}=mg[/tex]
And solving for v, we find the terminal speed:
[tex]v=\frac{mgR}{B^2L^2}=\frac{(0.0028)(9.8)(0.0011)}{(0.052)^2(0.21)^2}=0.253 m/s[/tex]
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 59.3 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy U that flows through area of 0.0225 m 2 over an interval of 12.3 s, assuming that the area is perpendicular to the direction of wave propagation.
Answers
Answer: S = 4.67 W/m², U = 1.29 J
Explanation:
Given
Time of flow, t = 12.3 s
Area of flow, a = 0.0225 s
Amplitude, E = 59.3 V/m
Intensity, S = ?
I = E² / cμ, where
μ = permeability of free space
c = speed of light
E = E(max) / √2
E = 59.3 / √2
E = 41.93 V/m
I = 41.93² / (2.99*10^8 * 1.26*10^-6)
I = 1758.125 / 376.74
I = 4.67 W/m²
Energy that flows through
U = Iat
U = 4.67 * 0.0225 * 12.3
U = 1.29 J
Therefore, the intensity is 4.67 W/m² and the energy is 1.29J
Answer:
A) Intensity = 4.664 W/m²
B) U = 1.29J
Explanation:
A) The intensity of the wave is related to a time-averaged version of a quantity called the Poynting vector, and is given by the formula:
I = (E_rms/cμo)
Where;
c = speed of light which has a value of 3 x 10^(8) m/s
μo = permeability of free space which has a constant value of 4π x 10^(-7) N/A²
E_rms is root mean square value of electric field
In the question, we are given maximum amplitude of the electric field. In this case, we would have to calculate the "root-mean-square" or "rms" value through the relationship:
E_rms = E_max/√2
Thus, E_rms = 59.3/√2 = 41.93 V/m
Thus, Intensity, I = (E_rms/cμo)= [41.93²/(3 x 10^(8) x 4π x 10^(-7))]
I = 4.664 W/m²
B) The formula for the energy flowing is given by the formula ;
U = IAt
Where;
I is intensity
A is area
t is time in seconds
Thus, U = 4.664 x 0.0225 x 12.3 = 1.29J
In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis.1)In which case is the torque due to the force about the rotation axis biggest?
a. Case 1b. Case 2c. Same
Answers
Explanation:
Case 1,
A force F is pushing perpendicular on an object a distance L/2 from the rotation axis. Torque is given by :
[tex]\tau_1=Fd\sin \theta\\\\\tau_1=\dfrac{FL}{2}\sin (90)\\\\\tau_1=\dfrac{FL}{2}\ ..........(1)[/tex]
Case 2,
The same force is pushing at an angle of 30 degrees a distance L from the axis. New torque is given by :
[tex]\tau_2=Fd\sin \theta\\\\\tau_2=FL\sin (30)\\\\\tau_2=\dfrac{FL}{2}\ ..........(2)[/tex]
From equation (1) and (2), it is clear that the force in both cases is same.
A cube of linear elastic material is again subjected to a vertical compressive stress s1 in the 1-direction, but is now constrained (ε ¼0) in both the 2 and the 3 directions. (a) Findexpressionsfortheinducedtransversestresses, s2 and s3 intermsof s1. Hence, derive an expression for the ‘effective modulus’ (s1/ε1) in this case. (b) Sketch the variation of effective modulus with n, and comment on the limiting values when n¼0 (foam) and n z 0.5 (rubber). (c) Explain why the rubber soles of running shoes are designed with some combination of air or gel pockets, partially foamed rubber, and a tread
Answers
Answer:
Explanation:
Solution is attached below
Why does the solar nebula theory imply planets are common? a. Planets formed from the spiral arms of matter encountered around the Sun. Most main-sequence stars pass spiral arms dozens of times in their lives. b. Planets formed from the Sun’s past binary partner star. Binary systems with different masses and evolution rates are common. c. Planets formed from the envelope of the Sun, ejected at the time of hydrogen fusion ignition. Stars commonly eject their envelopes. d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.
Answers
Answer:
d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.
Explanation:
A star passes through various stages before it becomes a full fledged star with its own planetary system. The same is with Sun as well. It was born out of a Nebula. Nebula is a cloud of dust and gases. The dust and gases start accumulating to form what we call as a protostar.
A lot of material at this stage is thrown out from the young star, this material forms a disk around the star. This disk is known as proto-planetary disk. The gases and dust of this disk then coalesce together to make planets and other objects of the star system.
We have observed such disks in the Orion nebula.
The solar nebula theory suggests that planets are formed from the disk of gas and dust that surrounds young stars. Such disks are commonly observed around stars, leading to the inference that planet formation is a universal occurrence.
Explanation:The solar nebula theory, or nebular hypothesis, explains the formation of our solar system from a nebular cloud of gas and dust. According to this theory, planets are formed from the disk of gas and dust that surrounds a young star, also known as protoplanetary disk. This leads to the implication that planets are common because such disks are commonly found around young stars.
When a star condenses from a nebula, it forms a hot, spinning disk of gas and dust around it. Over time, particles within this disk begin to collide and stick together, gradually forming planetesimals and eventually, planets. Since we observe many stars surrounded by these disks, we can infer that planet formation is a common process in the universe.
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A 0.065 kg ingot of metal is heated to 210◦C and then is dropped into a beaker containing 0.377 kg of water initially at 26◦C. If the final equilibrium state of the mixed system is 28.4 ◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg · ◦ C. Answer in units of J/kg · ◦ C.
Answers
Answer:
Explanation:
Given that,
Metal of mass
M = 0.065kg
Initial temperature of metal
θm = 210°C
The metal is drop into a beaker which contain liquid of mass
m = 0.377 kg
Initial temperature of water
θw = 26°C
The final mixture temperature is
θf = 28.14°C
Specific heat capacity of water
Cw = 4186 J/kg°C
Since the metal is hotter than the water, then the metal will lose heat, while the water will gain heat, we assume that no heat is loss by the beaker.
So,
Heat Loss = Heat gain
Now, heat loss by metal
H(loss) = M•Cm•∆θ
Where M is mass of meta
Cm is specific capacity of metal, which we are looking fro
So,
H(loss) = 0.065 × Cm × (θi - θf)
H(loss) = 0.065 × Cm × (210-28.14)
H(loss) = 11.821 •Cm
Now, Heat gain by water
H(gain) = m•Cw•∆θ
H(gain) = m•Cw•(θf - θi)
Where
m is mass of water and Cw is specific heat capacity of water
H(gain) = 0.377 ×4286 × (28.14-26)
H(gain) = 3457.86
So, H(loss) = Heat(gain)
11.821 •Cm = 3457.86
Cm = 3457.86/11.821
Cm = 292.52 J/Kg°C
The specific heat capacity of the metal ball is 292.52 J/Kg°C
Answer:
320.86J/kgC
Explanation:
To find the specific heat of the substance you take into account that the heat lost by the metal is gained by the water, that is:
[tex]Q_1=-Q_2[/tex]
Furthermore the heat is given by:
[tex]Q_1=m_1c(T_1-T)\\\\Q_2=m_2c(T_2-T)[/tex]
m1: mass of the metal
m2: mass of the water
c: specific heat
T: equilibrium temperature
T1: temperature of the metal
T2: temperature of water
By replacing all these values you can calculate c of the metal:
[tex]m_1c_1(T_1-T)=-m_2c_2(T_2-T)\\\\c_1(0.065kg)(210-28.4)\°C=-(0.377kg)(4186J/kg\°C)(26-28.4)\°C\\\\c_1=320.86\frac{J}{kg\°C}[/tex]
Hence, the specific heat of the metal is 320.86J/kgC
What is the magnitude of the rate of change of the magnetic field inside the coil? You will need to calculate this algebraically before you can get a number. Write an expression for the magnetic field due to the wire at the location of the coil. Use the approximate formula, since the wire is very long. Remember the chain rule, and remember that ⅆ ⅆ .
Answers
Answer:
The answer to this questions are, (a) 0.685 * 10^-4 T/sec (b) 0.946 * 10^-6V
Explanation:
Solution
Recall
radius r = 0.02m
N = 11 turns
Instant I = 3 amperes
Velocity v =3.3m/s
x = 0.17 m
(a) What is the magnitude of the rate of change of the magnetic field inside the coil
db/dt =μ₀T/2π x²
Thus,
= 4π * 10^-7* 3* 3.3/2π * 0.17²
=685.12* 10^-7
which is now,
0.685 * 10^-4 T/sec
(b) What is the magnitude of the voltmeter reading.
μ = N (db/dt)πr²
Note: this includes all 11 turns of the coil
Thus,
= 11 * 0.6585* 10^-4* 3.14 * (0.02)²
= 946 * 10^-7
which is = 0.946 * 10^-6V
Note: Kindly find an attached copy or document of the complete question of this exercise
The rate of change of a magnetic field in a coil can be determined using Faraday's law and the chain rule. Starting with the formula for the magnetic field created by current in a straight wire, we differentiate with respect to time to get the rate of change.
Explanation:The magnitude of the rate of change of the magnetic field inside a coil can be represented algebraically using Faraday's law of induction. Starting with the equation B = μ * I / 2λr (where I is the current, r is the distance to the wire, and μ is the permeability of free space), we can find the rate of change of the magnetic field using the chain rule.
Faraday's law is represented as E = - dΦ/dt, where E is the induced emf, Φ is the magnetic flux, and t is time. Since the magnetic flux Φ is the product of the magnetic field (B) and the area enclosed by the loop (A), we can express this as d/dt (B * A). By applying the chain rule, we can find the rate of change of the magnetic field in the coil.
Last, magnetic field B is determined by Ampère's law, often used in calculations involving magnetic fields around conductors and coils. For a long straight wire, field lines form concentric circles around the wire, following the right-hand rule.
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