After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
Suppose a van de Graaff generator builds a negative static charge, and a grounded conductor is placed near enough to it so that a 7.0 mu C of negative charge arcs to the conductor. Calculate the number of electrons that are transferred.
Answer:
# _electron = 4.375 10¹³ electrons
Explanation:
In this exercise it is indicated that 7.1 μC is transferred, let's use a direct ratio or rule of three. If an electron has a charge of 1.6 10⁻¹⁹ C, how many electrons have a charge of 7.0 10⁻⁶ C
# _electron = 7.0 10⁻⁶ C (1 electron / 1.6 10⁻¹⁹ C)
# _electron = 4.375 10¹³ electrons
If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound
Answer:
[tex]\alpha=-3.01dB[/tex]
Explanation:
From the question we are told that:
Sound level intensity
[tex]\triangle I=40dB-80dB[/tex]
Generally the equation for intensity level is mathematically given by
[tex]\alpha=10log_{10}(I/I_x)dB[/tex]
Where
I= Intensity measured
[tex]I_x=Threshold\ of\ audibility[/tex]
[tex]I_x= 10-12 W / m2[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]
[tex]\alpha=10 log10\frac{40}{80}[/tex]
[tex]\alpha=-3.01dB[/tex]
a nano second is what
Answer:
one thousand-millionth of a second.
A nanosecond is an SI unit of time equal to one billionth of a second, that is, ¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ¹⁄₁₀₀₀ microsecond.
How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1180 kHz
Answer:
254 m
Explanation:
Applying,
v = λf............... Equation 1
Where v = velocity of radio wave, λ = wave length, f = frequency
make λ the subject of the equation
λ = v/f............ Equation 2
From the question,
Given: f = 1180 kHz = 1180000 Hz
Constant: v = 3×10⁸ m/s
Substitite into equation 2
λ = 3×10⁸/1180000
λ = 2.54×10²
λ = 254 m
The velocity-time graph of a body is given. What quantities are represented by (a) slope of the graph and (b) area under the graph?
Answer:
a) acceleration
b) displacement
Explanation:
The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.
a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX
In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.
Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.
Also, acceleration = change in velocity/change in time.
Hence, the slope of a velocity-time graph = acceleration.
b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:
Area of a rectangle = length x breadth
= velocity x time (m/s x s)
Also, displacement = velocity x time (m)
Hence, the area under a velocity-time graph of a body would give the displacement of the body.
When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into place. How much does the nail's temperature increase (in °C) during this process? (The specific heat capacity of the nail is 450 J/kg-°C, and round to 3 significant digits.
Answer:
13.33 K
Explanation:
Given that,
Heat absorbed, Q = 30 J
Mass of nail, m = 5 g = 0.005 kg
The specific heat capacity of the nail is 450 J/kg-°C.
We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:
[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]
So, the increase in temperature is 13.33 K.
•. What is called the error due to the procedure and used apparatuses?
a. Random error
b. Index error
c. Systematic error
d. Parallax error.
Answer:
[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]
when blueshift occurs,the preceived frequency of the wave would be?
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
Why is the melting of ice a physical change?
A. It changes the chemical composition of water.
B. It does not change the chemical composition of water.
C. It creates new chemical bonds.
D. It forms new products.
E. It is an irreversible change that forms new products.
It does not change the chemical composition of water.
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies
Explanation:
https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 26.5 m/s2 with a beam of length 5.89 m , what rotation frequency is required
Answer:
The angular acceleration is 4.5 rad/s^2.
Explanation:
Acceleration, a = 26.5 m/s2
length, L = 5.89 m
The angular acceleration is
[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.
a. Does the temperature of the water bath go up or down?
b. Does the piston move in or out?
c. Does heat flow into or out of the gaseous mixture?
d. How much heat flows?
Helium gas at 20 °C is confined within a rigid vessel. The gas is then heated until its pressure is doubled. What is the final temperature of the gas?
Answer:
586 K
Explanation:
Let P is the initial pressure.
Initial temperature, T₁ = 20°C = 293 K
Final pressure, P₂ = 2P
We need to find the final temperature of the gas.
The relation between the pressure and the temperature is as follows
[tex]P\propto T\\\\or\\\\\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex]
Put all the values,
[tex]\dfrac{P}{2P}=\dfrac{293}{T_2}\\\\\dfrac{1}{2}=\dfrac{293}{T_2}\\\\T_2=2\times 293\\\\T_2=586\ K[/tex]
So, the final temperature of the gas is 586 K.
A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a density of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?
Answer:
[tex]N=0.37N[/tex]
Explanation:
Mass [tex]m=55g=>0.055kg[/tex]
Soapstone Density [tex]\rho_s=3000kg/m^2[/tex]
Whisky Density [tex]\rho_w=940kg/m^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=U+N[/tex]
Therefore
[tex]N=m*g-(\frac{m}{\rho_s})*\rho_w*g[/tex]
[tex]N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}[/tex]
[tex]N=0.37N[/tex]
Express the unit of force in terms of fundamental unit
Answer:
The fundamental unit of force is kg.m/s²
Explanation:
According to Newton's second law of motion, force is given as the product of mass and acceleration.
Mathematically, force can be expressed as; F = ma
where;
F is the force
M is mass of the object, unit of mass = kg
a is acceleration of the object, unit of acceleration = m/s²
Force = kg x m/s²
Force = kg.m/s² = Newton [N]
Therefore, the fundamental unit of force is kg.m/s²
A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2
and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?
Answer:
dont be lose because the person who lose will win the match
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
Explanation:
Given
Acceleration of the pebble is
At t=0, velocity is
considering horizontal motion
[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]
Velocity acquired during this time
[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]
Consider vertical motion
[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]
Net velocity is
[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]
Angle made is
[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]
When should a line graph be used?
A. When the independent variable is continuous and does not show a relationship to the dependent variable
B. When the independent variable is composed of categories and does not show a relationship
C. When the independent variable is continuous and shows a casual link to the dependent variable
D. When there is no independent variable
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 243 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle ? with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle ?.
(a) R = km
(b) ? = degrees
Answer:
a) [tex]R=126Km[/tex]
b) [tex]\theta=74.6\textdegree[/tex]
Explanation:
From the question we are told that:
1st segment
243km at Angle=30
2nd segment
178km West
Resolving to the X axis
[tex]F_x=243cos30+178[/tex]
[tex]F_x=33.44Km[/tex]
Resolving to the Y axis
[tex]F_y=243sin30+178sin0[/tex]
[tex]R=\sqrt{F_y^2+F_x^2}[/tex]
[tex]F_y=121.5Km[/tex]
Therefore
Generally the equation for Directional Angle is mathematically given by
[tex]\theta=tan^{-1}\frac{F_y}{F_x}[/tex]
[tex]\theta=tan^{-1}\frac{121.5}{33.44}[/tex]
[tex]\theta=74.6\textdegree[/tex]
Generally the equation for Magnitude is mathematically given by
[tex]R=\sqrt{F_y^2+F_x^2}[/tex]
[tex]R=\sqrt{33.44^2+121.5^2}[/tex]
[tex]R=126Km[/tex]
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a tube 4 mm in diameter and 20 cm long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.a. What is the flow speed in the urethra?b. If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this.)
Answer:
Explanation:
Given:
volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]
time taken for the discharge, [tex]t=30~s[/tex]
diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]
length of cylindrical urethra, [tex]l=0.2~m[/tex]
density of urine, [tex]\rho=1000~kg/m^3[/tex]
a)
we have volume flow rate Q:
[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]
where:
[tex]A=[/tex] cross-sectional area of urethra
[tex]v=[/tex] velocity of flow
[tex]A.v=\frac{V}{t}[/tex]
[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]
[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]
[tex]v=1.06~m/s[/tex]
b)
The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:
[tex]P=\rho.g.l[/tex]
[tex]P=1000\times 9.8\times 0.2[/tex]
[tex]P=1960~Pa[/tex]
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s. Calculate
(a) the angular acceleration,
(b) the time required to complete the 60 revolutions,
(c) the time required to reach the 10 rev/s angular speed, and
(d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.
Explanation:
Given:
[tex]\omega_0[/tex] = 10 rev/s = [tex]20\pi\:\text{rad/s}[/tex]
[tex]\omega[/tex] = 15 rev/s = [tex]30\pi\:\text{rad/s}[/tex]
[tex]\theta[/tex] = 60 rev = [tex]120\pi\:\text{rads}[/tex]
a) the angular acceleration [tex]\alpha[/tex] is given by
[tex]\alpha = \dfrac{\omega^2 - \omega_0^2}{2\theta}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(30\pi)^2 - (20\pi)^2}{240\pi} = 6.5\:\text{rad/s}^2[/tex]
b) [tex]t = \dfrac{\omega - \omega_0}{\alpha} = \dfrac{30\pi - 20\pi}{6.5} = 4.8\:\text{s}[/tex]
c) [tex]t = \dfrac{\omega - \omega_0}{\alpha}[/tex]
[tex]=\dfrac{20\pi - 0}{6.5} = 9.7\:\text{s}[/tex]
d)[tex]\theta = \frac{1}{2}\alpha t^2[/tex]
[tex]\:\:\:\:\:\:\:=\frac{1}{2}(6.5\:\text{rad/s}^2)(9.7\:\text{s})^2 = 305.8\:\text{rad}[/tex]
[tex]\:\:\:\:\:\:\:= 48.7\:\text{revs}[/tex]
If a car generates 22 hp when traveling at a steady 100 km/h , what must be the average force exerted on the car due to friction and air resistance
Answer:
The average force exerted on the car is 590.12 N.
Explanation:
Given that,
The power generated, P = 22 hp = 16405.4 W
Speed of the car, v = 100 km/h = 27.8 m/s
We need to find the average force exerted on the car due to friction and air resistance.
We know that,
Power, P = F v
Where
F is force exerted on the car
[tex]F=\dfrac{P}{v}\\\\F=\dfrac{16405.4}{27.8}\\\\F=590.12\ N[/tex]
So, the average force exerted on the car is 590.12 N.
Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Compare the time it takes each of them to reach the lake below.
a. Alice reaches the surface of the lake first
b. Tom reaches the surface of the lake first
c. Alice and Tom will reach the surface of the lake at the same time.
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
An object is moving from north to south what is the direction of the force of friction of the object
Answer:
North
Explanation:
Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.
Answer:
South To North
Explanation:
Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north
Magnets produce _________ in the spaces surrounding them
Answer:
magnetic field
Explanation:
