To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 meters from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom
The magnitude of the forces acting at the top are;
[tex]\mathbf{F_{Top, \ x}}[/tex] = 132.95 N
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0
The magnitude of the forces acting at the bottom are;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]\mathbf{ F_f}[/tex] = -132.95 N
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = 784.8 N
The known parameters in the question are;
The mass of the person, m₁ = 70.0 kg
The length of the ladder, l = 6.00 m
The mass of the ladder, m₂ = 10.0 kg
The distance of the base of the ladder from the house, d = 2.00 m
The point on the roof the ladder rests = A frictionless plastic rain gutter
The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder
The location of the point the person is standing = 3 meters from the bottom
g = The acceleration due to gravity ≈ 9.81 m/s²
The required parameters are;
The magnitudes of the forces on the ladder at the top and bottom
The strategy to be used;
Find the angle of inclination of the ladder, θ
At equilibrium, the sum of the moments about a point is zero
The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C
Taking moment about the point of contact of the ladder with the ground, B gives;
[tex]\sum M_B[/tex] = 0
Therefore;
[tex]\sum M_{BCW}[/tex] = [tex]\sum M_{BCCW}[/tex]
Where;
[tex]\sum M_{BCW}[/tex] = The sum of clockwise moments about B
[tex]\sum M_{BCCW}[/tex] = The sum of counterclockwise moments about B
Therefore, we have;
[tex]\sum M_{BCW}[/tex] = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81
[tex]\sum M_{BCCW}[/tex] = [tex]F_R[/tex] × √(6² - 2²)
Therefore, we get;
2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = [tex]F_R[/tex] × √(6² - 2²)
[tex]F_R[/tex] = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95
The reaction force on the wall, [tex]F_R[/tex] ≈ 132.95 N
We note that the magnitude of the reaction force at the roof, [tex]F_R[/tex] = The magnitude of the frictional force of bottom of the ladder on the floor, [tex]F_f[/tex] but opposite in direction
Therefore;
[tex]F_R[/tex] = [tex]-F_f[/tex]
[tex]F_f[/tex] = - [tex]F_R[/tex] ≈ -132.95 N
Similarly, at equilibrium, we have;
∑Fₓ = [tex]\sum F_y[/tex] = 0
The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;
[tex]\sum F_y[/tex] = -70.0 × 9.81 - 10 × 9.81 + [tex]F_{By}[/tex]
∴ The upward force acting at the bottom, [tex]F_{By}[/tex] = 784.8 N
Therefore;
The magnitudes of the forces at the ladder top and bottom are;
At the top;
[tex]\mathbf{F_{Top, \ x}}[/tex] = [tex]F_R[/tex] ≈ 132.95 N←
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0 (The surface upon which the ladder rest at the top is frictionless)
At the bottom;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]F_f[/tex] ≈ -132.95 N →
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = [tex]F_{By}[/tex] = 784.8 N ↑
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what is the dimensional formula of young modulas
Answer:
The dimensional formula of Young's modulus is [ML^-1T^-2]
Answer:
G.oogle : The dimensional formula for Young’s modulus is:
A. [ML−1T−2]A. [ML−1T−2]
B. [M0LT−2]B. [M0LT−2]
C. [MLT−2]C. [MLT−2]
D. [ML2T−2]
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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What is the primary purpose of politics?
Answer:
a politician is a person active in party politic or person holding os seeking an elected seat in government.politicans purpose, support,and create laws that govern the land and by extension ,it's people.
There are 5640 lines per centimeter in a grating that is used with light whose wavelegth is 455 nm. A flat observation screen is located 0.661 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
The minimum width of the screen is 34 cm.
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m
So, sinθ = mλ/d
Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.
For small angles sinθ ≈ tanθ
So, mλ/d = L/D
making L subject of the formula, we have
L = mλD/d
L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷ 1/5640 × 10⁻² m per line
L = 1 × 455 × 10⁻⁹ m × 0.661 m × 5640 × 10² line per m
L = 1696258.2 × 10⁻⁷ m
L = 0.16963 m
L ≅ 0.17 m
So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.
So, w = 2 × 0.17 m
w = 0.34 m
w = 34 cm
So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.
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g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye, at which boundary does most of the overall refraction occur?
Answer
Explanation
:giác mạc
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 21 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 61 cm. a. What is the wavelength of the sound
Answer:
The answer is "80 cm".
Explanation:
The distance of 21 cm between the speaker's effect of high strength but a spacing of 61 cm corresponds to a zero to zero intensity, that also is, the waves are all in phase with others [tex]\Delta \ x_1 = 21 \ cm[/tex] this is out of phase [tex]\Delta\ x_2 = 61\ cm[/tex]
[tex]\therefore\\\\\Delta\ x_2 -\Delta\ x_1 = \frac{\lambda}{2}\\\\\lambda= 2( \Delta\ x_2 -\Delta\ x_1)[/tex]
[tex]= 2 ( 61\ cm - 21\ cm)\\\\ = 2(40\ cm)\\\\= 80\ cm[/tex]
The current in resistor Y is..?
(A)
Explanation:
We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that
[tex]P = VI[/tex]
Since resistor Y dissipates 100 W of power, we can solve for the current as
[tex]I = \dfrac{P}{V} = \dfrac{100\:\text{W}}{100\:\text{V}} = 1.0\:\text{A}[/tex]
The current in resistor Y is
a)1.0 AA smokestack of height H = 50 m emits a pollutant in a 3 m/s wind. The plume is carried downwind by advection (wind speed U = 3 m/s) and is simultaneously dispersing vertically with a turbulent diffusion coefficient D. The vertical diffusion causes the plume to widen vertically over time, with halfâwidth (distance from centerline to edge) increasing as:
half width = 2 â2Dt
The plume reaches the ground some distance L downwind of the base of the smokestack (see sketch in book on page 203)
a. If L = 2 km, estimate the value of the turbulent diffusion coefficient D.
b. Under the same wind speed and turbulence conditions, what would be the value of L if the smokestack were twice as high?
Answer:
a) 0.46875
b) 8 km
Explanation:
Smokestack height ( H ) = 50 m
speed of pollutant / wind speed = 3 m/s
Half width = 2 [tex]\sqrt{2Dt }[/tex] = 50 m ---- ( 1 )
a) If L = 2 km
value of turbulent diffusion coefficient D
back to equation 1
50 = 2 √ 2 * D * ( 2000/3 )
2500 = 4 * 2 * D * ( 2000/3 )
D = 2500 / ( 8 * ( 2000/3 ) )
= 0.46875
where : time to travel ( t ) = Distance / speed = 2000 / 3
b) when the smoke stack = 50 * 2 = 100 m
L = 800 m = 8 km
attached below is the detailed solution
Is there any absolute rest or motion? Describe the types of motion with one example of each type
How can I solve the following statement?
What is the magnitude of the electric field at a point midway between a −8.3μC and a +7.8μC charge 9.2cm apart? Assume no other charges are nearby.
Answer:
The net electric field at the midpoint is 6.85 x 10^7 N/C.
Explanation:
q = − 8.3 μC
q' = + 7.8 μC
d = 9.2 cm
d/2 = 4.6 cm
The electric field due to the charge q at midpoint is
[tex]E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C[/tex] leftwards
The electric field due to the charge q' at midpoint is
[tex]E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C[/tex]
The resultant electric field at mid point is
E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?
Answer:
The charges are + 74.3 μC and - 74.3 μC
Explanation:
Let the charges be q and q'.
Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by
F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m
When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.
They now repel each other.
So, the magnitude of the force of repulsion is given by
F' = k[(q + q')/2][(q + q')/2]/r²
F' = k[(q + q')²/4r²
Since the magnitude of the force of attraction and repulsion are the same, we have that
F = F'
kqq'/r² = k[(q + q')²/4r²
qq' = (q + q')²/4
(q + q')² = 4qq'
q² + 2qq' + q'² = 4qq'
q² + 2qq' - 4qq' + q'² = 0
q² - 2qq' + q'² = 0
(q - q')² = 0
q - q' = 0
q = q'
Substituting q = q' into F, we have
F = kqq'/r²
F = kq²/r²
making q subject of the formula, we have
q² = Fr²/k
q = √(Fr²/k)
q = r√(F/k)
Substituting the values of the variables into the equation, we have
q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)
q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)
q = 0.189 m(0.3923 × 10⁻³ C/m)
q = 0.0743 × 10⁻³ C
q = 74.3 × 10⁻³ × 10⁻³ C
q = 74.3 × 10⁻⁶ C
q = 74.3 μC
Since q and q' initially attract, it implies that they initially had opposite charges.
So, q = 74.3 μC and q' = -74.3 μC
So, the charges are + 74.3 μC and - 74.3 μC
You are working on a project to make a more efficient engine. Your team is investigating the possibility of making electrically controlled valves that open and close the input and exhaust openings for an internal combustion engine. Determine the stability of the valve by calculating the force on each of its sides and the net force on the valve.
The valve is made of a thin but strong rectangular piece of non-magnetic material that has a current-carrying wire along its edges. The rectangle is 0.35 cm x 1.83 cm. The valve is placed in a uniform magnetic field of 0.15 T such that the field lies in the plane of the valve and is parallel to the short sides of the rectangle. The region with the magnetic field is slightly larger than the valve. When a switch is closed, a 1.7 A current enters the short side of the rectangle on one side and leaves on the opposite short side of the rectangle. At the suggestion of a colleague, who is hoping to ensure different currents along the sides of the valve, resistors have been included along the wire on each of the short sides of the valve. The value of the resistor on one side is twice that on the other side.
Answer:
The answer is "0.00466 N".
Explanation:
[tex]F=(B \times i) L\\\\[/tex]
therefore the smaller side is parallel to magnetic field
[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]
calculating the force on the layer side:
[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]
Therefore [tex]F_o[/tex] the net force on the rectangular loop [tex]= 0.00466 \ N[/tex]
An object moving with initial velocity 10 m/s is subjected to a uniform acceleration of 8 m/s ^² . The displacement in the next 2 s is: (a) 0m (b) 36 m (c) 16 m (d) 4 m
you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you or the friends you left behind back on earth
Your friend would have a better experience of this event, than you .
What is an eclipse?An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,
there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.
For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.
The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.
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calculate the value of 200°C in Kelvin
Answer:
473.15
Explanation:
which one of the following is a product of an acid base reaction? A. Base B. Acid C. Salt D. Fire
Answer:
salt
Explanation:
salt is a component for many acid base reactions
Which of the following would most likely produce the strongest magnetic
field?
A. A single moving electron
B. A stationary electric charge
C. A current in a straight wire
D. A current in a coil
Answer:
I current in a coil,,,,,,
Answer:D?
Explanation:
Sorry if i'm wrong....
When tightening a bolt, you push perpendicularly on a wrench a with force of 165 N at a distance of 0.140m from the center of the bolt. (A) How much torque are you exerting in newton x meters (relative to the center of the bolt)
Answer:
Part a)
23.1 Nm
..........
Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
The cells lie odjacent to the sieve tubes
Answer:
Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.the two ropes are used to vertically lower a 255 kg piano from exactly 4 m form a seocnd sotry window to the ground how much work is done by each of the three forces
Complete Question
The Question diagram is attached below
Answer:
a) [tex]W_{Fg}= 12500 Nm[/tex]
b) [tex]W_{T_1}= - 6339.3Nm[/tex]
c) [tex]W_{T_2}= - 3662.8Nm[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=255kg[/tex]
Distance [tex]d=4m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=F*d[/tex]
For [tex]F_g[/tex]
[tex]W_{Fg}=2500 x 5.3[/tex]
[tex]W_{Fg}= 12500 Nm[/tex]
For [tex]T_1[/tex]
[tex]W_{T_1}= - {1830 sin(60) x 4}[/tex]
[tex]W_{T_1}= - 6339.3Nm[/tex]
For [tex]T_2[/tex]
[tex]W_{T_2}= - {1295 sin(45) x 4}[/tex]
[tex]W_{T_2}= - 3662.8Nm[/tex]
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
[tex]v=0.9833\ c[/tex]
Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,
[tex]$\text{Density} = \frac{m}{lwh}$[/tex]
Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 [tex]kg/m^3[/tex]
So,
[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]
[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]
l = 0.024 m
Then for relativistic length contraction,
[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]
[tex]$\frac{v^2}{c^2}= 0.967$[/tex]
[tex]$\frac{v}{c}=0.9833$[/tex]
[tex]v=0.9833\ c[/tex]
Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
Which of the following would change mass as it accelerated? a bullet being shot out of a gun a roller skater pushing off a jet plane taking off a bowling ball slowing down
Answer:
Explanation:
A bullet being shot out of a gun tends to leave tiny amounts of the bullet behind due to friction between the bullet and the gun barrel.
A roller skater pushing requires the conversion of food chemical energy to muscle contraction energy. This conversion increases the body temperature and sweat is excreted to counteract the heat increase. The evaporation of the sweat causes a slight decrease in body mass.
A jet plane taking off consumes some of the fuel carried onboard to provide thrust. The products of combustion become part of the exhaust stream leaving the airplane rearward providing forward thrust.
A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 mmmm. The charge on the capacitor is 7.8 nCnC . A proton is released from rest next to the positive plate. Part A How long does it take for the proton to reach the negative plate
Answer:
t = 2.09 10⁻³ s
Explanation:
We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance
let's start with Newton's second law
∑ F = m a
the force is electric
F = q E
we substitute
q E = m a
a = [tex]\frac{q}{m} \ E[/tex]
a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]
a = 1.37 10³ m / s²
now we can use kinematics
x = v₀ t + ½ a t²
indicate that rest starts v₀ = 0
x = 0 + ½ a t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]
t = 2.09 10⁻³ s
A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.30 s V
(b) 3.0 s V
(c) 65 s V
Answer:
(a) emf = 0.507 V
(b) emf = 0.0507 V
(c) emf = 0.00234 V
Explanation:
Given;
number of turns of the coil, N = 40 turns
diameter of the coil, d = 11 cm
radius of the coil, r = 5.5 cm = 0.055 m
magnitude of the magnetic field, B = 0.4 T
The magnitude of the induced emf is calculated as;
[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]
(a) when the time, t = 0.3 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]
(b) when the time, t = 3.0 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]
(c) when the time, t = 65 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]
A 1050 kg car accelerates from 11.3 m/s to 26.2 m/s . What impulse does the engine give?
Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
