Answer:
The focal length of the lens is 2.5 cm
Explanation:
Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;
[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]
Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:
[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]
then, solving for the focal distance knowing that the object distance is 1.5 cm:
[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/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
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.
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.
The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
Question 2
A) A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released? (1 point)
The spring exerts a restoring force to the right and compresses even further
The spring exerts a restoring force to the left and returns to its equilibrium position
The spring exerts a restoring force to the right and returns to its equilibrium position
The spring exerts a restoring force to the left and stretches beyond its equilibrium position
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
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
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
If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
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.
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
The rectangular plate is tilted about its lower edge by a cable tensioned at a constant 600 N. Determine and plot the moment of this tension about the lower edge AB of the plate for the range 0 ≤ θ ≤ 90°
Answer:
Explanation:
From the figure , it is clear that moment of tension is balanced by moment of weight of plate about the line AB which is acting as axis . If W be the weight of plate ,
moment of tension about AB = moment of weight W about line AB
= W x 2.5 cosθ
moment of tension about AB = 2.5 W cosθ
here only variable is cosθ which changes when θ changes
So, moment of tension about AB varies according to cosθ.
When θ = 0
moment of tension about AB = 2.5 W x cos 0 = 2.5 W . It is the maximum value of moment of tension .
When θ = 90°
moment of tension about AB = 2.5 W cos 90 = 0
moment of tension about AB = 0
So graph of moment of tension about AB will vary according to graph of
cosθ . It has been shown in the file attached .
A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?
Explanation:
Using Equations of Motion :
[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]
Height = 0 * 4 + 4.9 * 16
Height = 78.4 m
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]
A father and his young son get on a teeter-totter. The son sits 2 m fromthe center, but the father has to sit closer to balance. Where does the father have to sit to balance the teeter-totter if he weighs 4 times as much as his son?
Answer:
The distance of the father from the center is [tex]d_f = \frac{1}{2} \ m[/tex]
Explanation:
From the question we are told that
The distance of the son from the center is [tex]d_s = 2 \ m[/tex]
Let the mass of the son be [tex]m_s[/tex]
then the mass of the father is [tex]m_f = 4m_s[/tex]
Now for the teeter-totter to be balanced the torque due to the weight of the father must be equal to the torque due to the weight the son, this is mathematically represented as
[tex]\tau_s = \tau_f[/tex]
Where [tex]\tau_s[/tex] is the torque of the son which is mathematically represented as
[tex]\tau_ s = m_s * d_s * g[/tex]
while [tex]\tau_f[/tex] is the torque of the father which is mathematically represented as
[tex]\tau_f = m_f * d_f * g[/tex]
=> [tex]\tau_f = 4 m_s * d_f * g[/tex]
So
[tex]4 m_s * d_f * g = m_s * d_s * g[/tex]
substituting values
[tex]4 * d_f * = 2[/tex]
=> [tex]d_f = \frac{1}{2} \ m[/tex]
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How fast does the electron move away from the metal
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.
You simultaneously shine two light beams, each of intensity I0, on an ideal polarizer. One beam is unpolarized, and the other beam is polarized at an angle of exactly 30.0∘ to the polarizing axis of the polarizer. Find the intensity of the light that emerges from the polarizer. Express your answer in term of I0 .
Answer:
The emerging intensity is equal to 0.75[tex]I_{o}[/tex]
Explanation:
The initial intensity of the light = [tex]I_{o}[/tex]
The angle of polarization β = 30°
We know that the polarized light intensity is related to the initial light intensity by
[tex]I[/tex] = [tex]I_{0} cos^{2}\beta[/tex]
where [tex]I[/tex] is the emerging polarized light intensity
inserting values gives
[tex]I[/tex] = [tex]I_{0} cos^{2}[/tex] 30°
[tex]cos^{2}[/tex] 30° = [tex](cos 30)^{2}[/tex] = [tex](\frac{\sqrt{3} }{2} )^{2}[/tex] = 0.75
[tex]I[/tex] = 0.75[tex]I_{o}[/tex]
Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answer:
They both have the same acceleration
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
please help !!!!!!!!!!
Answer:
Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.
Explanation:
As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.
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
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
A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides without friction down the inner surface toward the bottom of the bowl. At a certain point of its path the mass achieves a speed v = 3.57 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?
Answer:
The angle is [tex]\theta = 36.24 ^o[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.6 \ kg[/tex]
The radius is [tex]r = 1.1 \ m[/tex]
The speed is [tex]v = 3.57 \ m /s[/tex]
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
[tex]m * g * h = \frac{1}{2} * m * v^2[/tex]
=> [tex]h = \frac{1}{2 g } * v^2[/tex]
Here h is the vertical distance traveled by the mass which is also mathematically represented as
[tex]h = r * sin (\theta )[/tex]
So
[tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]
substituting values
[tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]
[tex]\theta = 36.24 ^o[/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
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 square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf
Answer:
The magnitude of the average induced emf is 90V
Explanation:
Given;
area of the square coil, A = 0.4 m²
number of turns, N = 15 turns
magnitude of the magnetic field, B = 0.75 T
time of change of magnetic field, t = 0.05 s
The magnitude of the average induced emf is given by;
E = -NAB/t
E = -(15 x 0.4 x 0.75) / 0.05
E = -90 V
|E| = 90 V
Therefore, the magnitude of the average induced emf is 90V
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
"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.
Sammy is 5 feet and 5.3 inches tall. What is Sammy's height in inches?
Answer:
[tex]\boxed{\sf 65.3 \ inches}[/tex]
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
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
Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is 4.82×10-11 T.A) Calculate the wavelength.B) Calculate the wave number.C) Calculate the angular frequency.
D) Calculate the electric-field amplitude.
Answer:
A
[tex]\lambda = 361.45 \ m[/tex]
B
[tex]k = 0.01739 \ rad/m[/tex]
C
[tex]w = 5.22 *10^{6} \ rad/s[/tex]
D
[tex]E = 0.01446 \ N/C[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 83 0 \ kHz = 830 *10^{3} \ Hz[/tex]
The magnetic field amplitude is [tex]B = 4.82*10^{-11} \ T[/tex]
Generally wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]\lambda = \frac{3.0*10^{8}}{ 830 *10^{3}}[/tex]
=> [tex]\lambda = 361.45 \ m[/tex]
Generally the wave number is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]k = \frac{2 * 3.142 }{ 361.45 }[/tex]
=> [tex]k = 0.01739 \ rad/m[/tex]
Generally the angular frequency is mathematically represented as
[tex]w = 2 * \pi * f[/tex]
=> [tex]w = 2 * 3.142 * 830*10^{3}[/tex]
=> [tex]w = 5.22 *10^{6} \ rad/s[/tex]
The the electric-field amplitude is mathematically represented as
[tex]E = B * c[/tex]
=> [tex]E = 4.82 *10^{-11} * 3.0*10^{8}[/tex]
=> [tex]E = 0.01446 \ N/C[/tex]
This question involves the concepts of wavelength, frequency, wave number, and electric field.
a) The wavelength is "361.44 m".
b) The wave number is "0.0028 m⁻¹".
c) The angular frequency is "5.22 x 10⁶ rad/s".
d) The electric field amplitude is "0.0145 N/C".
a)
The wavelength can be given by the following formula:
[tex]c=f\lambda[/tex]
where,
c = speed of light = 3 x 10⁸ m/s
f = frequency = 830 KHz = 8.3 x 10⁵ Hz
λ = wavelength = ?
Therefore,
[tex]3\ x\ 10^8\ m/s=(8.3\ x\ 10^5\ Hz)\lambda\\\\\lambda=\frac{3\ x\ 10^8\ m/s}{8.3\ x\ 10^5\ Hz}\\\\[/tex]
λ = 361.44 m
b)
The wave number can be given by the following formula:
[tex]wave\ number = \frac{1}{\lambda} = \frac{1}{361.44\ m}[/tex]
wave number = 0.0028 m⁻¹
c)
The angular frequency is given as follows:
[tex]\omega = 2\pi f = (2)(\pi)(8.3\ x\ 10^5\ Hz)[/tex]
ω = 5.22 x 10⁶ rad/s
d)
The electric field amplitude can be given by the following formula:
[tex]\frac{E}{B} = c\\\\c(B)=E\\\\E = (3\ x\ 10^8\ m/s)(4.82\ x\ 10^{-11}\ T)\\[/tex]
E = 0.0145 N/C
Learn more about wavelength and frequency here:
https://brainly.com/question/12924624?referrer=searchResults
