[tex]\star\:\:{\sf{\underline{\red{Diagram\:}}}} \\ [/tex]
[tex]\setlength{\unitlength}{1cm} \begin{picture}(1,0) \thicklines\qbezier(0,0)(0,0)(2,2)\qbezier(0,0)(0,0)(4,0)\qbezier(4,0)(2,2)(2,2)\put( - 0.29, 0.45){\line(2,1){1.6}}\put( - 0.29, 0.45){\vector(2, 1){1}}\put(1.2, 1.26){\line(1, 0){1.5}}\put( 1.2, 1.26){\vector(1, 0){0.9}}\put(2.7, 1.26){\line(1, 0){2.5}}\put(2.75,1.25 ){\line(6, -1){2.5}} \put(2.75,1.25 ){\line(5,-2){2.5}} \put(2.75,1.25 ){\line(4,-1){2.5}}\put(2.75,1.25 ){\line(3,-2){2.5}}\put(2.75,1.25 ){\line(5, - 3){2.5}}\put(2.75,1.25 ){\line(5,-4){2.5}}\put(5.5, -1.4 ){\line(0,1){3}}\put(7.2, -1.4 ){\line(0,1){3}} \put(5.7, 1.2 ){ $\bf Red$}\put(5.7, 0.8 ){ $\bf Orange$}\put(5.7, 0.4 ){ $\bf Yellow$}\put(5.7, 0){ $\bf Green$}\put(5.7, - 0.4 ){ $\bf Blue$}\put(5.7, - 0.8){ $\bf
Indigo$}\put(5.7, - 1.2 ){ $\bf Violet $}\end{picture}[/tex]
[tex]\\[/tex]
[tex]\star\:\:{\sf{\underline{\orange{What \:is\: the\: dispersion\: of \:light\:?\:}}}} \\ [/tex]
Passing of white light through a glass prism which splits into spectrum of colours.Those colours are to be said as VIBGYOR (violet, indigo, blue, green, yellow, orange, red).This process is called as dispersion of light[tex]\\[/tex]
[tex]\star\:\:{\sf{\underline{\blue{Examples \ of \ dispersion \ of \ light\:}}}} \\ [/tex]
Dewdrops in morning (mostly in winter season)Rainbow FormationCD (compact disc)DiamondOil fuel on road[tex]\\[/tex]
[tex]\star\:\:{\sf{\underline{\purple{Additional\: Information\:}}}} \\ [/tex]
"Dispersion of light" phenomenon is given by Issac Newton.Issac Newton conducted an experiment where he took a glass prism and passed white light rays through it.Violet colour has the highest dispersion of light.Red is the least deviated colour of light.
The decibel scale intensity for busy traffic is 80 dB. Two people having a loud conversation have a decibel intensity of 70 dB. What is the approximate combined sound intensity?
Given :
The decibel scale intensity for busy traffic is 80 dB.
Two people having a loud conversation have a decibel intensity of 70 dB.
To Find :
The approximate combined sound intensity.
Solution :
We know, intensity in decibel can be converted to W/m² by :
[tex]\beta(dB) = 10\ log_{10}( \dfrac{I}{I_o})[/tex]
Putting intensity in decibel scale, we get :
[tex]I(80\ dB ) =10^{-4}\ W/m^2\\\\I(70\ dB ) = 10^{-5}\ W/m^2[/tex]
Let, combine intensity is I .
I = I(80 dB) + I(70 dB)
[tex]I = 10^{-4} + 10^{-5} \ W/m^2\\\\I = 1.1 \times 10^{-4} \ W/m^2[/tex]
Therefore, the combined sound intensity is [tex]1.1 \times 10^{-4} \ W/m^2[/tex] .
why do feet smell and noses run?
Answer:
Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.
Explanation:
Hope this helps
You take a trip that covers 240 kilometers and takes 4 hours. Your average speed is
Answer:
Your average speed is 60 kilometers per hour
Explanation:
240 divided by 4 is 60
240/4=60
The image below shows four boxes that each contain a different sample of gas. The atoms of each gas are represented by dots, 1 2 3 4 Which box contains the gas with the greatest density?
A. 1
B. 2
C. 3
D. 4
Lab - Wave Properties in a Spring
11-05
The wave characteristics you will observe in this lab are common to all waves (water, light, sound,
etc.). Use your prior knowledge and the book to fill in the following blanks, then go in the hall and
perform the lab.
A wave is a disturbance that moves through (propagates) through empty space or through a
_____________. There are two types of waves. A _____________________ wave requires
matter to travel. List some examples of this type:
A _____________________ wave does not require a medium. Examples include:
In order to start and transmit a mechanical wave, a source of _____________ and an
_______________ medium are required. A single disturbance is referred to as a
_______________, and a series of disturbances is a wave __________.
The questions in bold are those you should observe directly. Others will be answered using the book.
A. TYPES OF MECHANICAL WAVES: In the hall, stretch the slinky on the floor until it is
stretched (but still loose). Practice sending single pulses down the slinky by popping your wrist
from the center to the side and back to the center. Then send a continuous wave train along as
your partner holds the other end still. A piece of ribbon should be tied to one coil. Watch the
motion of this ribbon (representing a particle) as the wave travels through the spring.
In this type of wave, the particles move (perpendicular, parallel)
to the direction the wave travels. This type of wave is called a __________________ wave.
Its pulses are called ________________ and ________________.
Now send a pulse by quickly pushing the spring forward and pulling
it back, as shown. This type of wave is called _______________. Watch the motion of the ribbon.
In this type, the particles move _____________ to the direction the wave travels. Its pulses
are called _____________ and _____________. Label each.
Note that all waves transfer _____________ without transferring _______________. In
mechanical waves, particles of the medium vibrate back and forth in simple harmonic motion while
the disturbance (or _____________) moves from one place to another.
B. WAVE SPEED
Send a large pulse, followed by a small one. Does one pulse catch up to the other? ______
(Hint: The person who sends these waves should watch how the waves look when they return. Make
sure that both pulses are large enough initially to make it back to the sender!) The size of the
pulse is called the __________________ of the wave. Did the size affect the speed? ______
Generate a single transverse pulse in the slinky, keeping the stretch constant. Using a stopwatch,
time the journey of the pulse from one end to the other and back again. Take the average of
several trials. _________
Without changing your positions on the floor (therefore keeping the _____________ the pulse
travels the same), pull the slinky tighter using only about 3/4 of the coils. This makes a completely
different medium through which the pulse will travel. Time the journey as before. ___________
Does the kind of medium affect the speed of the pulse? ___________
Lab – Wave Properties in a Spring ____________________
PHYSICSFundamentals
© 2004, GPB
11-06
C. WAVELENGTH AND FREQUENCY
Shake the slinky back and forth steadily to send a
transverse wave train while your partner holds the other end still. On the diagram, label wavelength
(- Greek letter lambda). The frequency of the wave depends on how fast you shake the slinky.
Shake it regularly but slowly, then regularly but rapidly.
Higher frequency waves are generated by shaking the spring (slowly, rapidly). High frequency
waves have (short, long) wavelengths, and low frequency waves have __________.
The speed of a wave in any medium is equal to the _______________ of the wave X
________________. This wave equation ___________________ shows that f and are
______________ proportional. Write the units for each of the variables in this equation.
The exercise involves filling in the gaps with the possible wave
properties that can be obtained from a spring.
How is the Wave Properties in a Spring Lab exercise correctly completed?The correctly completed exercise is presented as follows;
A wave is a disturbance that moves through a medium. There are two
types of waves. A mechanical wave requires matter to travel. List some
examples of this type: sound wave, water wave, spring waves.
A electromagnetic wave does not require a medium. Examples include: Light waves
In order to start and transmit a mechanical wave, a source of
disturbance and a physical medium are required. A single disturbance is
referred to as a pulse, and a series of disturbance is a wave train.
This type of wave is called transverse wave. Its pulses are called crest
and troughs.
Now send a pulse by quickly pushing the spring forward and pulling it
back, as shown. This type of wave is called longitudinal wave. Watch the
motion of the ribbon. In this type, the particles move parallel to the
direction the wave travels. Its pulses are called compression and
rarefactions. Note that all waves transfer energy without transferring
matter. In mechanical waves, particle of the medium vibrate back and
forth in simple harmonic motion while the disturbance (or energy)
moves from one place to another.
B. Wave speed
Does the pulse catch up to the other? yes. The size of the pulse is called
the amplitude of the wave.
Did the size of the pulse affect the speed? No.
The average time wave it takes the wave to travel
Without changing your positions therefore keeping the distance the
pulse travels the same), pull the slinky tighter using only about 3/4 of
coils. This makes a completely different medium through which the
pulse will travel. Time the journey as before time record. Does the kind
of medium affect the speed of the pulse? Yes
C. Wavelength and Frequency
High frequency waves have short wavelengths and low frequency waves
have long wavelengths.
The speed of a wave in any medium is equal to the frequency of the wave × the wavelength. This wave equation [tex]\underline{f = \dfrac{v}{\lambda } }[/tex] shows that f and λ are
inversely proportional. The units of the variables are;
Units of the frequency, f is hertz unit HzUnits of the velocity, v, is m/sUnits of the wavelength, λ, is meters (m)Learn more about waves here:
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A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?
Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
Bob Beamon's 1968 Olympic long jump set a world record which remains unbroken to this day. This amazing jump resulted from an initial velocity of 9.5 m/s at an angle of 40 degrees from the horizontal.
1. Calculate the initial horizontal velocity (V_ix) to two significant figures:
2. Calculate the initial vertical velocity (V_iy) to two significant figures:
3. Calculate the time needed to reach the highest point of the jump (t_1/2) to two significant figures:
4. Calculate the total time (t_TOT) needed to complete the jump to two significant figures:
5. Calculate the maximum height (h) reached during the jump to two significant figures:
6. Calculate the range (total horizontal distance) of his jump to two significant figures:
Please answer today! Thanks!
Answer:
1.) 7.3 m/s 2.) 6.1 m/s
Explanation:
To calculate the initial horizontal velocity with just degrees and velocity alone is pretty simple. The formula is Velocity*cos(degrees)
eg 9.5*cos(40)
2. To calculate the initial vertical velocity with just degrees and velocity alone is pretty simple. The formula Velocity*sin(degrees)
eg 9.5*sin(40)
which factor does not affect the strength of an electromagnet
Answer:
the placement of the ammeter in the circuit
Explanation:
An aluminum baking sheet with a mass of 225 g absorbs 2.4 x 104 J from an oven. If its temperature was initially 25 C, what will its new temperature be?
Answer:
The value is [tex]T_2 =416.9 \ K[/tex]
Explanation:
From the question we are told that
The mass of the aluminum baking sheet is [tex]m = 225 \ g = 0.225 \ kg[/tex]
The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]
The initial temperature is [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]
Generally the heat absorbed is mathematically represented as
[tex]Q = m * c_a * [T_2 - T_1][/tex]
Here [tex]c_a[/tex] is the specific heat capacity of aluminum with value [tex]c_a = 897 \ J / kg \cdot K[/tex]
So
[tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]
=> [tex]T_2 - 298 = 118.915[/tex]
=> [tex]T_2 =416.9 \ K[/tex]
A motorcycle skids for a distance of 2.0 m with the icy road pushing on its tires with force of 120 N as its
brakes are applied
What is the change in kinetic energy for the motorcycle?
Round the answer to two significant digits.
Answer:
-240
Explanation:
A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.
What is kinetic energy?The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.
Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.
When the body slows down from its present level to a condition of rest, the same quantity of energy is used.
Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.
As per the given information in the question,
Distance, d = 2.0 m
Friction, f = 120 N
The angle between displacement and friction force, θ = 180°
Now, the change in kinetic energy for the motorcycle = Work done by the friction.
K.E = f × d(cos θ)
= 120 (2.0 m)(cos 180°)
Δ K.E = -240 J
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A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 Hz. What driving frequency will set up a standing wave with five equal segments?
a) 360 Hz.b) 240 Hz.c) 600 Hz.d) 120 Hz.
Answer:
C) 600 Hz
Explanation:
The fundamental frequency can be related to the driving frequency by the expression below;
f(n) = n * f(1)
Where f(1)= fundamental frequency
f(n) = driving frequency
There are four equal segments in the standing wave , then our n= 4 and our f(n)=4, then we can get the fundamental frequency here
f(4) = 4× f(1)
480 = 4× f(1)
f(1) = 480/4
f(1)=120Hz
Hence, fundamental frequency is 120Hz
To calculate the driving frequency that will set up a standing wave with five equal segments?
n=5
f(n) = n× 120Hz
f(5) = 5×120Hz
= 600Hz.
Hence, the driving frequency that will set up a standing wave with five equal segments is 600Hz
An observer sitting on shore sees a canoe traveling 5.0 m/s east, and a sailboat traveling 15.0 m/s west. What is the velocity of the sailboat as observed on the canoe (relative to the canoe)?
Answer:
Vs/c = 20 [m/s]
Explanation:
This is a problem of relative velocities, as velocity is a vector we can use addition or subtraction of vectors for the solution.
We are asked for the velocity of the sailboat with respect to an observer located in the canoe.
[tex]V_{s/c}=V_{sailboat}-V_{canoe}\\V_{s/c}=15-(-5)\\V_{s/c}=20[m/s][/tex]
Why is the speed of the canoe negative?, it is negative because the canoe moves in the opposite direction to the sailboat.
A particular engine has a power output of 2 kW and an efficiency of 27%. If the engine expels 9085 J of thermal energy in each cycle, find the heat absorbed in each cycle. Answer in units of J.
Answer:
12445 J
Explanation:
Given that
Power output, P = 5 kW
efficiency of the engine, e = 27% = 0.27
Thermal energy expelled, Q(c) = 9085 J
Heat absorbed, Q(h) = ?
Using the formula
e = W/Q(h)
e = [Q(h) - Q(c)] / Q(h)
e = 1 - Q(c)/Q(h)
Now, substituting the values into the formula, we have
0.27 = 1 - 9085/Q(h)
9085/Q(h) = 1 - 0.27
9085/Q(h) = 0.73
Q(h) = 9085 / 0.73
Q(h) = 12445 J
Thus, the heat absorbed is 12445 J
A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?
a. 2.8
b. 1.7.
c. 1.3.
d. 4.0.
e. 1.0.
Answer:ratio of the radii of their orbits = 1.3 --- C
Explanation:
1- eV = to the kinetic energy of the electrons
and kinetic energy is given as
K.E= 1/2mv2
v = √(2E/m)----- equation 1
The force on the particles relating to the magnetic and circular motion ( centripetal force is given as
F = magnetic force = centripetal force
F= qvB = mv2/r
qvB = mv2/r
r = mv/qB ------ equation 2
We know from equation 1 that v = √(2E/m)
Therefore,
r = √(2mE)/qB------ equation 3
We can now say that the ratio of the two radii of their orbits can be calculated as
r1/r2 =(√(2mE1)/qB) /(√(2mE2)/qB
Where E1 = 500-eV and E2 = 300-eV (1- eV = to the kinetic energy of the electrons)
r1/r2 = (√(2m x500)/qB) /(√(2mx 300)/qB
Cancelling out common variables, we are left with
r1/r2 =[tex]\sqrt{500/300}[/tex]
r1/r2= 1.29 ≈ 1.3
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]
[tex]\\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]
[tex]\\[/tex]
☯ For left penetration (s₂)
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]
[tex]\\[/tex]
[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]
What happens to the molecules of water when it moves from a liquid to a gas?
A. Water molecules condense and move slower.
B. Water molecules spread out and move slower.
C. Water molecules spread out and move faster.
D. Water molecules condense and move faster.
its A or D but im not sure which one ik it moves fast
A pmdc has a stall torque of 10 and maximum mechanical power of 200. What is the maximum angular velocity?
Answer:
The maximum angular velocity is 20 rad/s
Explanation:
Given;
torque, τ = 10 N
maximum mechanical power, P = 200 J/s
The output power of the pmdc is given as;
P = τω
where;
P is the maximum mechanical power
ω is the maximum angular velocity
ω = P / τ
ω = (200) / (10)
ω = 20 rad/s
Therefore, the maximum angular velocity is 20 rad/s
A rock is at the top of a 20 meter tall hill. The rock has a mass of 10 kg. How much potential energy does it have?
Answer:
1960 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
PE = 10 × 9.8 × 20
We have the final answer as
1960 JHope this helps you
A speeding race car primarily contains potential energy.
:True
False
Calculate the mechanical advantage of a ramp if the box you are trying to move has a mass of 10 kilograms, the
board is 15 feet long and the height of the ramp is 5 feet .
3
300
150
45
Answer:
[tex]MA = 3[/tex]
Explanation:
Given
[tex]Box = 10kg[/tex]
[tex]Ramp\ Height = 5ft[/tex]
[tex]Ramp\ Length = 15ft[/tex]
Required
Determine the mechanical advantage
This is calculated as follows:
[tex]MA = \frac{Ramp\ Length}{Ramp\ Height}[/tex]
[tex]MA = \frac{15ft}{5ft}[/tex]
[tex]MA = 3[/tex]
Hence, the mechanical advantage is 3
How much of the day do we spend sleeping?
Answer:
roughly 8-10 hours for the average human being
Explanation:
A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?
Answer:
Angular speed = 27.78 rad/s (Approx)
Explanation:
Given:
Diameter = 21.6 cm
Speed = 3 m/s
Find:
Angular speed
Computation:
Radius = 21.6 / 2 = 10.8 cm = 0.108 m
Angular speed = v / r
Angular speed = 3 / 0.108
Angular speed = 27.78 rad/s (Approx)
A stone dropped from a bridge strikes the water 5.6 seconds later. What is the final velocity in meters/s?
A) 179.78 meters/s
B) 5.71 meters/s
C) 1.75 meters/s
D) 54.88 meters/s
Answer: 54.88 meters/s
Explanation:
The final velocity will be calculated by using the formula:
v = u + at
where,
v = final velocity
u = initial velocity = 0
a = 9.8
t = 5.6
Therefore, we slot the value back into the formula. This will be:
v = u + at
v = 0 + (9.8 × 5.6)
v = 0 + 54.88
v = 54.88 meters per second
Therefore, the final velocity is 54.88m/s
What differentiates galaxy groups from clusters?
A.
Clusters are bigger than groups.
B.
Clusters are more massive than groups.
C.
Clusters contain a hot intracluster medium, whereas groups do not.
D.
Clusters are collections of galaxy groups, whereas groups are collections of galaxies.
E.
Clusters don't gravitationally bind galaxies together, while groups bind galaxies gravitationally.
Answer:
A
Explanation:
Galaxy clusters are basically very large (>50 galaxies) groups
Answer:
The correct answer would be:
C.
Clusters contain a hot intracluster medium, whereas groups do not.
#PLATOFAM
Have a nice day!
A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the same momentum as the bus?
Answer:
bus momentum
p_bus= m_bus x v_bus
=18,200 x 16.5
basball momentum
pball=mball x vball
=0.142 x v
p_bus = pball
18200 x 16.5 = 0.142 x v
v=(18200 x 16.5)/0.142
v is the answer for baseball
Explanation:
⚠️not my answer tryna be honest here⚠️
The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶ m/s.
What is momentum?Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.
Given that one bus is having a mass of 18200 Kg and 13.2 m/s speed. The momentum is:
p = mv
=18200 kg × 13.5 m/s
= 245700 Kg m/s
To have a momentum of 245700 Kg m/s the base ball with 0. 142 g have to have a velocity = 245700 Kg m/s / 0.142 g
=1.7 × 10 ⁶ m/s
Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶ m/s
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Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal. How long is it in the air?
Given :
Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.
To Find :
How long is it in the air.
Solution :
We know, the formula of time of flight is :
[tex]T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds[/tex]
Therefore, the ball is in air for 3.06 seconds.
During a hockey game, a puck is given an initial speed of 10 m/s. It slides 50 m on the horizontal ice before it stops due to friction. What is the coefficient of kinetic friction between the pick and the ice.A) 0.12B) 0.10C) 0.11D) 0.090
Answer:
The value is [tex]\mu_k = 0.102[/tex]
Explanation:
From the question we are told that
The initial speed of the pluck is [tex]u = 10 \ m/s[/tex]
The distance it slides on the horizontal ice is [tex]s = 50 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is is the final velocity and the value is 0 m/s given that the pluck came to rest, so
[tex]0^2 = 10 ^2 + 2* a * 50[/tex]
=> [tex]a = - 1 \ m/s^2[/tex]
Here the negative sign show that the pluck is decelerating
Generally the force applied on the pluck is equal to the frictional force experienced by the pluck
So
[tex]F = F_f[/tex]
=> [tex]m * a = m* g * \mu_k[/tex]
=> [tex]1 = 9.8 * \mu_k[/tex]
=> [tex]\mu_k = 0.102[/tex]
ANSWER THIS FOR 16 POINTS!!!!!!!!!!
When hitting the ski slopes when does the skier has the most potential energy??
Answer:
As the ski jumper starts moving downhill, some of his potential energy changes into kinetic energy (KE). Kinetic energy moves him down the slope to the ramp. When the ski jumper takes off from the ramp, some of his kinetic energy is changed back into potential energy as he rises in the air.
Explanation: hope this helps
Answer:
at the top of the slope
Explanation:
Part A:
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?
Part B:
A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part C:
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part D:
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current I2?
Part E:
The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?
Part F:
A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1 in the primary coil?
Part G:
The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency e of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: e=100Pout/Pin.
Answer:
a) 0 V
b) 10 turns
c) 4000 turns
d) 12.5 A
e) 400 W
f) 0.5 A
g) 95.4%
Explanation:
A
0
B
To solve this, we would be using the simple relationship between voltage and number of turns
V1/V2 = N1/N2
500/25 = 200/N2
20 = 200/N2
N2 = 200/20
N2 = 10 turns
C
Here also, we would be using the relationship between current and the number of turns
I1/I2 = N2/N1
500/25 = N2/20
20 = N2/20
N2 = 20 * 20
N2 = 4000 turns
D
Like in the previous question, current and the number of turn relationship is used
N1/N2 = I2/I1
400/80 = I2/2.5
5 = I2/2.5
I2 = 5 * 2.5
I2 = 12.5 A
E
The power remains unchanged at 400 W
F
Power = Voltage * Current
P = VI
I = P/V
I = 60/120
I = 0.5 A
G
95.4%
The transformer is a device used to step up or step down voltage.
Part A;
Given that;
Es/Ep = Ns/Np
Es = voltage in the secondary coil
Ep = voltage in primary coil
Ns = Number of turns in secondary coil
Np = Number of coils in primary coil
Es = Ns/Np × Ep
Es = 200/100 × 1.5 V
Es = 3 V
Part B
Ns = Es/Ep × Np
Ns = 25/500 × 200
Ns = 10 turns
Part C
Ns/Np = Ip/Is
Ns = Ip/Is × Np
Ns = 500/25 × 200
Ns = 4000 turns
Part D
Ns/Np = Ip/Is
NsIs = NpIp
Is = NpIp/Ns
Is = 400 × 2.5/80
Is =12.5 A
Part E
The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.
Part F
Power supplied = 60 watts
Voltage of primary coil = 120 V
Since;
P = IV
I = P/V = 60/120 = 0.5 A
Part G
Since;
E = 100Pout/Pin
Pin = 120 V × 2 A = 240 W
Pout = 19.4 V × 11.8 A = 228.92 W
E = 100(228.92/240)
E = 95.4%
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At an amusemlec 27 collisions and powerent park, a 96.0 kg car moving with a speed v bounces elastically off a 135-kg bumper car at rest. If the final speed of the 135-kg car is 1.03 m/s, what is the initial speed of the 96.0-kg car?
By conservation of momentum :
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\96v + 0 = 96v_1 + 135\times 1.03\\\\96v = 96v_1 +139.05 \ \ \ \ \ \ .....1[/tex]
By conservation of energy :
[tex]\dfrac{m_1u_1^2}{2}+\dfrac{m_1u_2^2}{2} =\dfrac{m_2v_1^2}{2} +\dfrac{m_2v_2^2}{2}\\\\96v^2 + 0 = 96v_1^2 + 135(1.03)^2\\\\96v^2 = 96v_1^2 +143.22 \ \ \ \ \ ....2)[/tex]
Solving equation 1 and 2, we get :
v = 1.23 m/s
Therefore, the initial speed of 96 kg car is 1.23 m/s .