Answer:
The number of turns of the wire needed is 243 turns
Explanation:
Given;
self inductance of the solenoid, L = 1.8 mH
cross sectional area of the inductor, A = 1.6 x 10⁻³ m²
length of the inductor, l = 0.066 m
The self inductance of long solenoid is given by;
L = μ₀n²Al
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
n is number of turns per length
A is the area of the solenoid
l is length of the solenoid
[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]
The number of turns is given by;
N = nL
N = (3682.74)(0.066)
N = 243 turns
Therefore, the number of turns of the wire needed is 243 turns
Suppose a 225 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m. How high can it coast up the hill, if you neglect friction in m?
a) m = 180 kg
b) v = 29 m/s
c) h = 32 m
Answer:
It can coast uphill 6.2m
Explanation:
See attached file pls
g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride
Answer:
2.52 rpm
Explanation:
given that
diameter of the wheel, d = 25 m
Mass of the person, m = 75 kg
Weight experienced, N = 810 N
Since diameter is 25, radius then is 25/2 = 12.5 m
We all know that,
v = rw
Also, the passengers weight is equal to the centripetal acceleration, and thus
mg = mv²/r
Substitute for v, we have
mg = m/r * (rw)²
mg = mr²w²/r
g = rw²
If we make w the subject of formula, we have
w² = g/r
w = √(g/r)
mg = 810
75 * g = 810
g = 810 / 75
g = 1.08 m/s²
w = √(g/r)
w = √(1.08 / 12.5)
w = √0.0864
w = 0.294 rad/s
Since the question asked us in rpm, we convert to rpm
0.294 * (60 / 2π)
2.52 revolution per minute.
Suppose a space vehicle with a rest mass of 150 000 kg travels past the International Space Station at a constant speed of 2.6 x 108 m/s with respect to the I.S.S. When an observer on the I.S.S. measures the moving vehicle, her measurement of the space vehicle length is 25.0 m. Determine the relativistic mass of the space vehicle. Determine the length of the space vehicle as measured by an astronaut on the space vehicle.
Answer:
m = 300668.9 kg
L₀ = 12.47 m
Explanation:
The relativistic mass of the space vehicle is given by the following formula:
[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]
where,
m = relativistic mass = ?
m₀ = rest mass = 150000 kg
v = relative speed = 2.6 x 10⁸ m/s
c = speed of light = 3 x 10⁸ m/s
Therefore
[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]
m = 300668.9 kg
Now, for rest length of vehicle:
L = L₀√(1 - v²/c²)
where,
L = Relative Length of Vehicle = 25 m
L₀ = Rest Length of Vehicle = ?
Therefore,
25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]
L₀ = (25 m)(0.499)
L₀ = 12.47 m
What portion of the difference in the angular speed before and after you increased the mass can be accounted for by frictional losses
Answer:
As the mass increases, the moment of inertia(I) increases, therefore, the angular momentum(L) increases too.
Explanation:
friction can be defined as resistance in motion of bodies in relative to one another
momentum is the product of mass and velocity
torque is the time rate of change in momentum
τ = [tex]\frac{dL}{dt}[/tex]
where L = Iω = mvr
I = moment of inertia
ω= angular frequency
if there is no external force(torque) acting on the system, then
[tex]\frac{dL}{dt}[/tex] = 0
dL = 0 = constant
moment of inertia I depends on the distribution of mass on the axis of rotation.
as the mass increases, the angular momentum(L) increases
angular frequency, ω, remains constant
A charged particle moves into a region of uniform magnetic field B (pointing out of the page), goes through half a circle, and exits the region. The particle is either a proton or an electron. It spends 130 ns in the region. (a) What is the magnitude of B
The figure is missing, so i have attached it
Answer:
Magnitude of B = 0.252 T
Explanation:
From the image, considering the point at which it enters the field-filled region, the velocity vector is pointing downwards. The field points out of the page so that; (v→) × (B→) points leftward, points leftward which indeed seems to be the direction it is pushed. Therefore q > 0 and thus it's a proton.
The equation for the period since it goes through half circle is;
T = 2t = 2πm/(e|B|)
Where;
m is mass of proton = 1.67 × 10^(-27) kg
e is electron charge = 1.60 x 10^(-19) Coulombs.
|B| is magnitude of magnetic field
t = 130 ns = 130 × 10^(-9) s
Making |B| the subject, we have;
|B| = πm/et
Thus, plugging in all relevant values, we have;
|B| = π(1.67 × 10^(-27))/(1.60 x 10^(-19) × 130 × 10^(-9)) = 0.252 T
When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery
Answer:
The internal resistance is [tex]r = 0.5 \ \Omega[/tex]
Explanation:
From the question we are told that the resistance of
The resistance of the resistor is [tex]R = 20.0\ \Omega[/tex]
The voltage is [tex]V = 12.0 \ V[/tex]
The magnitude of the voltage fall is [tex]e = 0.300\ V[/tex]
Generally the current flowing through the terminal due to the voltage of the battery is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
substituting values
[tex]I = \frac{12.0 }{20 }[/tex]
[tex]I = 0.6 \ A[/tex]
The internal resistance of the battery is mathematically represented as
[tex]r = \frac{e}{I}[/tex]
substituting values
[tex]r = \frac{0.300}{ 0.6 }[/tex]
[tex]r = 0.5 \ \Omega[/tex]
The internal resistance of the battery is 0.5 ohms.
To calculate the internal resistance of the battery, we use the formula below
Formula:
(V/R)r = V'............. Equation 1Where:
V = Voltage across the terminal of the batteryR = Resistance connected across the batteryr = internal resistance of the batteryV' = voltage drop of the battery.Make r the subject of the equation
r = V'R/V............ Equation 2From the question,
Given:
V = 12 VR = 20 ohmsV' = 0.3 VSubstitute these values into equation 2
r = (0.3×20)/12r = 6/12r = 0.5 ohms.Hence, The internal resistance of the battery is 0.5 ohms.
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It takes 144 J of work to move 1.9 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates
Answer:
151.58 V
Explanation:
From the question,
The work done in a circuit in moving a charge is given as,
W = 1/2QV..................... Equation 1
Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.
make V the subject of the equation
V = 2W/Q.................. Equation 2
Given: W = 144 J. Q = 1.9 C
Substitute into equation 2
V = 2(144)/1.9
V = 151.58 V
A block with a mass of 0.28 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.0 N on the block. When the block is released, it oscillates with a frequency of 1.2 Hz. How far was the block pulled back before being released?
Answer:
Explanation:
For spring
[tex]n=\sqrt{\frac{k}{m} }[/tex]
where n is frequency of oscillation and k is force constant and m is mass
Putting the values
[tex]1.2=\sqrt{\frac{k}{.28} }[/tex]
k = .4032 N/m
F= k x
where F is force , k is force constant and x is extension
Putting the given values
1 = .4032 x
x = 2.48 m
A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baking while the top element is used to broil foods. When only the bottom element is active and glowing red hot, what heat transfer mechanisms carry most of the heat to the food in the oven?
Answer:
Convection and Radiation mechanisms carry most of the heat
Explanation:
This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.
A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and the pump efficiency is 85%, the power required by the pump is most nearly:
Answer:
The power required by the pump is nearly 230.588 kW
Explanation:
Flow rate of the pump Q = 1 m^3/s
the head flow H = 20 m
specific weight of water γ = 9800 N/m^3
efficiency of the pump η = 85%
First note that specific gravity of water is the product of the density of water and acceleration due to gravity.
γ = ρg
where ρ is density. For water its value is 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
The power to lift this water at this rate will be gotten from the equation
P = ρgQH
but ρg = γ
therefore,
P = γQH
imputing values, we'll have
P = 9800 x 1 x 20 = 196000 W
But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.
we can say that
196000 W is 85% of the power of the pump power needed, therefore
196000 = 85% of [tex]P_{p}[/tex]
where [tex]P_{p}[/tex] is the power of the pump needed
85% = 0.85
196000 = 0.85[tex]P_{p}[/tex]
[tex]P_{p}[/tex] = 196000/0.85 = 230588.24 W
Pump power = 230.588 kW
What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?
Answer:
The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]
Explanation:
From the question we are told that
The order of maximum diffraction is m = 2
The wavelength is [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]
The angle is [tex]\theta = 62.5^o[/tex]
Generally the condition for constructive interference for diffraction grating is mathematically represented as
[tex]dsin \theta = m * \lambda[/tex]
where d is the distance between the lines on a diffraction grating
So
[tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]
substituting values
[tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]
[tex]d = 1.747 *10^{-6} \ m[/tex]
You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6 cm distributed uniformly along its 33 cm length.
A. How much current is needed?
_____________ A
B. What is the magnitude of the magnetic field inside the solenoid?
________________T
C. What is the energy density (energy/volume) inside the solenoid?
________________ kJ/m^3
Answer:
(a) The current needed is 56.92 A
(b) The magnitude of the magnetic field inside the solenoid is 0.134 T
(c) The energy density inside the solenoid is 7.144 kJ/m³
Explanation:
Given;
energy stored in the magnetic field of solenoid, E = 8.1 J
number of turns of the solenoid, N = 620 turns
diameter of the solenoid, D = 6.6 cm = 0.066 m
radius of the solenoid, r = D/2 = 0.033 m
length of the solenoid, L = 33 cm = 0.33 m
Inductance of the solenoid is given as;
[tex]L= \frac{\mu_o N^2 A}{l}[/tex]
where;
A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
[tex]L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H[/tex]
(A). How much current needed
Energy stored in magnetic field of solenoid is given as;
[tex]E = \frac{1}{2} LI^2\\\\[/tex]
Where;
I is the current in the solenoid
[tex]E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A[/tex]
(B) The magnitude of the magnetic field inside the solenoid
B = μ₀nI
where;
n is number of turns per unit length
B = μ₀(N/L)I
B = (4π x 10⁻⁷)(620/0.33)(56.92)
B = 0.134 T
(C) The energy density (energy/volume) inside the solenoid
[tex]U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3[/tex]
Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field
Answer:
case 1 of physics is the answer
Which is true about refraction from one material into a second material with a greater index of refraction when the incident angle is, say, 30º? At the interface, the ray bends toward the normal.
Answer:
Explanation:
Refraction is defined as the bending of light rays as an incident ray pass from one medium to another. If the incident ray is passing from the media with low refractive index to a greater refractive index, the refracted ray tends to bend away from the normal.
Refractive index is the ratio of the sin of angle of incidence to the sine of angle of refraction.
n = sin i/sin r
For us to have a greater index of refraction, the denominator must be lesser than the numerator. This means that the angle of refraction must be smaller and if the angle of refraction must get smaller, this means that the refracted ray must bend towards the normal
A negatively charged object is located in a region of space where the electric field is uniform and points due north. The object may move a set distance d to the north, east, or south. Write the three possible movements by the change in electric potential energy (Ue) of the object.
Answer:
the three possible movements by the change in electric potential energy (Ue) of the object are NORTH EAST SOUTH
Explanation:
This is because When the object moves south, the force is in the direction of the displacement, and positive work is done with decreasing electric potential energy.
The opposite is true if the particle moves north—that is, negative work is done with increasing electric potential energy.
No work is done and the electric potential energy is constant if the motion is perpendicular to the electric field.
You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass windows of the display cases. Your friend says this is because coolness is transferred from the display case to your hand. What do you think?
Answer:
I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows
Explanation:
In thermodynamics, coolness can not be transferred, only heat can be transferred
Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.
A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.
Answer:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
1. weight
2. gravitational potential energy to kinetic energy.
Explanation:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
work done by toboggan = weight × distance
W = mg and the distance is down the icy slope
By using law of conservation of energy, energy can neither be created nor destroyed, but can be conserve from one form to another in a closed system.
Toboggan converts gravitational potential energy (mgh) to kinetic energy(¹/₂mv²)
A trash compactor can compress its contents to 0.350 times their original volume and 4 times denser than their original density. Neglecting the mass of air expelled, what factor is the old density of the rubbish
Answer:
2.8
Explanation:
Using p = m/v; (old density)
p' = m/v (new density)
=m/0.350 V
p'/p = (m/0.350V)/(m/v) = 1/0.350 = 2.86
If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B
Answer:
0.9m/s²
Explanation:
See attached files
Suppose I am viewing light through a camera lens (i.e. a circular aperture). If I want a wider field of view I should _____ the diameter of the lens.
Answer:
Increase
Explanation:
Because For a given focal length, a lens with a larger front element will generally be faster. That is, it'll have a larger maximum aperture, allowing a shorter exposure time, But a larger aperture requires larger elements to maintain the same angle of view
a positively charged ion, due to a cosmic ray, is headed through earth's atmosphere toward the center of Earth. Due to Earth's magnetic field, the ion will be delfected:
Answer:
East direction
Explanation:
Given that
Charge on the particle is positive.
Moving towards the center of earth .
We know that N(north ) pole in magnetic fields work as source of magnetic lines and S(South ) pole works and sink for magnetic lines.
Therefore due to the earth magnetic fields , the positive ions will deflect towards East direction.
Thus the answer will be East direction.
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and the axis of the second is rotated θ = 65° from the vertical.
Required:
a. What is the intensity of the light after it passes through the first polarizer in W/m2?
b. What is the intensity of the light after it passes through the second polarizer in W/m2?
Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Answer:
D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity
Explanation:
Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.
What do Equations 1 and 2 predict will happen to the single-slit diffraction pattern (intensity, fringe width, and fringe spacing) as the slit width is increased.
Equation 1:
Sinθ = mλ/ω
Equaiton 2:
I= Io [Sinθ (πωλ/πωλ/Rλ)
Answer:
the firtz agrees with the expression for the shape of the curve of diracion of a slit
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
where a is the width of the slit, t is the angle from the center of the slit, l is the wavelength and m is an integer that corresponds to the maximum diffraction.
the previous equation qualitatively describes the curve of the diffraction phenomenon the equation takes the form
I = I₀ [(sin ππ a y / R λ) / π a y / Rλ]²
I = I₀ ’[sin π a y /Rλ]²
I₀ ’= I₀ / (π a y /Rλ)²
By reviewing the two expressions given
equation 1
w sin θ = m λ
where w =a w is the slit width
we see that the firtz agrees with the expression for the shape of the curve of diracion of a slit
Equation 2
the squares are missing
A Buchner funnel uses _______ when separating a(n) _______ by filtration.
Explanation:
A Buchner funnel uses perforatet glass plate when separating a(n) solide from liquid by filtration.
[tex]hope \: this \: helps[/tex]
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that places them in series across normal household voltage. If each one is an incandescent bulb of fixed resistance, which statement about these bulbs is correct?
Answer:
When they are connected in series
The 50 W bulb glow more than the 100 W bulb
Explanation:
From the question we are told that
The power rating of the first bulb is [tex]P_1 = 100 \ W[/tex]
The power rating of the second bulb is [tex]P_2 = 50 \ W[/tex]
Generally the power rating of the first bulb is mathematically represented as
[tex]P_1 = V^2 R[/tex]
Where [tex]V[/tex] is the normal household voltage which is constant for both bulbs
So
[tex]R_1 = \frac{V^2}{P_1 }[/tex]
substituting values
[tex]R_1 = \frac{V^2}{100}[/tex]
Thus the resistance of the second bulb would be evaluated as
[tex]R_2 = \frac{V^2}{50}[/tex]
From the above calculation we see that
[tex]R_2 > R_1[/tex]
This power rating of the first bulb can also be represented mathematically as
[tex]P_ 1 = I^2_1 R_1[/tex]
This power rating of the first bulb can also be represented mathematically as
[tex]P_ 2 = I^2_2 R_2[/tex]
Now given that they are connected in series which implies that the same current flow through them so
[tex]I_1^2 = I_2^2[/tex]
This means that
[tex]P \ \alpha \ R[/tex]
So when they are connected in series
[tex]P_2 > P_1[/tex]
This means that the 50 W bulb glows more than the 100 \ W bulb
Wind gusts create ripples on the ocean that have a wavelength of 5.00 cm and propagate at 2.00 m/s. What is their frequency
Answer:
f = 40Hz
Explanation:
v=f x wavelength
f =v / wavelength
f = 2/5 x 10-²= 40 Hz
f = 40Hz
f = 40Hz
What is frequency?In physics, the term frequency refers to the number of waves that pass a fixed point in unit time.
It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
Wavelength (λ) - The wavelength of light is defined as the distance between the crests or troughs of a wave motion.
The wave equation: v = fλ
As per question,
Wavelength = 5.00 cm
v = 2.00 m/s.
v=fλ
f =v / λ
f = 2/5 x 10⁻² = 40 Hz
f = 40Hz
Therefore,
The frequency is 40Hz.
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A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?
Answer:
31
Explanation:
The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.
Answer:
27°
Explanation:
The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)
So theta = arcsin(0.45)
=27°
The angle between the wire and the magnetic field is 27°.
Calculation of the angle:Since The magnetic force per meter on a wire is measured to be only 45 %
So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field
Therefore,
theta = arcsin(0.45)
=27°
Hence, The angle between the wire and the magnetic field is 27°.
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A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m