Answer:
A. Attractive
B. ( μ₀I² ) / ( 2πd )
Explanation:
A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.
B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -
[tex]B_1[/tex] = μ₀I / 2πd
The force experienced by wire 2 should thus be -
[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )
= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )
= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )
Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...
( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution
An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it
Answer:
V = 451.47 volts
Explanation:
Given that,
Distance, d = 0.21 m
Initial speed, u = 0
Time, t = 33.3 ns
Let v is the final velocity. Using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0
So,
[tex]d=\dfrac{1}{2}(v-u)t[/tex]
[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]
Now applying the conservation of energy i.e.
[tex]\dfrac{1}{2}mv^2=qV[/tex]
V is voltage
[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]
So, the voltage is 451.47 V.
1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.
Answer:
3.26m
Explanation:
See attached file
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
Learn more:
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A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely immersed in water, the scale reads 15.3 N.
A) What is the volume of the block?
B) What is the density of the block?
Answer:
A) [tex]V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}[/tex]
B) [tex] d = 4181.49 kg/m^{3} = 4.18 g/cm^{3} [/tex]
Explanation:
A) Using the Archimedes' force we can find the weight of water displaced:
[tex] W_{d} = W_{a} - W_{w} [/tex]
Where:
[tex]W_{a}[/tex]: is the weight of the block in the air = 20.1 N
[tex]W_{w}[/tex]: is the weight of the block in the water = 15.3 N
[tex] W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N [/tex]
Now, the mass of the water displaced is:
[tex] m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg [/tex]
The volume of the block can be found using the mass of water displaced and the density of the water:
[tex]V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}[/tex]
B) The density of the block can be found as follows:
[tex] d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3} [/tex]
I hope it helps you!
Which of the following represents a concave mirror? +f,-f,-di,+di
Answer:
fully describes the concave mirror is + f
Explanation:
A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f
This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di
Answer:
+f
Explanation:
because you have to be really dumb to get an -f
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.
Answer:
4.9x10^-6T
Explanation:
See attached file
Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra
Answer:
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Explanation:
Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:
[tex]v = v_{o} + g\cdot t[/tex]
[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.
[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.
Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:
[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]
[tex]v = -196.14\,\frac{m}{s}[/tex]
[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]
[tex]y-y_{o} = -1961.4\,m[/tex]
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?
Answer:
The angle is [tex]\theta = 15.48^o[/tex]
Explanation:
From the question we are told that
The distance of the dartboard from the dart is [tex]d = 3.66 \ m[/tex]
The time taken is [tex]t = 0.455 \ s[/tex]
The horizontal component of the speed of the dart is mathematically represented as
[tex]u_x = ucos \theta[/tex]
where u is the the velocity at dart is lunched
so
[tex]distance = velocity \ in \ the\ x-direction * time[/tex]
substituting values
[tex]3.66 = ucos \theta * (0.455)[/tex]
=> [tex]ucos \theta = 8.04 \ m/s[/tex]
From projectile kinematics the time taken by the dart can be mathematically represented as
[tex]t = \frac{2usin \theta }{g}[/tex]
=> [tex]usin \theta = \frac{g * t}{2 }[/tex]
[tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]
[tex]usin \theta = 2.23[/tex]
=> [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]
[tex]\theta = tan^{-1} [0.277][/tex]
[tex]\theta = 15.48^o[/tex]
A small barge is being used to transport trucks across a river. If the barge is 10.00 m long by 8.00 m wide and sinks an additional 3.75 cm into the river when a loaded truck pulls onto it, determine the weight of the truck and load.
Answer: Weight truck+load = 29.4×[tex]10^{3}[/tex] N
Explanation: When an object floats in a fluid, there is an upward force, caused by the liquid, acting on the object that opposes the weight of the immersed object. This force is called Buyoyant Force and is determined by:
B = d*V*g
where
d is density of the fluid;
V is volume of liquid displaced due to the immersed object;
g is acceleration due to gravity;
For the truck, the system is in equilibrium, which means buyoyant force is equal weight. Then:
Volume displaced is
V = 10*8*0.0375
V = 3 [tex]m^{3}[/tex]
Density of water: 1000kg/[tex]m^{3}[/tex]
[tex]F_{P} = F_{B}[/tex]
[tex]F_{P}[/tex] = 1000*3*9.8
[tex]F_{P}[/tex] = 29.4×[tex]10^{3}[/tex] N
The weight of the truck and the load is 29.4×[tex]10^{3}[/tex] Newtons
A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense.
(a) For points OUTSIDE the loop, the magnetic field caused by this current:________.
a. points straight up.
b. circles the loop in a clockwise direction.
c. circles the loop in a counterclockwise direction.
d. points straight down.
e. is zero.
(b) For points INSIDE the loop, the magnetic field caused by this current:________.
a. circles the loop in a counterclockwise direction.
b. points straight up.
c. points straight down.
d. circles the loop in a clockwise direction.
e. is zero
Answer:
D &B
Explanation:
Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force
A 43.0-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.05R.
Required:
a. What is the speed of the car at the top of the vertical loop?
b. What is the magnitude of the normal force acting on the car at the top of the vertical loop?
Answer:
A.) 909 cm/s
B.) 33075 N
Explanation:
A.) Given that the
Mass M = 43 g
Height h = 4.05 R
Radius r = R
At the top of the loop, the maximum potential energy P.E = mgh
Substitutes m and h into the formula where g = 9.8 m/s^2 = 9610.517 cm/s^2
P.E = 43 × 9610.517 × 4.05R
P.E = 1673671.536R J
According to conservative of energy
The maximum P.E = maximum K.E
But K.E = 1/2mv^2
1673671.536R = 1/2mv^2
Substitutes for mass m into the formula
1673671.536R = 1/2× 4.05R × v^2
The R will cancel out
Cross multiply
4.05 v^2 = 3347343.072
V^2 = 3347343.072 / 4.05
V^2 = 826504.4622
V = sqrt( 826504.4622)
V = 909 cm/s
B.) At the top of the loop, the centripetal force = the sum of the normal force N and the weight W of the car. That is,
MV^2/R = N + W
Make N the subject of formula
N = mv^2/ R - W
Where W = mg
Substitute all the parameters into the formula
N = (4.05R × 909^2) /R - 4.05R × 9610.517
N = 3346438.05 - 38922.59
N = 3307515 N
A sinusoidal electromagnetic wave emitted by a mobile phone has a wavelength of 34.8 cm and an electric-field amplitude of 5.70×10−2 V/m at a distance of 210 m from the phone.
Calculate
(a) the frequency of the wave;
(b) the magnetic-field amplitude;
(c) the intensity of the wave.
Answer:
a) [tex] f = 8.62 \cdot 10^{8} Hz [/tex]
b) [tex] B = 1.9 \cdot 10^{-10} T [/tex]
c) [tex] I = 4.30 \cdot 10^{-6} W/m^{2} [/tex]
Explanation:
a) The frequency (f) of the wave can be found as follows:
[tex] f = \frac{c}{\lambda} [/tex]
Where:
c: is the speed of light = 3x10⁸ m/s
λ: is the wavelength = 34.8 cm
[tex] f = \frac{3 \cdot 10^{8} m/s}{0.348 m} = 8.62 \cdot 10^{8} Hz [/tex]
b) The magnetic-flied amplitude (B) is:
[tex] B = \frac{E}{c} [/tex]
Where:
E: is the electric field amplitude = 5.70x10⁻² V/m
[tex] B = \frac{E}{c} = \frac{5.70 \cdot 10^{-2} V/m}{3 \cdot 10^{8} m/s} = 1.9 \cdot 10^{-10} T [/tex]
c) The intensity of the wave (I) is the following:
[tex] I = \frac{E*B}{2\mu_{0}} [/tex]
Where:
μ₀: is the permeability of free space = 1.26x10⁻⁶ m*kg/(s²A²)
[tex] I = \frac{E*B}{2\mu_{0}} = \frac{5.70 \cdot 10^{-2} V/m*1.9 \cdot 10^{-10} T}{2*1.26 \cdot 10^{-6} m*kg/((s^{2}A^{2})} = 4.30 \cdot 10^{-6} W/m^{2} [/tex]
I hope it helps you!
The frequency of the wave is [tex]8.62\times 10^8\rm\;Hz[/tex], the magnetic-field amplitude is [tex]1.9\times 10^{-10}\rm\;T[/tex], and the intensity of the wave is [tex]4.298\rm\;W/m^2[/tex].
Given information:
A mobile phone emits electromagnetic radiation.
The wavelength of the wave is [tex]\lambda=34.8[/tex] cm.
The electric-field amplitude is [tex]5.70\times10^{-2}[/tex] V/m.
Phone is at a distance of 210 m.
The speed of the electromagnetic wave is [tex]c=3\times 10^8[/tex] m/s.
(a)
Now, the frequency of the wave will be calculated as,
[tex]f=\dfrac{c}{\lambda}\\f=\dfrac{3\times 10^8}{0.348}\\f=8.62\times 10^8\rm\;Hz[/tex]
(b)
The magnetic-field amplitude can be calculated as,
[tex]B=\dfrac{E}{c}\\B=\dfrac{5.70\times10^{-2}}{3\times 10^8}\\B=1.9\times 10^{-10}\rm\;T[/tex]
(c)
[tex]\mu_0[/tex] is the permeability of the vacuum. [tex]\mu_0=1.26\times10^{-6} \rm\;\frac{kg-m}{(A^2s^2)}[/tex]
The intensity of the wave can be calculated as,
[tex]I=\dfrac{BE}{2\mu_0}\\I=\dfrac{1.9\times10^{-10 }\times5.7\times10^{-2}}{2\times1.26\times10^{-6}}\\I=4.298\rm\;W/m^2[/tex]
Therefore, the frequency of the wave is [tex]8.62\times 10^8\rm\;Hz[/tex], the magnetic-field amplitude is [tex]1.9\times 10^{-10}\rm\;T[/tex], and the intensity of the wave is [tex]4.298\rm\;W/m^2[/tex].
For more details, refer to the link:
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During a visit to the beach, you get in a small rubber raft and paddle out beyond the surf zone. Tiring, you stop and take a rest. Describe the movement of your raft during your rest. How does this movement differ, if at all, from what you would have experienced if you had stopped paddling while in the surf zone?
Answer:
The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach
Explanation:
The movement in the two parts is very different, when the surf zone has passed it is in a deeper part of the water where the seabed does not rise much, therefore due to the movement of the waves there is an upward oscillatory movement and descending, in this movement there is no horizontal displacement.
When it is within the southern zone, there is a rapid rise of the sea floor, which generates a horizontal movement, having a traveling wave, therefore your movement is more complicated, you can have some oscillating movement on the axis and, but in addition to this you have a horizontal movement that reaches you towards the beach, forming a Traveling wave.
The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach
What explains why a prism separates white light into a light spectrum?
A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.
Answer:
I think the answer probably be B
What explains why a prism separates white light into a light spectrum ?
C. The different colors that make up a white light have different refractive indexes in glass.
✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).
✔ That's why purple is more deflected so is lower than red radiation.
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
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
352 m/s
Explanation:
The velocity of sound in air is approximated as:
v ≈ 331 + 0.6 T
where v is the velocity in m/s and T is the temperature in Celsius.
At T = 35:
v = 331 + 0.6 (35)
v = 352 m/s
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through
Answer:
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Explanation:
The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:
[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]
Where:
[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.
[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.
[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.
The final speed of the block is cleared:
[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]
[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]
If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:
[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]
[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If of the light passes through this combination, what is the angle between the transmission axes of the two filters
Answer:
The angle between the transmission axes of the filters is 65°
Explanation:
The complete question is
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 18% of the light passes through this combination, what is the angle between the transmission axes of the two filters.
From Malus law,
[tex]I = I_{0} cos^{2} \beta[/tex] ....1
where [tex]I[/tex] is the intensity of the polarized light,
[tex]I_{o}[/tex] is the intensity of the incident light
β the angle between the transmission axes of the two filters
Since the intensity is reduced to 18% or 0.18 of its initial value, this means that
[tex]cos^{2} \beta[/tex] = 0.18
substituting into the equation above, we have
[tex]I = 0.18I_{0}[/tex] ....2
equating the two equations, we have
[tex]I_{0}cos^{2} \beta[/tex] = [tex]0.18I_{0}[/tex]
[tex]cos^{2}\beta[/tex] = [tex]\frac{0.81I_{0} }{I_{0} }[/tex] = 0.18
[tex]cos \beta[/tex] = [tex]\sqrt{0.18}[/tex] = 0.424
[tex]\beta[/tex] = [tex]cos^{-1} 0.424[/tex] = 64.9 ≅ 65°
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic friction between the rod and rails is 0.200, what vertical magnetic field is required to keep the rod moving at a constant speed?
Answer:
The magnetic field is [tex]B = 0.0764 \ T[/tex]
Explanation:
From the question we are told that
The mass of the metal is [tex]m = 0.210 \ kg[/tex]
The current is [tex]I = 11.0 \ A[/tex]
The distance between the rail(length of the rod ) is [tex]d = 0.490 \ m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.200[/tex]
Generally the magnetic force is mathematically represented as
[tex]F_b = B * I * d[/tex]
Given that the rod is moving at a constant velocity, it
=> [tex]F_b = F_k[/tex]
Where [tex]F_k[/tex] is the kinetic frictional force which is mathematically represented as
[tex]F_k = \mu_k * m * g[/tex]
So
[tex]B * I * d = \mu_k * m * g[/tex]
=> [tex]B = \frac{\mu_k * m * g}{I * d }[/tex]
substituting values
=> [tex]B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }[/tex]
=> [tex]B = 0.0764 \ T[/tex]
Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
What is the wavelength λλlambda of the wave described in the problem introduction? Express the wavelength in terms of the other given variables and constants
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda= \frac{2 \pi }{k}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]
The magnetic field is [tex]\= B = B_0 sin (kx -wt) \r k[/tex]
From the above equation
and k is the wave number which is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]\lambda= \frac{2 \pi }{k}[/tex]
Where [tex]\lambda[/tex] is the wavelength
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
You have explored constructive interference from multi-layer thin films. It is also possible for interference to be destructive, a phenomenon exploited in making antireflection coatings for optical elements such as eyeglasses. In order to allow the lenses to be thinner (and thus lighter weight), eyeglass lenses can be made of a plastic that has a high index of refraction (np = 1.70). The high index causes the plastic to reflect light more effectively than does glass, so it is desirable to reduce the reflection to avoid glare and to allow more light to reach the eye. This can be done by applying a thin coating to the plastic to produce destructive interference.
a. Consider a plastic eyeglass lens with a coating of thickness d with index nc . Light with wavelength is incident perpendicular to the lens. If nc < n p , then determine an equation for d in terms of the given variables (and an integer m) in order for there to be destructive interference between the light reflected from the top of the coating and the light reflected from the coating/lens interface.
b. Repeat part a assuming that nc > n p .
c. Choose a suitable value for nc and calculate a value for d that will result in destructive interference for 500 nm light. Note that materials to use for coatings that have nc < 1.3 or nc > 2.5 are difficult to find.
d. Does the index of refraction n p of the eyeglass lens itself matter? Explain.
Answer:
a) d sin θ = m λ₀ / n
b) d sin θ = (m + ½) λ₀ / n
c) d = 2,439 10⁻⁷ m
Explanation:
For the interference these rays of light we must take as for some aspects,
* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2
* when the ray penetrates the lens the donut length changes by the refractive index
λ = λ₀ / n
now let's write the destructive interference equation for these lightning bolts
d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n
d sin θ = m λ₀ / n
b) now nc> np
in this case there is no phase change in the reflected ray and the equation for destructive interference remains
d sin θ = (m + ½) λ₀ / n
c) select the value of nc = 2.05 of the ZnO
we calculate the thickness of the film (d)
d = m λ / (n sin 90)
in this type of interference the observation is normal, that is, the angle is 90º)
d = 1 500 10-9 / (2.05 1)
d = 2,439 10⁻⁷ m
d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference
Suppose that a 0.275 m radius, 500 turn coil produces an average emf of 11800 V when rotated one-fourth of a revolution in 4.42 ms, starting from its plane being perpendicular to the magnetic field.
Required:
Find the magnetic field strength needed to induce an average emf of 10,000 V.
Answer:
The magnetic field strength : 0.372 T
Explanation:
The equation of the induced emf is given by the following equation,
( Equation 1 ) emf = - N ( ΔФ / Δt ) - where N = number of turns of the coil, ΔФ = change in the magnetic flux, and Δt = change in time
The equation for the magnetic flux is given by,
( Equation 2 ) Ф = BA( cos( θ ) ) - where B = magnetic field, A = area, and θ = the angle between the normal and the magnetic field
The area of the circular coil is a constant, as well as the magnetic field. Therefore the change in the magnetic flux is due to the angle between the normal and the magnetic field. Therefore you can expect the equation for the change in magnetic flux to be the same as the magnetic flux, but only that there must be a change in θ.
( Equation 3 ) ΔФ = BA( Δcos( θ ) )
Now as the coil rotates one-fourth of a revolution, θ changes from 0 degrees to 90 degrees. The " change in cos θ " should thus be the following,
Δcos( θ ) = cos( 90 ) - cos( 0 )
= 0 - 1 = - 1
Let's substitute that value in the third equation,
( Substitution of Δcos( θ ) previously, into Equation 3 )
ΔФ = BA( - 1 ) = - BA
Remember the first equation? Well if the change in the magnetic flux = - BA, then through further substitution, the emf should = - N( - BA ) / Δt. In other words,
emf = - N( - BA ) / Δt,
emf = NBA / Δt,
B = ( emf )Δt / NA
Now that we have B, the magnetic field strength, isolated, let's solve for the area of the circular coil and substitute all known values into this equation.
Area ( A ) = πr²,
= π( 0.275 )² = 0.2376 m²,
B = ( 10,000 V )( 4.42 [tex]*[/tex] 10⁻³ s ) / ( 500 )( 0.2376 m² ) = ( About ) 0.372 T
The magnetic field strength : 0.372 T
A nano-satellite has the shape of a disk of radius 0.80 m and mass 8.50 kg.
The satellite has four navigation rockets equally spaced along its edge. Two
navigation rockets on opposite sides of the disk fire in opposite directions
to spin up the satellite from zero angular velocity to 14.5 radians/s in 30.0
seconds. If the rockets each exert their force tangent to the edge of the
satellite (the angle theta between the force and the radial line is 90
degrees), what was is the force of EACH rocket, assuming they exert the
same magnitude force on the satellite?
Answer:
Explanation:
moment of inertia of satellite I = 1/2 m R²
m is mass and R is radius of the disc
I = 0.5 x 8.5 x 0.8²
= 2.72 kg m²
angular acceleration α = change in angular velocity / time
α = (14.5 - 0) / 30
α = .48333
Let force of each rocket = F
torque created by one rocket = F x R
= F x .8
Torque created by 4 rockets = 4 x .8 F = 3.2 F
3.2 F = I x α
3.2 F = 2.72 x .48333
F = 0 .41 N
4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?
Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
To know more about Earthquake:
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Which type of reaction is shown in this energy diagram?
Answer:
Option C
Explanation:
The graph shows endothermic reaction because the reactants are lower in energy and the products are higher is energy. Endothermic reactions absorb energy having products with higher energy.
Answer:
C
Explanation:
In an endothermic reaction, the energy-time graph shows reactants are at a lower energy level than the products.
What happens to a bar of metal when it's heated?
A.It gets longer.
B.The effect depends on the density of the bar.
C.It stays the same length.
D.It gets shorter
Hey There!!
Your answer will be A. It get longer.
Because, The kinetic energy of its atoms increase. They vibrate faster. This means that each atom will take up more space due to its movement so the material will expand. Some metals expand more than others due to differences in the forces between the atoms / molecules. . . .
Hope This Helps <3!!
