Answer:
[tex]v=1.24\times 10^8\ m/s[/tex]
Explanation:
Given that,
The refractive index of benzene is 2.419
We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,
[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]
So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.2 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.8 cm .
Required:
What is the magnitude of the charge (in nC) on each bead?
Answer:
The magnitude of the charge is 54.9 nC.
Explanation:
The charge on each bead can be found using Coulomb's law:
[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]
Where:
q₁ and q₂ are the charges, q₁ = q₂
r: is the distance of spring stretching = 4.8x10⁻² m
[tex]F_{e}[/tex]: is the electrostatic force
[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]
Now, we need to find [tex]F_{e}[/tex]. To do that we have that Fe is equal to the spring force ([tex]F_{k}[/tex]):
[tex] F_{e} = F_{k} = -kx [/tex]
Where:
k is the spring constant
x is the distance of the spring = 4.8 - 4.0 = 0.8 cm
The spring constant can be found by equaling the sping force and the weight force:
[tex] F_{k} = -W [/tex]
[tex] -k*x = -m*g [/tex]
where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²
[tex] k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m [/tex]
Now, we can find the electrostatic force:
[tex] F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N [/tex]
And with the magnitude of the electrostatic force we can find the charge:
[tex]q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC[/tex]
Therefore, the magnitude of the charge is 54.9 nC.
I hope it helps you!
The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.
Given the following data:
Original length = 4.0 cm to m = 0.04 mMass = 1.8 grams to kg = 0.0018New length = 5.2 cm to m = 0.052.Final length = 4.8 cm to m = 0.048 m.Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m
Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:
First of all, we would determine the spring constant of this lightweight spring by using this formula:
[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]
Spring constant, K = 1.47 N/m.
For the electrostatic force:
[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]
F = 0.01176 Newton.
Coulomb's law of electrostatic force.
Mathematically, the charge in an electric field is given by this formula:
[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]
Substituting the given parameters into the formula, we have;
[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]
Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]
Charge, q = 55.21 nC.
Read more on electric field here: https://brainly.com/question/14372859
A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be 43 cmcm outside the aquarium.
Required:
What is the actual distance between the can and the aquarium?
Answer:
The actual distance is [tex]d_a = 0.3233\ m[/tex]
Explanation:
From the question we are told that
The distance of the can is d = 43 cm = 0.43 m
Generally the actual distance is mathematically represented as
[tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]
Where [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is
[tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]
So
[tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]
[tex]d_a = 0.3233\ m[/tex]
Experiments are performed with ultracold neutrons having velocities of 7.54 m/s. (a) What is the wavelength (in nm) of such a neutron
Answer:
λ = 52.5 nm
Explanation:
De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression
p = h / λ
where the moment
p = mv
λ = h / mv
let's calculate
λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54
λ = 5.25 10⁻⁸ m
Let's reduce anm
λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)
λ = 52.5 nm
a body accelerate uniformly from rest at the rate of 3meters per seconds for 8 sec . calculate the distance covered by the body during the acceleration
SOL
Answer:
96 m
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²
Δx = 96 m
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film on water?
2. What do you think causes the colors on the artwork panels on the side of HLS2 (Health Sciences building) which change with time of day and the angle from which you view them?
Explanation:
1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.
2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.
Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.
We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.
These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.
Learn more: https://brainly.com/question/17177523
When water freezes, it expands about nine percent. What would be the pressure increase inside your automobile engine block if the water in there froze? The bulk modulus of ice is 2.0 × 109 N/m2, and 1 ATM = 1.01 × 105 N/m2.
Answer:
The pressure increase inside the automobile engine block is 1782.18 ATM
Explanation:
Given;
the change in volume of water, ΔV = 9%
the bulk modulus of ice, K = 2 x 10⁹ N/m²
Bulk modulus is given by;
[tex]K = -V\frac{dP}{dV}[/tex]
for pressure increase in the automobile engine block, when the water in there froze;
[tex]dP = K(\frac{dV}{V} )\\\\dP = K(\frac{0.09V}{V} )\\\\dP = 0.09K\\\\dP = 0.09 (2*10^9)\\\\dP = 1.8 *10^{8} \ N/m^2\\\\dP = 1782.18 \ ATM[/tex]
Therefore, the pressure increase inside the automobile engine block is 1782.18 ATM
The pressure increase inside your automobile engine block will be 1782.18 atm. The force involved vertical to the surface of an object per unit area is pressure.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
ΔV is the change in volume of water = 9%
K is the bulk modulus of ice = 2 x 10⁹ N/m²
dP is the change in the presure=?
The bulk modulus is found as;
[tex]\rm K=-V \frac{dp}{dv} \\\\[/tex]
The change in the presure is obtained as;
[tex]\rm dP = K\frac{dV}{V} \\\\ \rm dP = K\frac{0.09V}{V} \\\\ \rm dP = 0.09 K \\\\ \rm dP = 0.09 \times 2 \times 10^9 \\\\ \rm dP = 1.8 \times 10^8 \\\\ \rm dP =1782.18 \ atm[/tex]
Hence the pressure increase inside your automobile engine block will be 1782.18 atm.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
The cylinder is displaced 0.17 m downward from its equilibrium position and is released at time t = 0. Determine the displacement y and the velocity v when t = 3.1 s. The displacement and velocity are positive if downward, negative if upward. What is the magnitude of the maximum acceleration?
Complete Question
The image of this question is shown on the first uploaded image
Answer:
a
[tex]d =0.161 \ m[/tex]
b
[tex]v = - 0.054 \ m/s[/tex]
c
[tex]a = 6.12 \ m/s^2[/tex]
Explanation:
From the question we are told that
The maximum displacement is A = 0.17 m
The time considered is [tex]t = 3.1 \ s[/tex]
The spring constant is [tex]k = 137 \ N \cdot m[/tex]
The mass is [tex]m = 3.8 \ kg[/tex]
Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as
[tex]d = A cos (w t )[/tex]
Where [tex]w[/tex] is the angular frequency which is mathematically evaluated as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
substituting values
[tex]w = \sqrt{\frac{137}{ 3.8} }[/tex]
[tex]w =6[/tex]
So the displacement is at t
[tex]d = 0.17 cos (6 * 3.1 )[/tex]
[tex]d =0.161 \ m[/tex]
Generally the velocity of a SHM(simple harmonic motion) is mathematically represented as
[tex]v = - Asin (wt)[/tex]
substituting values
[tex]v = - 0.17 sin ( 6 * 3.1 )[/tex]
[tex]v = - 0.054 \ m/s[/tex]
Generally the maximum acceleration is mathematically represented as
[tex]a = w^2 * A[/tex]
substituting values
[tex]a_{max} = 6^2 * (0.17)[/tex]
substituting values
[tex]a = 6^2 * (0.17)[/tex]
[tex]a = 6.12 \ m/s^2[/tex]
In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.600 mm and the screen is 2.00 m from the slits. Find the distance along the screen between adjacent bright fringes.
Answer:
The distance is [tex]y = 0.00162 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]
The slit spacing is [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 2.0 \ m[/tex]
Generally the distance along the screen between adjacent bright fringes is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]
[tex]y = 0.00162 \ m[/tex]
The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.
Answer:
T = 80√3 N ≈ 139 N
W = 160 N
Explanation:
Sum of forces on B in the x direction:
∑F = ma
80 N sin 60° − T sin 30° = 0
T = 80 N sin 60° / sin 30°
T = 80√3 N
T ≈ 139 N
Sum of forces on B in the y direction:
∑F = ma
80 N cos 60° + T cos 30° − W = 0
W = 80 N cos 60° + T cos 30°
W = 40 N + 120 N
W = 160 N
The elastic limit of an alloy is 5.0×108 N/m2. What is the minimum radius rmin of a 4.0 m long wire made from the alloy if a single strand is designed to support a commercial sign that has a weight of 8000 N and hangs from a fixed point? To stay within safety codes, the wire cannot stretch more than 5.0 cm.
Answer:
4.5x 10^ -9m
Explanation:
See attached file
Answer:
The radius is [tex]r_{min} = 0.00226 \ m[/tex]
Explanation:
From the question we are told that
The elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]
The length is [tex]L = 4.0 \ m[/tex]
The weight of the commercial sign is [tex]F_s = 8000 \ N[/tex]
The maximum extension of the wire is [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]
Generally the elastic limit of an alloy (stress) is is mathematically represented as
[tex]\sigma = \frac{ F_s }{ A }[/tex]
Where A is the cross-sectional area of the wire which is mathematically represented as
[tex]A = \pi r^2[/tex]
here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign
So
[tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]
=> [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]
substituting values
[tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]
[tex]r_{min} = 0.00226 \ m[/tex]
An isolated system consists of two masses. The first, m1, has a mass of 1.90 kg, and is initially traveling to the east with a speed of 6.71 m/s. The second, m2, has a mass of 2.94 kg, and is initially traveling to the west with an unknown initial speed. The two masses collide head-on in a completely inelastic collision that stops them both. Calculate the initial kinetic energy of m2.
Answer:
m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J
Can anyone provide me the answer with explanation?
Answer:
the answer to your question us c honey
Answer:
C
Explanation:
This is so because different materials vary in resistance and conductance of current, heat. Metals are good conductors while none metals like rubber, plastic, glass etc are good insulators or resistors.
1.) When the acceleration is zero, what can you say about the velocity of an object?
Answer:
it is either constant or zero
Explanation:
3 QUESTIONS PLEASE ANSWER!
Answer:
1. A
2. C
3. D
Explanation:
At what minimum angle will you get total internal reflection of light traveling in diamond and reflected from ethanol? °
Answer:
34°
Using the relation
θᶜ = sin^-1(n₂/n₁),
where n1= the refractive index of light is propagating from a medium
And n2 = refractive index of medium into which light is entering
So we know that
refractive index of diamond at 589nm = 2.41= n₁
refractive index of ethanol at 589nm and 20°C = 1.36= n₂
Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°
Explanation:
"Two waves of the same frequency have amplitudes 1.00 and 2.00. They interfere at a point where their phase difference is 60.0°. What is the resultant amplitude?"
Answer:
The resultant amplitude of the two waves is 2.65.
Explanation:
Given;
amplitude of the first wave, A₁ = 1
amplitude of the second wave, A₂ = 2
phase difference of the two amplitudes, θ = 60.0°.
The resultant amplitude of two waves after interference is given by;
[tex]A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2Cos \theta} \\\\A = \sqrt{1^2 + 2^2 + 2(1)(2)Cos 60} \\\\A= 2.65[/tex]
Therefore, the resultant amplitude of the two waves is 2.65.
In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?
Answer:
Δy = 1 10⁻⁴ m
Explanation:
In double-slit experiments the constructive interference pattern is described by the equation
d sin θ = m λ
In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle
tan θ = y / L
as the angles are small,
tan θ = sin θ / cos θ = sin θ
substituting
sin θ = y / L
d y / L = m λ
y = m λ / d L
let's apply this formula for each wavelength
λ = 450 nm = 450 10⁻⁹ m
m = 5
d = 5.0 mm = 5.0 10⁻³ m
y₁ = 5 450 10⁻⁹ / (5 10⁻³ 1.4)
y₁ = 3.21 10⁻⁴ m
we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m
y₂ = 5 590 10⁻⁹ / (5 10⁻³ 1.4)
y₂= 4.21 10⁻⁴ m
the separation of these two lines is
Δy = y₂ - y₁
Δy = (4.21 - 3.21) 10⁻⁴ m
Δy = 1 10⁻⁴ m
how does a system naturally change over time
Answer:
The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.
Explanation:
Answer:
Decrease in entropy
Explanation:
Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.
In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.
On the other hand, in open system both the matter and energy move into and out of the system.
Using a conventional two-slit apparatus with light of wavelength 605 nm, 34 bright fringes per centimeter are observed on a screen 3.1 m away. What is the slit separation
Answer:
d = 6.38 x 10⁻³ m = 6.38 mm
Explanation:
Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:
Fringe Spacing = Δx = 1 cm/34
Δx = 0.0294 cm = 2.94 x 10⁻⁴ m
But, the formula for fringe spacing in a double slit experiment is:
Δx = λL/d
where,
λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m
L = Distance between screen and slits = 3.1 m
d = slit separation = ?
Therefore,
2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d
d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)
d = 6.38 x 10⁻³ m = 6.38 mm
Consider 2 converging lenses of focal lengths 5 mm (objective) and 50 mm.(eyepiece) An object 0.1 mm in size is placed a distance of 5.2 mm from the objective.
1. What is the size and location of the image from the objective? What is the linear magnification of this objective?
2. Treat the image from the objective as an object for the eyepiece. If the eyepiece creates an image at infinity, how far apart are the two lenses?
3. What is the angular magnification of the pair of lenses?
Answer:
1) q₁ = 12.987 cm , b) L = 17.987 cm , c) m = 179.87
Explanation:
We can solve the geometric optics exercises with the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's apply this equation to our case
1) f = 5mm = 0.5 cm
p₁ = 5.2 mm = 0.52 cm
h = 0.1 mm = 0.01 cm
1 / q₁ = 1 / f- 1 / p
1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923
1 / q₁ = 0.077
q₁ = 12.987 cm
2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece
p₂ = 5 cm
The absolute thing that goes through the two lenses is
L = q₁ + p₂
L = 12.987 +5
L = 17.987 cm
3) This lens configuration forms the so-called microscope, whose expression for the magnifications
m = -L / f_target 25 cm / f_ocular
m = - 17.987 / 0.5 25 / 5.0
m = 179.87
Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
Required:
What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?
Complete Question
Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
Required:
What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?
Answer:
The velocity is [tex]v = 80.82 \ m/s[/tex]
Explanation:
From the question we are told that
The magnitude of charge one is [tex]q_1 = 3.25 nC = 3.25 *10^{-9} \ C[/tex]
The magnitude of charge two [tex]q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C[/tex]
The distance of separation is [tex]d = 58.0 \ cm = 0.58 \ m[/tex]
Generally the electric potential of the electron at the midway point is mathematically represented as
[tex]V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }[/tex]
substituting values
[tex]V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }[/tex]
[tex]V = 1.8103 *10^{-8} \ V[/tex]
Now when the electron is 10 cm = 0.10 m from charge 1 , it is (0.58 - 0.10 = 0.48 m ) m from charge two
Now the electric potential at that point is mathematically represented as
[tex]V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}[/tex]
substituting values
[tex]V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}[/tex]
[tex]V_1 = 3.67*10^{-8} \ V[/tex]
Now the law of energy conservation ,
The kinetic energy of the electron = potential energy of the electron
i.e [tex]\frac{1}{2} * m * v^2 = [V_1 - V]* q[/tex]
where q is the magnitude of the charge on the electron with value
[tex]q = 1.60 *10^{-19} \ C[/tex]
While m is the mass of the electron with value [tex]m = 9.11*10^{-31} \ kg[/tex]
[tex]\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}[/tex]
[tex]v = \sqrt{6532.4}[/tex]
[tex]v = 80.82 \ m/s[/tex]
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately what? Group of answer choices
Answer:
0.05cos10t
Explanation:
X(t) = Acos(wt+φ)
The oscillation angular frequency can be calculated using below formula
w = √(k/M)
Where K is the spring constant
But we were given body mass of 5.0 kg
We know acceleration due to gravity as 9.8m)s^2
The lenghth of spring which stretches =10 cm
Then we can calculate the value of K
k = (5.0kg*9.8 m/s^2)/0.10 m
K= 490 N/m
Then if we substitute these values into the formula above we have
w = √(k/M)
w = √(490/5)
= 9.90 rad/s=10rads/s(approximately)
Its position as a function of time can be calculated using the below expresion
X(t) = Acos(wt+φ)
We were given amplitude of 5 cm , if we convert to metre = 0.05m
w=10rads/s
Then if we substitute we have
X(t)=0.05cos(10×t)
X(t)= 0.05cos10t
Therefore,Its position as a function of time=
X(t)= 0.05cos10t
The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?
1. I = 0 always.
2. I = constant, not equal to 0.
3. I increases, then is constant.
4. I instantly jumps up, then slowly decreases.
5. None of the above.
Answer:
The current in the ammeter is zero.
(1) is correct option.
Explanation:
Given that,
The capacitor is charged.
We need find the current after closed switched
We know that,
When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.
Then, the ammeter reads zero.
Hence, The current in the ammeter is zero.
(1) is correct option.
An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)
Answer:
10.4⁰ and 169.6⁰Explanation:
The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;
q is the charge on the electron
v is the velocity of the electron
B is the magnetic field strength
θ is the angle that the velocity of the electron make with the magnetic field.
Given parameters
F = 1.40*10⁻¹⁶ N
q = 1.6*10⁻¹⁹C
v = 3.94*10³m/s
B = 1.23T
Required
Angle that the velocity of the electron make with the magnetic field
Substituting the given parameters into the formula:
1.40*10⁻¹⁶ = 1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ
sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶
sinθ = 1.40/7.75392
sinθ = 0.1806
θ = sin⁻¹0.1806
θ₁ = 10.4⁰
Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁
θ₂ = 180-10.4
θ₂ = 169.6⁰
Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰
When an LRC series circuit is driven at resonance, which of the following statements about the circuit are correct? (There may be more than one correct choice.)a) The impedance of the circuit has its minimum value.b) The inductive reactance and the capacitive reactance are exactly equal to each other.c) The impedance of the circuit is zero.d) The inductive reactance and the capacitive reactance are both zero.e) The impedance of the circuit has its maximum value.
Answer:
Answers a) and b) should be marked as correct.
Explanation:
Recall that the resonance in an LRC circuit occurs when the current through the circuit is at its maximum, and such takes place when the impedance (Z) of the circuit reaches its maximum. This means that the impedance (see formula below) is at its minimum value:
[tex]Z=\sqrt{R^2+(\omega\,L-\frac{1}{\omega\,C})^2 }[/tex]
as per the impedance expression above, such happens when the term in parenthesis inside the root which contains the inductive reactance ([tex]\omega\,L[/tex]) and the capacitive reactance ([tex]1/\omega\,C[/tex]) have the same value.
Therefore, answers:
a) "The impedance of the circuit has its minimum value."
and
b) "The inductive reactance and the capacitive reactance are exactly equal to each other."
are correct answers.
(a) The impedance of the circuit has its minimum value.
(b) The inductive reactance and the capacitive reactance are exactly equal to each other
LRC series circuit consists of inductor, resistor and capacitor is series.
The impedance of the circuit is calculated as follows;
[tex]Z = \sqrt{R^2 + (X_C -X_L)^2}[/tex]
where;
R is the resistance[tex]X_C[/tex] is the capacitive reactance[tex]X_L[/tex] is the inductive reactanceThe impedance of the circuit is minimum when the capacitive reactance is equal to the inductive reactance.
[tex]X_C = X_L \\\\Z = \sqrt{R^2 \ + (0)^2} \\\\Z = R[/tex]
Learn more here:https://brainly.com/question/17330472
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing Photo 2.
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
[tex]E=\frac{F}{Q}[/tex]
But the force F
[tex]F= \frac{kQ1Q2}{r^2}[/tex]
But the electric field intensity due to a point charge Q at a distance r meters away is given by
[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?
Answer:
a) Connect a series resistance of 8,47 ohms
b)14,16 [A]
c) r = 10,96 ohms
Explanation:
My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower
In America
P = V*I
1700 (w) = 120*I
I = 1700/120 [A]
I = 14,16 [A] current needed for the blower
In Europe
120 (v) (the drop of voltage I need) when a current of 14,16 passes through to series resistor is
V = I*R 120 = 14,16* R R = 8,47 ohms
c) P = I*r²
1700 (w) = 14,16 (A) * r²
r² = 120,06
r = 10,96 ohms
Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?
Answer:
The separation of the two slits is 0.456 mm.
Explanation:
Given the wavelength of light = 519 nm
The indifference pattern = 4.6 m
Adjacent bright fringes = 5.2 mm
In the interference, the equation required is Y = mLR/d
Here, d sin theta = mL
L = wavelgnth
For bright bands, m is the order = 1,2,3,4
For dark bands, m = 1.5, 2.5, 3.5, 4.5
R = Distance from slit to screen (The indifference pattern)
Y = Distance from central spot to the nth order fringe or fringe width
Thus, here d = mLR/Y
d = 1× 519nm × 4.6 / 5.2mm
d = 0.459 mm
A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?
Answer:
Pressure, P = 1250 Pa
Explanation:
Given that,
Weight of a brick, F = 50 N
Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm
We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.
A = 40 cm × 10 cm = 400 cm² = 0.04 m²
So,
Pressure,
[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]
So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.
The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.
The given data in the problem is;
W is the weight of a brick = 50 N
The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm
A is the area,
The area is found as;
A=40 cm × 10 cm = 400 cm² = 0.04 m²
The pressure is the ratio of the force and area
[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]
Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
