The total horizontal distance traveled by the tennis ball during the entire time it is in the air is approximately: 23 m
What is horizontal distance?It is the line drawn horizontally from the viewer's eye. If an item is above the horizontal level, the angle of elevation is the angle created between the line of sight and horizontal level.
As we know,
H = [u(y)]²/2 g
or, H = (6.1)²/20
or, H = 1.86 m ≈ 1.9 m
Total max height = 1.9 m + 9.8 = 11 m
H = [u(y)]²/2g
or, 11 = [u(y)]²/20
or, u(y) = 14.832 m/s
u(x) = 7.6 m/sec
Now, R = u²sin 2θ/g
or, R = [2 u(x) u(y)] /g
or, R = [2×(7.6 )×( 14.832 )] / 10
or, R = 22.54 m
or, R = 23 m
Correct option: A
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for universes with no dark energy ( ) composed of pressureless matter , what does this imply for an upper limit on , if is equal to 70 km/s/mpc? curvature is determined by the value of , you mustn't simply assume it is flat, obviously
The value of the Hubble constant, denoted H0, represents the expansion rate of the universe. It is defined as the ratio of the speed at which a galaxy is receding from an observer to its distance from the observer. In a universe with no dark energy and composed of pressureless matter, the expansion rate is determined by the density of matter in the universe.
If the Hubble constant is equal to 70 km/s/Mpc, this implies that the expansion rate of the universe is 70 km/s per megaparsec of distance. It does not directly provide information about the curvature of the universe, as the curvature depends on the amount and distribution of matter and energy in the universe, as well as the expansion rate. To determine the curvature of the universe, you would need to know the density of matter in the universe and the expansion rate. If the density of matter in the universe is high enough, the expansion rate will be slowed down and the universe will be closed and have positive curvature. If the density of matter in the universe is low enough, the expansion rate will be accelerated and the universe will be open and have negative curvature. If the density of matter in the universe is just right, the expansion rate will remain constant and the universe will be flat. Without knowing the density of matter in the universe, it is not possible to determine the curvature of the universe based on the value of the Hubble constant alone.
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How much work would it take to push two protons very slowly from a separation of 2e-10 m. (a typical atomic distance) to 3e-15 m. (a typical nuclear distance)?
If the protons are both released from rest at the closer distance, how fast are they moving when they reach their original separation?
The amount of work done to push the protons is 7.67 X 10⁻¹⁴ J. If the protons are both released from rest at the closer distance, then the speed at which they reach their original separation is 6.77 X 10⁶ m/s.
Initial distance = d = 2 X 10⁻¹⁰ m
Final distance = d' = 3 X 10⁻¹⁵ m
Mass of both protons = 1.6 X 10⁻²⁷ Kg each
Charge of both protons = q = 1.6 X 10⁻¹⁹ C each
The constant for proton energy = 1/(4πε) = 9 X 10⁹ Nm²/C²
The potential change changes when their is a change in the work done, thus, potential energy initially = U =
= U = 1/(4πε) X (q₁q₂ / d)
= U = ( 9 X 10⁹ X 1.6 X 10⁻¹⁹ X 1.6 X 10⁻¹⁹ ) / 2 X 10⁻¹⁰
= U = 11.52 X 10⁻¹⁹ J
Now, finial potential energy = U' =
= U' = 1/(4πε) X (q₁q₂ / d')
= U' = ( 9 X 10⁹ X 1.6 X 10⁻¹⁹ X 1.6 X 10⁻¹⁹ ) / 3 X 10⁻¹⁵
= U' = 7.68 X 10⁻¹⁴ J
Thus, the work done to push the two protons is the different between final potential energy and the initial potential energy = W =
= W = U' - U
= W =11.52 X 10⁻¹⁹ - 7.68 X 10⁻¹⁴
= W = 7.67 X 10⁻¹⁴ J
Now, the speed of protons while coming back to the original positions = V;
Work done is also equal to the sum of kinetic energy's, hence,
= W = KE + KE'
= W = (1/2)mv² + (1/2)mv'²
= W = 2 X (1/2)mV²
= V = √ (W / m)
= V = √ (7.67 X 10⁻¹⁴ / 1.6 X 10⁻²⁷)
= V = 6.77 X 10⁶ m/s
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the total potential energy of a system of multiple springs is defined to be the sum of the potential energies stored in each of the springs. The blacks all have the same mass nctonles the instant shown, each block is ele leoxk on a horizontal, rhecks n dea th spring constant k. At Case A the four block- m, and all of the released from rest a a ro the right of its equilibrium position nt indicated by the dashed line). me that each spring is at its equilibrium Case B a case B, assume length when the block is at its equilibrium tank the cases according to magnitude of the net Ran or on the block at the instant shown, from largest to smallest. (Hint: In case C, how far was the point was displaced a distance A?) Explain Case C Case D b Use your answers above to rank the cases according to the time it takes the block n reaun its equilibriam position
Sort the scenarios based on how long it takes the block n to reach equilibrium T[tex]_{c}[/tex]>T[tex]_{A}[/tex]>T[tex]_{B}[/tex]=T[tex]_{D}[/tex].
F[tex]_{net}[/tex]=kx₁=kx₂
x₁+x₂=A
[tex]\frac{F}{k}[/tex]+[tex]\frac{F}{k}[/tex]=A
A=[tex]\frac{2F}{k}[/tex]
F[tex]_{net}[/tex]=[tex]\frac{kA}{2}[/tex]
point of connecting spring displaced by A/2
Ranking ⇒[(case-B)=(case-D)>(case-A)>(case-C)]
Part-(b)
Now, time period T=2π[tex]\sqrt{\frac{m}{k_{eq} } }[/tex]
case-A [tex]k_{q}[/tex]=k,
case-B [tex]k_{eq}[/tex]=2k
case-C [tex]k_{eq}[/tex]=k/2
case-D [tex]k_{eq}[/tex]=2k
so,T[tex]_{c}[/tex]>T[tex]_{A}[/tex]>T[tex]_{B}[/tex]=T[tex]_{D}[/tex]
A process is in equilibrium when the observable properties, such as color, temperature, pressure, concentration, etc., do not change.
Equilibrium, which literally means "balance," denotes that the reactants and products involved in a chemical process are in balance with one another. The equilibrium condition can also be seen in some physical processes, such as the melting point of ice at 0 degrees Celsius, when both ice and water are present.
The equilibrium is known as a physical equilibrium in the case of physical processes like the melting of solids, the dissolution of salt in water, etc., whereas the equilibrium connected to chemical reactions is known as a chemical equilibrium.
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for lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of italian ham. the slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 n/m . the slices of ham are dropped on the plate all at the same time from a height of 0.250 m . they make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (shm). you may assume that the collision time is extremely small. Express your answer numerically in meters and take free-fall acceleration to be g= 9.80m/s^2
What is the period of oscillation T of the scale?
Express your answer numerically in seconds.
The amplitude of oscillation of the scale is 0.0581 m after the slices of ham land on the plate.
The amplitude of oscillation of a body performing oscillation is the maximum distance traveled from its mean position.
The peaks in the sinusoidal graph are the amplitude of oscillation which describes, how much distance the body oscillates from the mean position on either side. Within any oscillation system, the magnitude of variation in the oscillation variable of the body with each oscillation is termed oscillation amplitude.
The frequency of the pendulum s oscillations does not depend on amplitude or mass. It depends only upon the acceleration due to gravity and the length of the pendulum.
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The spring in the figure (Figure 1)/ (a) is compressed by length delta x . It launches the block across a frictionless surface with speed v0. The two springs in the figure (Figure 2) (b) are identical to the spring of the figure (a). They are compressed by the same length delta x and used to launch the same block. What is the block's speed now?
The speed from two springs is √2 times the speed with one spring.
The rate of change in position of an object in any course. Speed is measured because the ratio of distance to the time wherein the distance became included. Speed is a scalar amount because it has handiest route and no significance.
A quick-moving object has a high speed and covers a notably big distance in a given amount of time, whilst a sluggish-shifting item covers a relatively small amount of distance in the same amount of time.
The primary unit or the S.I. unit of speed is m/s or ms?¹.
Calculation :
The energy stored in the spring is 0.5*k*∆x²; this must equal the kinetic energy of the released block. 0.5*m*v0²:
0.5*m*v0² = 0.5*k*∆x²
v0 = ∆x*√[k/m]
Two springs will store twice the energy, so the stored energy in the two springs is k*∆x²
0.5*k*∆x² = k*∆x²
v0 = ∆x*√[2*k/m]
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Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our Sun, but a much smaller diameter.
If you weigh 715 N on the Earth, what would you weigh on the surface of a neutron star that has the same mass as our Sun and a diameter of 20.0 km?
Express your answer with the appropriate units.
We would weigh 9.68x1013 N on the surface of a neutron star with the same mass as our Sun and a diameter of 20.0 km if we were to weigh 715 N on Earth.
According to the information provided, a neutron star has the same mass as the Sun, making its mass (m2) = 1.99x10^30 kg.
We must first determine our mass using the formula Fg = mg.
Given force (F) = 715N
mass (m) = 715/9.8 = 72.95kg
The distance from a planet's centre to its surface, as determined by the force of gravity equation, is equal to r.
Since the issue gives us the diameter, we can utilise it to find the radius, which is given by r = d/2.
Given distance (d) = 20km
r = 20/2 = 10km = 10000m
Now, using the formula for the force of gravity, we can determine our weight at the star: F = Gm1m2/r^2
G = gravitational constant = 6.67×10-11 N* m^2/kg^2
F = (6.67×10-11)x72.95x 1.99×10^30/10^8 = 9.68x10^13N
Hence the weight on surface of neutron star = 9.68x10^13N
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The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a = nb in the figure (Figure 1)). How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum. Express your answer in terms of n and c.
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The speed at which an Empire ship must travel for its markings to be confused with those of a Federation ship is sqrt(c^2 - 1/n^2) * c, where n is the ratio of the major axis to the minor axis of the ellipse and c is the speed of light in a vacuum.
The appearance of an object changes when it moves at a significant fraction of the speed of light due to the phenomenon of relativistic length contraction. This effect causes the length of the object to appear shorter to an observer, and also causes any features on the surface of the object to be distorted.
In the case of the Empire ship and the Federation ship, the appearance of the symbols on their hulls would be affected by relativistic length contraction. If the Empire ship is moving fast enough, its ellipse-shaped symbol may appear to an observer as a circle, similar to the symbol on the Federation ship.
To determine the speed at which this would occur, we can use the equation for relativistic length contraction: L' = L / sqrt(1 - v^2/c^2), where L' is the length of the object as observed by the observer, L is the actual length of the object, v is the speed of the object, and c is the speed of light in a vacuum.
By setting L' equal to the length of the minor axis of the ellipse (b) and L equal to the length of the major axis (a), and solving for v, we can determine the speed at which the Empire ship must travel for its markings to be confused with those of the Federation ship. This speed is given by the equation v = sqrt(c^2 - b^2/a^2) * c, where n is the ratio of the major axis to the minor axis of the ellipse and c is the speed of light in a vacuum.
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3. A ballon of weight 300 N is moving upward. Volume of
ballon is 30m³ and displacing air of density 1.2kg/m³ 11.A
then the upward force acting in the ballon is?
The upward force acting in the ballon is -352.8N
What is upward force?This refers to a force exerted by a fluid that opposes the weight of a partially or fully immersed object. In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. the upward force is also known as upthrust. It operates under Archimedes' principle.
The formula for calculating upward force:
Fb = -ρ g V
Where,
Fb = buoyant force
ρ = fluid density
g = acceleration due to gravity
V = fluid volume
From the question:
V= 30m³
ρ = 1.2kg/m³
W = 300 N
g = 9.8m/s
To calculate buoyant force, Fb:
Fb = -ρ g V
Fb = - 1.2kg/m³ * 9.8m/s * 30m³
Fb = -352.8N
Note: Negative upward force suggests that the immersed object is denser than the fluid displaced which results in the sinking of the object. Neutral buoyancy takes place when the weight of the immersed object is equivalent to the fluid displaced. The dive taken by the scuba diver is a perfect example of neutral buoyancy.
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a block of mass 3 kg slides down an inclined plane at an angle of p with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). the pulley can be approximated as a disk. the coefficient of kinetic friction on the plane is 0.4. what is the acceleration of the block?
The acceleration of the block is 4.16 m/s^2 when the block slides down on inclined plane at an angle 45 degrees.
Acceleration is a dynamic characteristic of an object. Acceleration (a) can be obtained via inertial force (F) on a mass (m) subjected to acceleration (a) of the moving object:
F = - m a
The inertial force can be measured either through strain (if deformation is min) or through the deformation of elastic element. Equal increments of speed gained in equal time intervals,Distance increases in each time interval.The unit of acceleration in the System International: m/s2.
Acceleration shows how fast an object changes its velocity. If an object is moving with constant velocity, acceleration is equal zero: Positive and Negative Acceleration.Positive acceleration: velocity increases with the time and Negative acceleration: velocity decreases with the time.
Given,
m=3kg
angle = 45 degrees
Uk= 0.2
by the equation ,
mgsinФ-f-T=ma
3* 9.81*sin45-UkN-T=3a----(1)
T=a----(2)
Now,N= mgcos45
= 3*9.81/sqrt2
solve eqn (1) and put value in (2)
a=4.16 m /s^2
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as a jet is taking off, it is accelerating down the runway with acceleration , with . you are sitting inside the jet and holding a simple pendulum consisting of a bob of mass attached to a string of length . the pendulum comes to rest in the equilibrium position where the string makes an angle with respect to the vertical. when the bob is in the equilibrium position, which of the following free-body force diagrams on the bob is correct in a reference frame fixed to the accelerating jet.
A jet plane's acceleration is a critical phase of flight that necessitates intense focus on the part of both pilots.
On the ground, things can get hectic and demand a lot of situational awareness, especially when there is little traffic separation, a runway is wet, visibility is poor, and there are heavy winds. Nevertheless, you start to relax a little and take in the beauty of flying as the engines begin to accelerate and you take off and break through the clouds. We have put up this essay with all you need to know so that you may master the takeoff phase in complete confidence.
You must commit to memory three call-outs for the acceleration and takeoff phases of your flight. You will first check your instruments at 80 knots. When you push the nose of your aircraft upward and rotate in V1, you are too quick to abort the takeoff.
This was merely a succinct justification. In order to help you mentally prepare, we will go over every important stage of a jet plane's acceleration.
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Which of the following describes the independent variable?
It is called the respondent variable.
It is plotted on the vertical axis when graphed.
It refers to a quantity changed by the experimenter.
It is something that remains controlled during trials.
The one that describes the independent variable is it is something that remains controlled during trials. The correct option is D.
What is an independent variable?A variable that is independent is precisely what it sounds like. It is a stand-alone variable that is unaffected by the other variables you are attempting to assess.
On the abscissa, the independent variable is plotted (also called the x-axis or horizontal axis), It is not the respondent variable because it does not respond to changes.
Therefore, the correct option is D, It is something that remains controlled during trials.
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in the circuit at right, the bulbs are identical, and the battery is ideal. rank the brightness of the bulbs in the ircuit at right
Brightest resistor is the one with the most current. The current via bulb C is the highest since it receives the same amount of current as the battery.
What is an example of a resistor?Digital potentiometers and rheostats are variable resistors. Electrical parts called rheostats are constructed from resistive thin wire around for a ceramic core. They were once employed to regulate the quantity of power utilized by ovens, heaters, and motors.
How do resistors function?By decreasing the voltage and obstructing the passage of electricity, the resistance absorbs the electrical energy, which is then released as heat. The heat dissipation in electronic circuits today is typically a few hundredths of a watt.
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Part A Which of the following statements are true? Check all that apply. View Available Hint(s) Mechanical energy is conserved because no dissipative forces perform work on the ball. The forces of gravity and the spring have potential energies associated with them. No conservative forces act in this problem after the ball is released from the spring gun.
The statement that is true is Mechanical energy is conserved because no dissipative forces perform work on the ball. The correct option is A.
What is mechanical energy?Mechanical energy, also known as kinetic energy or potential energy, is the energy that an object possesses when it is in motion or the energy that an object stores due to its location.
Renewable energy is also fueled by mechanical energy. In order to efficiently produce electricity or convert energy, many sources of renewable energy depend on mechanical energy.
Therefore, the correct option is A, Mechanical energy is conserved because no dissipative forces perform work on the ball.
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9. A 0.25 kg steel ball experiences a net force of 1.15 N as it rolls down a ramp. What is the
acceleration of the ball?
a. 0.29 m/s²
b. 0.90 m/s²
c. 1.4 m/s²
d. 4.6 m/s²
Answer:
4.6 option d
Explanation:
from second Newton law of motion
F=ma
to find acceleration
a= F/m
a= 1.15/0.25
a = 4.6
For finding the distances to globular clusters, Harlow Shapley used
A. eclipsing binaries
B. planetary nebulae
C. Type I supernovae
D. RR Lyrae variables
E. Population I classical Cepheidsd
Harlow Shapley used RR Lyrae variables.
RR Lyrae variables are periodic variable stars, normally found in globular clusters. they're used as preferred candles to degree galactic distances, supporting the cosmic distance ladder. This elegance is named after the prototype and brightest example, RR Lyrae.
RR Lyrae is a variable, horizontal branch star with periods starting from a few hours to 2 days, and optical brightnesses that commonly vary from zero. They lie within the instability strip of the Hertzsprung-Russell diagram and suffer instabilities that purpose their length to periodically trade.
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a 2000 hz sound wave passes through a wall with two narrow openings 30 cm apart. if sound travels on average 352 m/s, find the following. (a) what is the angle of the first order maximum?
Angle of the first order maximum is 35.1°.
Given,
frequency f= 2000 Hz
sound travels average 352 m/s
The speed of the sound wave is 352 meters per second, and the Frequency is 2000. The slate is separated by 30 cm. The position of the first order needs to be found. There is a maximum for interference.
a) wavelength = v/f = 352/2000
λ = d sinФ
Ф = si[tex]n^-^1[/tex]( (352/2000)/.3)= 35.4°
The speed of sound can be written as 52 meter per second if the Frequency is 2000 times one over the separation. The first order's maximum is 35.4 degrees.
b) then λ/d
=(352/2000)/.3 = 2.85E-2/d
d=0.0496 m
c) (352/3000)/.3 = (352/f)/1.0E-6
f= 9.0E8 Hz
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The acceleration function (in m/s2) and the initial velocity are given for a particle moving along a line.
a(t) = t + 4, v(0) = 3, 0 ≤ t ≤ 11
The velocity function is v(t)=1/2 t^2+4t + 3
The acceleration function is given by a(t) = t + 4 where v(0) = 3. The acceleration function a(t) is a known function of time. The velocity function is integral of the acceleration function plus a constant of integration. The time derivative of the velocity function is acceleration, hence d/dt v(t)=a(t)
Taking indefinite integral of both sides,
∫d/dt v(t)dt=∫a(t)dt+C1
Where C1 is the constant of integration. Since
∫d/dt v(t)dt=v(t)
Hence,
v(t)=∫a(t)dt + C1
Inserting the acceleration function, a(t) = t + 4
v(t)=∫(t +4)dt + C1
v(t)=1/2 t^2+4t + C1
In order to find C1, let v = 3 when t = 0 as given
3=1/2 (0)^2+4(0) + C1
3=0 + C1
C1=3
Hence,
v(t)=1/2 t^2+4t + 3
Note: The question is incomplete. The complete question probably is: The acceleration function (in m/s-) and the initial velocity are given for a particle moving along a line a(t) = t + 4, v(0) = 3, 0 ≤ t ≤ 11 (a) Find the velocity at time t.
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put the following events in order the describe the effects of light exposure on the retina. start with rhodopsin absorbing light, and finish with ganglion cells firing action potentials.
The order of events that that describes effects of light exposure on the retina are;
Rhodopsin absorbs lightCis-retinal isomerizes to trans-retinalOpsin triggers signaling cascade in outer segment of rod cellsGlutamate release from rod cells decreasesBipolar cells detect changes in glutamate releaseGanglion cells fire action potentialsWhat is the effect of light on the retina?Light acts directly on the retina in mammals to perform two important functions; visual function via rod and cone photoreceptor cells and non-image forming tasks such as circadian rhythm synchronization to a 24 hours solar cycle, pineal melatonin suppression, and pupil light reflexes.
This process takes place first in the rhodopsin and trans-retina isomerizes from cis-retinal. Opsin initiates a signaling cascade in rod cell outer segments where rod cell glutamate release is decreased. Changes in glutamate release are detected by bipolar cells and lastly action potentials are emitted by ganglion cells.
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Listed following is a set of statements describing individual stars or characteristics of stars. Match these to the appropriate object category.a. Red Giant or Supergiant Starts.b. Main-sequence starsc. White dwarfs1. Very cool but very luminous2. Found in the upper right of the H-R diagram3. The majority of stars in our galaxy4. The Sun, for example5. A very hot and very luminous star6. not much larger in radius than Earth7. Very hot but very dim
The five fundamental features of a star are its brightness, color, surface temperature, size, and mass.
Supergiant or Red Giant Stars are located in the upper right corner of the H-R diagram and are extremely cool but bright.
The majority of the stars in our galaxy are main-sequence stars, such as the Sun, which is an extremely hot and brilliant star.
White dwarfs are tremendously hot, but they are also quite dim. Their radii are not much larger than Earth's.
The largest and hottest stars are those in the 'O' class, with temperatures over 30,000 °C. The smallest and coldest stars are those in the 'M' class, with temperatures below 3,000 °C. The stars in the sky are not all the same color when you look attentively at them.
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A rocket fires a projectile at a speed of 0.95c while traveling past the Earth. An earthbound scientist measures the projectile's speed to be 0.88c. What was the rocket's speed as a fraction of c?
The speed of rocket travelling past the earth is 0.34c.
Consider the earth frame of reference as S and rocket's frame S moving along the direction.
Lorentz transformation equations for addition of relativistic velocities is
u' = u - v / 1 - (uv/c²)
Here,
u = observed speed of the projectile on earth frame of reference
u' = speed of the projectile leaving the rocket
c = speed of the light
If we rearrange the equation for v, we get
u'[1- (uv/c²)] = u - v
u' - (uu'v/c²) = u - v
v - (uu'v/c²) = u - u'
v = u-u' / [1-(uu'/c²)]
The rocket speed is given as
v = u-u' / [1-(uu'/c²)]
Substituting the values,
v = 0.90-0.95u' / [1-(0.90*o.95/c²)]
v = - 0.34c
Our assumptions contradicts that the rocket moving along the same direction of the projectile.
Therefore, the speed of rocket travelling past the earth is 0.34c.
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The speed of rocket travelling past the earth is 0.34c.
Consider the earth frame of reference as S and rocket's frame S moving along the direction.
Lorentz transformation equations for addition of relativistic velocities is
u' = u - v / 1 - (uv/c²)
Here,
u = observed speed of the projectile on earth frame of reference
u' = speed of the projectile leaving the rocket
c = speed of the light
If we rearrange the equation for v, we ge
u'[1- (uv/c²)] = u - v
u' - (uu'v/c²) = u - v
v - (uu'v/c²) = u - u'
v = u-u' / [1-(uu'/c²)]
The rocket speed is given as
v = u-u' / [1-(uu'/c²)]
Substituting the values,
v = 0.90-0.95u' / [1-(0.90*o.95/c²)]
v = - 0.34c
Our assumptions contradicts that the rocket moving along the same direction of the projectile.
Therefore, the speed of rocket travelling past the earth is 0.34c.
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eudora ran from her home to her secret laboratory at an average speed of 12\text{ km/h}12 km/h12, start text, space, k, m, slash, h, end text. she then took one of her jetpacks and flew to her school at an average speed of 76\text{ km/h}76 km/h76, start text, space, k, m, slash, h, end text. eudora traveled a total distance of 120120120 kilometers, and the entire trip took 222 hours.
Eudora used her jetpack for around 1.5 hours of flying and 0.5 hours of jogging.
Given: Eudora moved at an average pace of 12 km/h as she sprinted from her house to her underground laboratory. She then hopped into one of her jetpacks and soared 76 km/h to her school. Eudora covered 120 kilometres in all, and the journey took her two hours.
To determine: How much time Eudora spent running and how much time she spent using her jetpack to fly
Solution:
If she utilised the jet pack for x seconds, then
76x + 12 (2-x) = 120
76x + 24 – 12x = 120
x = 96/64
x= 1.5
She spent 0.5 hours running, thus when we simplify, x = 1.5 hours.
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Q1: A pupil fires an elastic band gun. Just before it is released the band is at its most stretched point. If the band is extended by 0.36m and the band is fired with 412J of energy, what is the spring constant of the band?
The spring constant of the given band is equal to 6.36 × 10³ N/m.
What is Hooke's law?Hooke's law can be demonstrated as a law that states that the force (F) required to extend or compress a spring by displacement linearly w.r.t. that distance along a straight line.
The spring force can be represented in the form of mathematical equation:
F = -kx,
where k is a spring constant and x is the deformation of the spring. The negative sign shows that the spring returns to its original position.
The energy stored in the spring when it is extended is given by:
E = (1/2)kx²
Given, the energy stored in the band when it is extended, E = 412 J
The distance by which the band is extended, x = 0.36 m
412 = (1/2)×k (0.36)²
k = 6.36 × 10³ N/m
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as light reflected off an object reaches your eye, it passes through several structures before it reaches the retina. describe the functions of the following structures:.
Eye's cornea, are in front, bends and concentrates light rays. The pupil serves as the eyeball's entry for light, while the lens, a transparent behind pupil that may change form, aids in focusing images on the retina.
What else do you call reflecting?The verbs cogitate, deliberate, reason, conjecture, and think are some frequent synonyms for reflect. All of these phrases refer to "using one's faculties of conceptualization, judgment, or inference," but reflect denotes thoughtful reflection on a memory.
What comes when you suspect about this at all?You can enjoy each small victory if you take the time to think back on life you did well and your triumphs. You are able to see how much of your selflessness and life accomplishments you have accomplished. Without introspection, it's all too simple to lose sight of these things and concentrate only on our gaps.
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Jane, looking for Tarzan, is running at top speed (4.5 m/s) and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward?
Does the length of the vine affect your answer?
Jane could swing 1.03316 m high and the length of the vine does not affect the answer .
In the question ,
it is given that ,
the top speed of Jane is = 4.5 m/s .
she grabs a vine that is hanging vertically from a tall tree in the jungle .
The kinetic energy of running Jane can be converted into gravitational potential energy, and
the highest point is reached when all the Kinetic Energy has been changed to Potential energy .
we know , KE = 1/2mv²
PE = mgh
So Equating KE = PE
and solving for h:
we get , h = v²/2g
So , h = 4.5²/(2*9.8)
= 1.03316 m
also we can conclude that , the length of the vine does not affect the answer .
Therefore , the Jane could swing 1.03316 m high .
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Problems 2-3 refer to the following situation. A lake has a depth of
27.5 m
. The lake is filled with water, with has a density of
1000 kg/m^3
. Since the top of the lake is open to the air, the pressure at the top of the lake is the atmospheric pressure,
P 0​ =1.0×10^5
. Problem 2: What is the pressure at the bottom of the lake? a.
1.7×10^5 Pa
b.
3.7×10^5 Pa
c.
5.7×10^5 Pa
d.
7.7×10^5 Pa
Problem 3: An underwater vehicle is at the bottom of the lake. The vehicle has a circular window of radius
0.175 m
. What is the force on this window, due to the water? a.
1.6×10^4 N
b.
3.6×10^4 N
c.
5.6×10^4 N
d.
7.6×10^4 N
A pressure at the bottom of the lake of: P = 3.7 x 10^5 Pa So the answer to problem 2 is 3.7 x 10^5 Pa the answer to problem 3 is 3.6 x 10^4 N.
To find the pressure at the bottom of the lake, you can use the formula for hydrostatic pressure: P = P0 + rho * g * h Where P is the pressure at the bottom of the lake, P0 is the atmospheric pressure at the top of the lake, rho is the density of the water, g is the acceleration due to gravity, and h is the depth of the lake. Plugging in the given values, you get: P = 1.0 x 10^5 Pa + 1000 kg/m^3 * 9.8 m/s^2 * 27.5 m This gives a pressure at the bottom of the lake of: P = 3.7 x 10^5 Pa. To find the force on the circular window, you can use the formula for the force exerted by a fluid on a surface: F = P * A Where F is the force on the window, P is the pressure at the bottom of the lake, and A is the area of the window. The area of the window can be calculated using the formula for the area of a circle: A = pi * r^2 Where A is the area of the window, pi is the mathematical constant pi (approximately 3.14), and r is the radius of the window. Plugging in the values, you get: F = 3.7 x 10^5 Pa * (pi * (0.175 m)^2) This gives a force on the window of: F = 3.6 x 10^4 N So the answer to problem 3 is b. 3.6 x 10^4 N.
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The proper time between two events is measured by clocks at rest in a reference frame in which the two events: A. occur at the same time B. occur at the same coordinates C. are separated by the distance a light signal can travel during the time interval D. occur in Boston E. satisfy none of the above
The proper time between two events is measured by clocks at rest in a reference frame in which the two events occur at the same coordinates. (Option B)
In physics and astronomy, a frame of reference or reference frame refer to an abstract coordinate system whose origin, orientation, and scale are specified by a set of reference points which are geometric points whose position is identified both mathematically (with numerical coordinate values) and physically (signaled by conventional markers). Hence, the proper time between two events is measured by clocks at rest in a reference frame in which the two events occur at the same coordinates.
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Which of the following atmospheric characterization techniques allow for 2D mapping of an exoplanet's atmospheric temperature and circulation patterns?
Measuring phase curves
Direct imaging
Measuring transits at different wavelengths
Measuring Doppler (velocity) curves
Astronomers can learn important details about the atmosphere, surface temperature, and size of transiting planets.
A transiting planet that is the size of a brown dwarf often only dims the brightness of Sun-like stars by a few percent or less. The transit technique has been used to find the majority of known exoplanets. When a planet passes in front of an observer and a star, it is said to be transiting. Transits in our solar system can be seen from Earth as Venus or Mercury pass in front of the Sun. The atmosphere of a transiting exoplanet can be seen backlit by the star when it passes its host star. As exoplanets pass in front of and behind their stars, one can do further atmosphere observations.
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Suppose a 36 kg squid (not including water) at rest at the surface of the water brings in and quickly ejects 3.0 kg of water to achieve a takeoff speed of 3.5 m/s; these are typical numbers. At what speed does the squid eject the water? If we ignore lift and drag forces as the squid flies through the air, what is the maximum horizontal range that the squid can achieve before splashing down?
The speed with which the squid ejects the water is 42 m/s. If we ignore lift and drag forces, the maximum horizontal range that the squid can achieve before splashing down is about 1.25 m
From the conservation of momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(36 kg) (3.5 m/s) - (3 kg) v = 0
(3 kg)v= (36 kg) (3.5 m/s)
v = 42 m/s
Hence, required speed is 42 m/s.
The maximum horizontal range is,
R = u ² sin 20/g
Rmax = u²sin 2 (45°)/g
= (3.5 m/s) ² sin 90° / 9.8
=1.25 m
The maximum horizontal range is 1.25 m.
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Classical physics defines the vacuum as a state of absence: a vacuum is said to exist in a region of space if there is nothing in it. In the quantum field theories that describe the physics of elementary particles, the vacuum becomes somewhat more complicated. Even in empty space, particles can appear spontaneously as a result of fluctuations of the vacuum. For example, an electron and a positron, or antielectron, can be created out of the void. Particles created in this way have only a fleeting existence; they are annihilated almost as soon as they appear, and their presence can never be detected directly. They are called virtual particles in order to distinguish them from real particles, whose lifetimes are not constrained in the same way, and which can be detected. Thus it is still possible to define the vacuum as a space that has no real particles in it.
The Vacuum - Its Fluctuations and Decay.
Vacuum is not the problem. The reason space is almost a complete vacuum is not because it is attracted to you, but because it is almost empty.
This emptiness results in extremely low pressure. This is an approach to theoretical physics and quantum mechanics that views the underlying physical vacuum as a superfluid or Bose-Einstein condensate.
A vacuum is a space devoid of matter, or a space in which the pressure is so low that the particles in the space do not affect the processes taking place therein. This is a state well below normal atmospheric pressure and is measured in units of pressure. There is no perfect vacuum because it is impossible to remove all air or particles from space. However, partial vacuums are common.
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find ds in terms of dv and approximate the change in the surface when the volume change from 1,000 m3 to 999 m3 .
Terms of dv and approximate the change in the surface when the volume change from 1,000 m3 to 999 m3 is 2.9 M2.
What is volume?Volume is a measure of the amount of three-dimensional space occupied by a solid, liquid, gas, or plasma. It is measured in cubic meters, liters, or milliliters. Volume is an important measurement when dealing with liquids and other substances since it determines the amount of material that can fit in a given container. Volume also has an effect on pressure and temperature, as the pressure and temperature of a substance increase when its volume decreases.
The change in the surface of a given volume can be found by taking the derivative of the volume in terms of the surface. The derivative of V (volume) with respect to S (surface) is S = 3√V. Therefore, the change in surface when the volume changes from 1000 m3 to 999 m3 is approximately 3√999 - 3√1000 = 2.9 m2.
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