Answer:
3.18 m/s
Explanation:
Given that
Initial speed of the ball, u = 20 m/s
Angle of inclination, θ = 45°
Distance from the ball, h = 50 m
Using equations of projectile to solve this, we have
We start by finding the time of flight, T
T = 2Usinθ/g
T = (2 * 20 * sin45)/9.8
T = (40 * 0.7071) / 9.8
T = 28.284/9.8
T = 2.89 s
Next we find the Range, R
R = u²sin2θ/g
R = (20² * sin 90) / 9.8
R = (400 * 1) / 9.8
R = 400/9.8 = 40.82 m
Distance the gk must cover
40.82 - 50 m
-9.18 m or 9.18 m in the opposite direction.
Speed of the GK = d/t
9.18 / 2.89 = 3.18 m/s
a car brakes and stops at 10 [m]. While stopping, the friction force of the wheels on the pavement is 400 [N]. Calculate the work done.
Explanation:
Work = force × distance
W = (400 N) (10 m)
W = 4000 J
How would the interference pattern produced by a diffraction grating change if the laser light changed from red to blue?
Answer
fringe separation l distance between maxima decreases
Explanation:
Because the wavelength of blue light is smaller than that if red light
A light bulb is completely immersed in water. Light travels out in all directions from the bulb, but only some light escapes the water surface. What happens to the fraction (f) of light that escapes the water's surface as the bulb is moved deeper into the water?
Answer:
The fraction of light that escapes the water surface as the water moves deeper into the water will decrease.
Explanation:
The speed of light in water is small compared to the speed of light in air, and a larger part of the light energy is absorbed in water than in air. When the bulb is immersed in water, some of the light energy is absorbed by the mass of water. When the light bulb is further moved deeper into the water, the fraction of light that escapes decreases, because more mass of water is made available to absorb more of the light energy from the bulb.
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary
Answer:
1.32 is the turns ratio
Explanation:
Note that The transformer steps up the voltage from 12000 V to 390000V
12000 V is the primary and in the secondary it is 390000 V in old transformer
If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils
the formula is
n₂ / n₁ = voltage in secondry / voltage in primary
n₂ / n₁ = 390000 / 12000
ratio of turns in old transformer is 32.5
ratio of turns in new transformer
n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )
= 42.9
T he ratio of turns in the new secondary compared with the old secondary
n₃ / n₂ = 42.9 / 32.5
= 1.32
It takes 144 J of work to move 1.9 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates
Answer:
151.58 V
Explanation:
From the question,
The work done in a circuit in moving a charge is given as,
W = 1/2QV..................... Equation 1
Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.
make V the subject of the equation
V = 2W/Q.................. Equation 2
Given: W = 144 J. Q = 1.9 C
Substitute into equation 2
V = 2(144)/1.9
V = 151.58 V
Need help understanding this. If anyone help, that would be greatly appreciated!
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
a) acceleration:
ā=v^2/r
ā=(15m/s)^2/27m
ā=225/27 m/s^2
ā=8.333 m/s^2
force:
F=mā. where the is equal to v^2/r
F=1000kg*8.3 m/s^2
F=8333.3 N
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the electric field between the two plates?
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
A 5 kg block is sliding on a horizontal surface while being pulled by a child using a rope attached to the center of the block. The rope exerts a constant force of 28.2 N at an angle of \theta=θ = 30 degrees above the horizontal on the block. Friction exists between the block and supporting surface (with \mu_s=\:μ s = 0.25 and \mu_k=\:μ k = 0.12 ). What is the horizontal acceleration of the block?
Answer:
The horizontal acceleration of the block is 4.05 m/s².
Explanation:
The horizontal acceleration can be found as follows:
[tex] F = m \cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]
[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]
Where:
a: is the acceleration
F: is the force exerted by the rope = 28.2 N
θ: is the angle = 30°
[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12
m: is the mass = 5 kg
g: is the gravity = 9.81 m/s²
[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]
Therefore, the horizontal acceleration of the block is 4.05 m/s².
I hope it helps you!
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. The number of fringes per unit length on the screen can be doubled:______.
a. if the distance between the slits is doubled.
b. if the wavelength is changed to λ = λ/2.
c. if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ.
d. if any of the above occurs.
e. only if the width of the slits is changed to w = w/2
Answer:
d. if any of the above occurs
Explanation:
That is The number of fringes per unit length on the screen can be doubled if
if the distance between the slits is doubled.
if the wavelength is changed to λ = λ/2.
And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
Exercise 1 - Questions 1. Hold the grating several inches from your face, at an angle. Look at the grating that you will be using. Record what details you see at the grating surface. 0 Words 2. Hold the diffraction grating up to your eye and look through it. Record what you see. Be specific. 0 Words 3. Before mounting the diffraction grating, look through the opening that you made for your grating. Record what you see across the back of your spectroscope.
Answer:
1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
2)If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum
3) must be able to see the well-collimated light emission source
Explanation:
1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.
These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.
The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.
2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.
If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.
3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.
The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.
2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the
upthrust of the liquid on the cube.
Answer:
vpg = 0.064 N
Explanation:
Upthrust = Volume of fluid displaced
upthrust liquid on the cube g=10ms−2
vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N
vpg = 0.064 N
hope it helps.
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary coil, and an input voltage of 120 V. Randomized Variables Δ 33%
Part (a) What is the voltage output Vs, in volts, of the transformer used for to charge the batteries? Grade Summar Deductions Potential sin tan) ( Submissions Attempts remain coso cotan) asin) acos() atan acotan)sinh( cosh)tanhcotanh0 % per attempt detailed view END Degrees Radians DEL CLEAR Submit Hint I give up! Hints:% deduction per hint. Hints remaining:I Feedback: 1% deduction per feedback. - 쇼 33%
Part (b) what input current ,. İn milliamps, is required to produce a 3.2 A output current? 33%
Part (c) What is the power input, in watts?
Answer:
a) 0.72 V
b) 19.2 mA
c) 2.304 Watts
Explanation:
A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.
number of primary turns = [tex]N_{p}[/tex] = 500 turns
input voltage = [tex]V_{p}[/tex] = 120 V
number of secondary turns = [tex]N_{s}[/tex] = 3 turns
output voltage = [tex]V_{s}[/tex] = ?
using the equation for a transformer
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
substituting values, we have
[tex]\frac{V_{s} }{120 } = \frac{3 }{500} }[/tex]
[tex]500V_{p} = 120*3\\500V_{p} = 360[/tex]
[tex]V_{p}[/tex] = 360/500 = 0.72 V
b) by law of energy conservation,
[tex]I_{P}V_{p} = I_{s}V_{s}[/tex]
where
[tex]I_{p}[/tex] = input current = ?
[tex]I_{s}[/tex] = output voltage = 3.2 A
[tex]V_{s}[/tex] = output voltage = 0.72 V
[tex]V_{p}[/tex] = input voltage = 120 V
substituting values, we have
120[tex]I_{p}[/tex] = 3.2 x 0.72
120[tex]I_{p}[/tex] = 2.304
[tex]I_{p}[/tex] = 2.304/120 = 0.0192 A
= 19.2 mA
c) power input = [tex]I_{p} V_{p}[/tex]
==> 0.0192 x 120 = 2.304 Watts
A dust particle on a phonograph record rotates at a speed of 45 revolutions per minute if the particle is 10 cm from the axis of rotation. Find. 1) its linear speed and linear acceleration.
Explanation:
ω = 45 rev/min × (2π rad/rev) × (1 min / 60 s) = 4.71 rad/s
r = 10 cm = 0.10 m
1) The linear speed is:
v = ωr
v = (4.71 rad/s) (0.10 m)
v = 0.471 m/s
2) The linear acceleration in the tangential direction is 0.
The linear acceleration in the radial direction is:
a = v² / r
a = (0.471 m/s)² / (0.10 m)
a = 2.22 m/s²
Ohm’s Law
pls answer this photos
Answer:
Trial 1: 2 Volts, 0 %
Trial 2: 2.8 Volts, 0%
Trial 3: 4 Volts, 0 %
Explanation:
Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:
V = IR
TRIAL 1:
V = IR
V = (0.1 A)(20 Ω)
V = 2 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2 - 2)/2| x 100%
% Difference = 0 %
TRIAL 2:
V = IR
V = (0.14 A)(20 Ω)
V = 2.8 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2.8 - 2.8)/2.8| x 100%
% Difference = 0 %
TRIAL 3:
V = IR
V = (0.2 A)(20 Ω)
V = 4 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(4 - 4)/4| x 100%
% Difference = 0 %
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low-frequency sound is originating from.
(a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 20 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear?
(b) Assume the same person was scuba diving and a low-frequency sound source was to the right of the scuba diver. How long (in ) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s? S
(c) What is significant about the time interval of the two situations?
Answer:
(a) 0.59 ms
(b) 0.15 ms
(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.
Explanation:
(a) distance between ears = 20 cm = 0.2 m
speed of sound generated = 340 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{340}[/tex]
= 5.8824 × [tex]10^{-4}[/tex]
= 0.59 ms
The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.
(b) distance between ears = 20 cm = 0.2 m
speed of sound in water = 1530 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{1530}[/tex]
= 1.4815 × [tex]10^{-4}[/tex]
= 0.15 ms
The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.
(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.
A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds
B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds
C) The significance about the time interval of the two situations is that;
Transmission of sound varies with different mediums.
A) We are given;
Distance between the two ears; d = 20 cm = 0.2 m
Speed of sound; v = 340 m/s
Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;
t = d/v
t = 0.2/340
t = 0.588 × 10⁻³ seconds
B) We are now told that the speed of sound in water is 1530 m/s. Thus;
t = 0.2/1530
t = 0.131 × 10⁻⁵ seconds
C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.
Read more at; https://brainly.com/question/18451537
A brick of mass M has been placed on a rubber cushion of mass m. Together they are sliding to the right at constant velocity on an ice-covered parking lot. (a) Draw a free-body diagram of the brick and identify each force acting on it. (b) Draw a free-body diagram of the cushion and identify each force acting on it. (c) Identify all of the action–reaction pairs of forces in the brick–cushion–planet system.
A) The free-body diagram of the forces acting on the brick is attached.
B) The free-body diagram of the forces acting on the rubber cushion is attached.
C) The action and reaction forces of the entire brick–cushion–planet system has been enumerated below.
A) The brick has a Mass M placed on top of a rubber cushion of mass m.
This means that there will be a normal force acting acting upwards on the brick and also a gravitational force acting downward. These forces are denoted as;
Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Gravitational force acting on brick = Mg
Find attached the free body diagram.
B) The forces acting on the cushion will be;
Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Gravitational force of earth acting on cushion = mg
Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
C) The action pairs of forces are;
i) Force; Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Reaction Force; Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
ii) Action Force; Gravitational force acting on brick = Mg
Reaction; Gravitational force of brick acting on the earth
iii) Action Force; Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Reaction; Force of rubber cushion on parking lot pavement
iv) Action Force; Gravitational force of earth acting on rubber cushion = mg
Reaction Force; Gravitational force of rubber cushion on the earth.
Read more at; https://brainly.com/question/17747931
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit is:
Answer:
Explanation:
The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point
Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima is π radian .
a positively charged ion, due to a cosmic ray, is headed through earth's atmosphere toward the center of Earth. Due to Earth's magnetic field, the ion will be delfected:
Answer:
East direction
Explanation:
Given that
Charge on the particle is positive.
Moving towards the center of earth .
We know that N(north ) pole in magnetic fields work as source of magnetic lines and S(South ) pole works and sink for magnetic lines.
Therefore due to the earth magnetic fields , the positive ions will deflect towards East direction.
Thus the answer will be East direction.
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string
Answer:
The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 3.00 \ m[/tex]
The mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]
The tension on the string is [tex]T = 500 \ N[/tex]
The velocity of the pulse is mathematically represented as
[tex]v = \sqrt{ \frac{T}{\mu } }[/tex]
Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as
[tex]\mu = \frac{m}{l}[/tex]
substituting values
[tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]
[tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]
Thus
[tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]
[tex]v = 547.7 m/s[/tex]
The time taken is evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{3}{547.7}[/tex]
[tex]t = 5.48 *10^{-3} \ s[/tex]
To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Answer:
Vi = 2 m/s
Explanation:
First we find the force applied to the car by wall to stop it. We use Hooke's Law:
F = kx
where,
F = Force = ?
k = spring constant = 4 x 10⁶ N/m
x = compression = 3.18 cm = 0.0318 m
Therefore,
F = (4 x 10⁶ N/m)(0.0318 m)
F = 127200 N
but, from Newton's Second Law:
F = ma
a = F/m
where,
m = mass of car = 2010 kg
a = deceleration = ?
Therefore,
a = 127200 N/2010 kg
a = 63.28 m/s²
a = - 63.28 m/s²
negative sign due to deceleration.
Now, we use 3rd equation of motion:
2as = Vf² - Vi²
where,
s = distance traveled = 3.18 cm = 0.0318 m
Vf = Final Speed = 0 m/s
Vi = Initial Speed = ?
Therefore,
2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²
Vi = √4.02 m²/s²
Vi = 2 m/s
A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
The mutual inductance is [tex]M = 0.000406 \ H[/tex]
Explanation:
From the question we are told that
The number of turns per unit length is [tex]N = 1800[/tex]
The radius is [tex]r = 0.0165 \ m[/tex]
The number of turns of the solenoid is [tex]N_s = 210 \ turns[/tex]
Generally the mutual inductance of the system is mathematically represented as
[tex]M = \mu_o * N * N_s * A[/tex]
Where A is the cross-sectional area of the system which is mathematically represented as
[tex]A = \pi * r^2[/tex]
substituting values
[tex]A = 3.142 * (0.0165)^2[/tex]
[tex]A = 0.0008554 \ m^2[/tex]
also [tex]\mu_o[/tex] is the permeability of free space with the value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554[/tex]
[tex]M = 0.000406 \ H[/tex]