Answer:
Explanation:
a ) A good conducting material conducts heat easily from one point to another . So the metal spoon which is good conductor will become hot easily by conducting heat from hot water to tip of the spoon . So even after touching the far end of spoon , we can feel the heat of the hot water in the cup . For other material like wood or plastic we have to hold the part which is very close to water to feel heat of water .
b ) The best conductor will be that which touches hot even when we touch its farthest end . Hence metal spoon will be best conductor . Plastic pen will be worst conductor because even touching its part close to hot water , we do not feel much heat .
c ) metal spoon > glass rod > wooden pencil > plastic pen .
At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
Answer:
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.
Explanation:
The energy stored in the inductor is given as
E₁ = ½LI²
The rate at which energy is stored in the inductor is
(dE₁/dt) = (d/dt) (½LI²)
Since L is a constant
(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)
(dE₁/dt) = LI (dI/dt)
Rate of Energy dissipated in a resistor = Power = I²R
(dE₂/dt) = I²R
When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field
(dE₁/dt) = (dE₂/dt)
OK (dI/dt) = I²R
L (dI/dt) = IR
Current in a this kind of series setup of inductor and resistor at any time, t, is given as
I = (V/R) (1 - e⁻ᵏᵗ)
k = (1/time constant) = (R/L)
(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ
L (dI/dt) = IR
L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)
V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)
e⁻ᵏᵗ = 1 - e⁻ᵏᵗ
2 e⁻ᵏᵗ = 1
e⁻ᵏᵗ = (1/2) = 0.5
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = -0.69315
- kt = -0.69315
kt = 0.69315
k = (1/time constant)
Time constant = 36.0 ms = 0.036 s
k = (1/0.036) = 27.78
27.78t = 0.69315
t = (0.69315/27.78) = 0.02495 = 24.95 ms
Hope this Helps!!!
Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.
Answer:
Explanation:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4*10^6[2*2*10^-2/7*10^14]^1/2
=3*10^-2=3 cm
now we find the velocity f the electron strike the plate
v^2-(4*10^6)^2=2*7*10^14*2*10^-2
v^2=16*10^12+28*10^12
v^2=44*10^12
speed after hits =>V=6.6*10^6 m/s
Suppose you have a box of identical resistors. Each resistor in the box has a resistance of 20 Ohms. Explain how to connect them together so the total resistance of the network is 25 Ohms. Use the minimum number of resistors in your design.
Answer:
By using 4 resistors in parallel and followed by a resistor in series.
Explanation:
Let consider 5 resistors with a resistance of 20 ohms each, the total resistance is equal to 25 ohms when the first four resistance are settled in parallel and is followed by a resistor is series. The calculations are presented below:
1st Stage (4 resistors in parallel)
[tex]R_{eq,1} = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2} }+\frac{1}{R_{3}}+\frac{1}{R_{4}}}[/tex]
If [tex]R_{1} = R_{2} = R_{3} = R_{4} = 20\,\Omega[/tex], then:
[tex]R_{eq,1} = \frac{1}{\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega} }[/tex]
[tex]R_{eq,1} = \frac{1}{\frac{4}{20\,\Omega} }[/tex]
[tex]R_{eq,1} = 5\,\Omega[/tex]
2nd Stage (1 resistor in series)
[tex]R_{5} = 20\,\Omega[/tex]
Now, the equivalent resistance is:
[tex]R_{eq,2} = R_{eq,1} + R_{5}[/tex]
[tex]R_{eq,2} = 5\,\Omega + 20\,\Omega[/tex]
[tex]R_{eq,2} = 25\,\Omega[/tex]
Describe the relationship between the density of electric field lines and the strength of the electric field?
Answer:
The greater the density of the electric field lines the stronger the electric field and vice versa
Explanation:
Electric field can be defined as the region where an electric force is experienced by a charged body. A charged body experiences a force whenever it is positioned close to another charged body.
An electric field may be described in terms of lines of force which represent the direction of a small positive charge placed at that point assuming that the charge is so small that it does not change appreciably in the presence of another charge. Arrows on the lines of force indicate the direction of the electric field.
The lines of force are indicated in such a way that the strength of the electric field is shown by the number or density of electric field lines crossing a unit area perpendicular to the lines. Hence, the greater the density of the electric field lines the stronger the the electric field and vice versa
Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the earth observer, while ship B moves at 0.60c relative to the same observer. What speed does the captain of ship A report for the speed of ship B
Answer:
0.80 c
Explanation:
The computation of speed is shown below:-
Here, The speed of the captain ship A report for speed of the ship B which is
[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]
where
[tex]S_A[/tex] indicates the speed of the ship A
[tex]S_B[/tex] indicates the speed of the ship B
and
C indicates the velocity of life
now we will Substitute 0.40c for A and 0.60 for B in the equation which is
[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]
after solving the above equation we will get
0.80 c
So, The correct answer is 0.80c
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
Three capacitors with capacitances:
C1=6.00μF,
C2=3.00μF,
C3=5.00μF.
The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge Q2 on the second capacitor is 40.0 μC.
A. What is the charge Q1 on capacitor C1?
B. What is the charge on capacitor C3?
Express your answer in microcoulombs to three significant figures.
C. What is the applied voltage, Vab?
Express your answer in volts to three significant figures.
Answer:
A. 80.0microColoumbs
B.120.0 microcoloumbs
C.37.3v
Explanation:
See attached file
a. The charge Q1 on capacitor C1 should be 80.0 μC.
b. The charge on capacitor C3 should be 120 μC.
c. The applied voltage should be 37.5 v.
Calculation of the charge and voltage:a. We know that
v = q/c
Q1/c1 = Q2/C2
Q1 = (C1/C2) Q2
= (6.00/3.00) * 40
= 80.0
b.
Q3 = Q1 + Q2
= (80 + 40)
= 120
c.
The voltage should be
= 80/ 6+ 120/5
= 37.5 V
hence,
a. The charge Q1 on capacitor C1 should be 80.0 μC.
b. The charge on capacitor C3 should be 120 μC.
c. The applied voltage should be 37.5 v.
Learn more about voltage here: https://brainly.com/question/14466758
what is the preferred method of using percentage data by using a circle divided into sections
Answer:
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole
Explanation:
pie charts are a useful way to organize data in order to see the size of components relative to the whole.
Suppose the same magnitude force is applied at the same point as in the example, and the torque is found to have the same magnitude but in the opposite direction of the torque found there. What are the components of the force?
Answer:
-i - 7j
Explanation:
The computation of components of the force is shown below:-
torque T = r cross F
T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)
Now we will cross multiplying vectorilly
T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0
T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0
T = 28 × (-k) + 5 × (k) = -23k
net torque |T| = 23 N - m
direction >> negative k
Or Simply we can do by the below method
r × f
28 - 5 = 23 K
-i - 7j
How fast must a 2500-kg elephant move to have the same kinetic energy as a 67.0-kg sprinter running at 15.0 m/s
Answer:
2.45 m/s
Explanation:
kinetic energy = 1/2 * m * v^2
then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2
by solving for x, X = 2.45 m/s
A tennis player tosses a tennis ball straight up and then catches it after 1.91 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction ---Select--- (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m
Explanation:
(a) The acceleration is 9.8 m/s² downwards.
(b) At the maximum height, the velocity is 0 m/s.
(c) v = at + v₀
0 m/s = (-9.8 m/s²) (1.91 s) + v₀
v₀ = 18.7 m/s
(d) Δy = vt − ½ at²
Δy = (0 m/s) (1.91 s) − ½ (-9.8 m/s²) (1.91 s)²
Δy = 17.9 m
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct
Complete question:
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?
A. the interior field points in a direction parallel to the exterior field
B. There is no electric field on the interior of the conducting sphere.
C. The interior field points in a direction perpendicular to the exterior field.
D. the interior field points in a direction opposite to the exterior field.
Answer:
B. There is no electric field on the interior of the conducting sphere.
Explanation:
Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).
Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of free charges in the spherical conductor in response to any field until its neutralization.
Option B is correct.
There is no electric field on the interior of the conducting sphere.
11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75 m away from the pivot point on the opposite side. The second child's mass is _____ kg.
Answer:
40 kgExplanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light.
Answer:
The wavelength is [tex]\lambda = 6.28 *10^{-7}=628 nm[/tex]
The frequency is [tex]f = 4.78 Hz[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 0.048 \ mm = 4.8 *0^{-5}\ m[/tex]
The distance from the screen is [tex]D = 6.50 \ m[/tex]
The distance between fringes is [tex]Y = 8.5 \ cm = 0.085 \ m[/tex]
Generally the distance between the fringes for a two slit interference is mathematically represented as
[tex]Y = \frac{\lambda * D}{d}[/tex]
=> [tex]\lambda = \frac{Y * d }{D}[/tex]
substituting values
[tex]\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }[/tex]
[tex]\lambda = 6.27 *10^{-7}=628 nm[/tex]
Generally the frequency of the light is mathematically represented as
[tex]f = \frac{c}{\lambda }[/tex]
where c is the speed of light with values
[tex]c = 3.0 *10^{8} \ m/s[/tex]
substituting values
[tex]f = \frac{3.0*10^8}{6.28 *10^{-7}}[/tex]
[tex]f = 4.78 Hz[/tex]
A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released from the equilibrium position with an upward velocity of 6 meters per second. Find an equation of the motion.
Answer:
x(t) = -3sin2t
Explanation:
Given that
Spring force of, W = 720 N
Extension of the spring, s = 4 m
Attached mass to the spring, m = 45 kg
Velocity of, v = 6 m/s
The proper calculation is attached via the image below.
Final solution is x(t) = -3.sin2t
A trolley going down an inclined plane has an acceleration of 2cm/s^2 What will be its velocity
3s after the start.
Answer:
[tex]V_{f}[/tex] = 6 cm/s
Explanation:
Given:
Acceleration = a = 2 cm/s²
Time = t = 3s
Initial Velocity = [tex]V_{i}[/tex] = 0 cm/s
Required:
Velocity = [tex]V_{f}[/tex] = ?
Formula:
a = [tex]\frac{V_{f}-V_{i}}{t}[/tex]
Solution:
2 = [tex]\frac{V_{f}-0}{3}[/tex]
=> [tex]V_{f}[/tex] = 2*3
=> [tex]V_{f}[/tex] = 6 cm/s
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true?
A) The potential at the center of the sphere is zero.
B) The potential is lowest, but not zero, at the center of the sphere.
C) The potential at the center of the sphere is the same as the potential at the surface.
D) The potential at the center is the same as the potential at infinity.
E) The potential at the surface is higher than the potential at the center.
What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.
Answer:
[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]
Explanation:
Assume the stone consists of basalt, which has a density of 3.0 g/cm³.
[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the
Answer:
for the first interference m = 1 y = 2,839 10-3 m
for the second interference m = 2 y = 5,678 10-3 m
Explanation:
The double slit interference phenomenon, for constructive interference is described by the expression
d sin θ = m λ
where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.
In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles
tan θ = y / L
with the angle it is small,
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
the distance between the central maximum and an interference line is
y = m λ L / d
let's reduce the magnitudes to the SI system
λ = 546 nm = 546 10⁻⁹ m
d = 0.25 mm = 0.25 10⁻³ m
let's substitute the values
y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³
y = m 2,839 10⁻³
the explicit value for a line depends on the value of the integer m, for example
for the first interference m = 1
the distance from the central maximum to the first line is y = 2,839 10-3 m
for the second interference m = 2
the distance from the central maximum to the second line is y = 5,678 10-3 m
If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical?
Complete Question:
A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.
(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its
speed is 3.7 m/s, what angle does the cord make with the vertical?
(Check attached image for the diagram.)
Answer:
(a) The ball’s speed, v = 2.06 m/s
(b) The angle the cord makes with the vertical is 50.40⁰
Explanation:
If the ball is revolved in a horizontal plane, it will form a circular trajectory,
the radius of the circle, R = Lsinθ
where;
L is length of the string
The force acting on the ball is given as;
F = mgtanθ
This above is also equal to centripetal force;
[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]
(b) when the speed is 3.7 m/s
[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]
[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]
Therefore, the angle the cord makes with the vertical is 50.40⁰
Newton’s first law says that if motion changes, then a force is exerted. Describe a collision in terms of the forces exerted on both objects.
Answer:
In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
Explanation:
In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.
Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
If, the limits of the visible spectrum are approximately 3000 A.U. and 5000 A.U. respectively. Determine the angular breadth of the first order visible spectrum produced by a plane diffraction grating having 12000 lines per inch when light is incident normally on the grating.
Answer:
θ₁ = 0.04º , θ₂ = 0.00118º
Explanation:
The equation that describes the diffraction pattern of a network is
d sin θ = m λ
where the diffraction order is, in this case they indicate that the order
m = 1
θ = sin⁻¹ (λ / d)
Trfuvsmod ls inrsd fr ll red s SI units
d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm
the distance between two lines we can look for it with a direct proportions rule
If there are 4724 lines in a centimeter, the distance for two hundred is
d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm
let's calculate the angles
λ = 300 10-9 m
θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)
θ₁ = sin⁻¹ (7.08 10-4)
θ₁ = 0.04º
λ = 5000
θ₂ = sin-1 (500 10-9 / 4,2337 10-4)
θ₂ = 0.00118º
A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, the mass is released from rest at x = 10.0 cm. ( That is, the spring is stretched by 10.0 cm.) (a) Determine the frequency of the oscillations. (b) Determine the maximum speed of the mass. Where dos the maximum speed occur? (c) Determine the maximum acceleration of the mass. Where does the maximum acceleration occur? (d) Determine the total energy of teh oscillating system. (e) Express the displacement as a function of time.
Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by:
[tex]v_{max}=\omega A=2\pi f A[/tex] (2)
A: amplitude of the oscillations = 10.0cm = 0.10m
[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by:
[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]
[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is:
[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]
The total energy is 0.1J
(e) The displacement as a function of time is:
[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?
Answer:
Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.
Explanation:
Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.
A skier goes down a slope and detaches from the ground moving in the horizontal direction with a speed of 25m / s. The slope has an inclination of 35 °
a) At what point does the skier make contact again with the ground?
Answer:
107 m down the incline
Explanation:
Given:
v₀ₓ = 25 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
-Δy/Δx = tan 35°
Find: d
First, find Δy and Δx in terms of t.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (0 m/s) t + ½ (-10 m/s²) t²
Δy = -5t²
Δx = v₀ₓ t + ½ aₓ t²
Δx = (25 m/s) t + ½ (0 m/s²) t²
Δx = 25t
Substitute:
-(-5t²) / (25t) = tan 35°
t/5 = tan 35°
t = 5 tan 35°
t ≈ 3.50 s
Now find Δy and Δx.
Δy ≈ -61.3 m
Δx ≈ 87.5 m
Therefore, the distance down the incline is:
d = √(x² + y²)
d ≈ 107 m
A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.70 s. Find the force constant of the spring.
Answer:
109.32 N/m
Explanation:
Given that
Mass of the hung object, m = 8 kg
Period of oscillation of object, T = 1.7 s
Force constant, k = ?
Recall that the period of oscillation of a Simple Harmonic Motion is given as
T = 2π √(m/k), where
T = period of oscillation
m = mass of object and
k = force constant if the spring
Since we are looking for the force constant, if we make "k" the subject of the formula, we have
k = 4π²m / T², now we go ahead to substitute our given values from the question
k = (4 * π² * 8) / 1.7²
k = 315.91 / 2.89
k = 109.32 N/m
Therefore, the force constant of the spring is 109.32 N/m
Point X is midway between the charges. In what section of the line will there be a point where the resultant electric field is zero?
Answer:
I believe the answer is in fact section (VW) on the line where the electric field result will be zero.
Explanation:
The direction of the electric field due to a positive charge is away from it and the direction of the electric field due to a negative one is towards it.
There are two nearby point sources, each emitting a light wave of frequency f. When the frequency f is increased, how will the distance between troughs of constructive interference change?
Answer:
An increase in frequency will cause an increase in the number of lines per centimeter and a smaller distance between each consecutive line. That is the distance between each trough
Explanation:
This is because higher frequency light source should produce an interference pattern with more lines per centimeter in the pattern and a smaller spacing between lines.
can I get help please?
Flower bed is filled with five types of flowers. Which placement of the flowers represents the highest entropy?
Answer:
B
Explanation:
ANSEWER :B IN ROWS ONLY