Answer:
The kinetic energy of the system decrease
Ki = 5.78 KJ
Kf = 4.55 KJ
Explanation:
For answer this question we will use the law of the conservation of the angular momentum so,
Li = Lf
Where Li is the inicial momentum of all the system, and Lf is the final momentum of the system.
also, the angular momentum L can be calculated in two ways
L = IW
where I is the momentum of inertia and the W is the Angular velocity.
or,
L = MVD
where M is the mass, V is the lineal velocity and the D is the lever arm.
Therefore,
Li = Ld ( merry-go-round) + Lp ( person )
Lf = Ls
Where Ld is the angular momentum of the merry go round, Lp is the angular momentum of the person and Ls is the angular momentum of the sistem (merry-go-round + person)
so,
[tex]L_d=I_dW_d[/tex]
Ld = [tex]\frac{1}{2}M_dR^{2}W_d[/tex]
Ld = [tex]\frac{1}{2}(155) (2.63)^{2}(0.718*2\pi)[/tex]
Ld = 2418.43
and,
[tex]L_p=M_pV_pD[/tex]
Lp = (59.4)(3.34)(2.63)
Lp = 521.78
then,
Lf = Ls
L_d=I_sW_s
Lf = [tex](\frac{1}{2}(155)(2.63)^{2}+(59.4)(2.63^2))(W_s)[/tex]
[tex]Lf = 946.92W_s[/tex]
so, solving for Ws
Lf = Li
[tex]946.92W_s = 521,78 + 2418.43[/tex]
Ws = 3.1 rad/s
Finally, the inicial and the final Kinetic energy
Ki = [tex]\frac{1}{2}I_d(W_d)^2 + \frac{1}{2}M_p(V_p)^2[/tex]
Ki = 5786.284 J = 5.78 KJ
Kf = [tex]\frac{1}{2}I_s(W_s)^2[/tex]
Kf = 4549.97 J = 4.55 KJ
Then, The kinetic energy of the system decrease because Kf < Ki
Most of the water vapor and carbon dioxide in earth's atmosphere is found A. In the upper part of the atmosphere. B.In the thin layer right in the middle of the atmosphere C.In the lower part of the atmosphere. D.In the thick layer right in the middle of the atmosphere
As the gases ( O₂ & CO₂ ) are most abundant here and supports life.
Hope it helps!
Happy Learning!!
:D
Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 75 N and that the pressured water provides a frictionless support. How long will it take her to rotate the sphere one time, starting from rest?
Answer:
The time taken to rotate the sphere one time is, t = 22 s
Explanation:
Given data,
The mass of the sphere, m = 8200 kg
The radius of the sphere, r = 90 cm
= .9 m
The force applied by the girl, F = 75 N
The moment of inertia of the sphere is,
I = 2/5 mr²
= (2/5) 8200 x (.9)²
= 2657 kg·m²
The torque,
τ = I α
75 x 0.9 = 2657 x α
α = 0.0254 rad/s²
The angular displacement,
θ = ½αt²
2π = ½ x 0.0254 rad/s² x t²
t = 22 s
Hence, the time taken to rotate the sphere one time is, t = 22 s
The intensity of a sunspot is found to be 3 times smaller than the intensity emitted by the solar surface. What is the approximate temperature of this sunspot if the temperature of the solar surface is 5800 K?
Answer:
4400 K approximately
Explanation:
Stefan-Boltzmann law establish that the power radiated from a black body in terms of its temperature which is proportional to [tex]T^4[/tex]
[tex]i = aT^4[/tex] with a Stefan-Boltzmann constant
Also we know that [tex]i_{sunspot}[/tex] is three times [tex]i_{sun}[/tex]
[tex]i_{sun}=3i_{sunspot}[/tex] or [tex]\frac{i_{sun}}{i_{sunspot}} = 3[/tex]
Using the Stefan-Boltzmann we can write
[tex]\frac{i_{sun}}{i_{sunspot}} = 3 = \frac{aT_{sun}^4}{aT_{sunspot}^4}[/tex]
solving for [tex]T_{sunspot}[/tex]
[tex]T_{sunspot} = \left(\frac{T_{sun}^4}{3}\right)^{1/4}[/tex]
Replacing the value of [tex]T_{sun}[/tex] (5800 K) it is obtained that [tex]T_{sunspot}[/tex] is 4407.05
Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of 0.01 rad/s2 about the center O. In the figure, the magnitude of the linear acceleration of P, when it reaches the y-axis, is closest to:
a..063
b..075
c..072
d..069
e..066
Answer:
e). [tex]a = 0.066 m/s^2[/tex]
Explanation:
As we know that wheel is turned by 90 degree angle
so the angular speed of the wheel is given as
[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]
now we have
[tex]\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})[/tex]
[tex]\omega = 0.177 rad/s[/tex]
now the centripetal acceleration of the point P is given as
[tex]a_c = \omega^2 R[/tex]
[tex]a_c = (0.177)^2(2)[/tex]
[tex]a_c = 0.063 m/s^2[/tex]
tangential acceleration is given as
[tex]a_t = R\alpha[/tex]
[tex]a_t = 2(0.01)[/tex]
[tex]a_t = 0.02 m/s^2[/tex]
now net acceleration is given as
[tex]a = \sqrt{a_t^2 + a_c^2}[/tex]
[tex]a = \sqrt{0.02^2 + 0.063^2}[/tex]
[tex]a = 0.066 m/s^2[/tex]
The question involves the calculation of linear acceleration of a point on a wheel undergoing angular acceleration. The linear acceleration involves both tangential and radial components, and these are combined to provide the total acceleration. The closest answer is 0.066 m/s^2.
Explanation:The subject of this question is physics, specifically the concepts of angular acceleration and linear acceleration. Given an angular acceleration, we can find the linear acceleration by using the formula a = rα, where a is linear acceleration, r is the radius, and α is angular acceleration. Substituting the given values, we get a = 2.0 m * 0.01 rad/s2 = 0.02 m/s2. This value is the tangential acceleration of point P.
Since the point P is on the y-axis, the linear acceleration, a, is given by the radial acceleration which is equal to ω2r, where ω is the angular velocity and r is the radius of the circle. But we also have ω2 = 2αr, where ρ is the angular acceleration and r is the radius. So, the radial acceleration is (2αr)2r = 4α2r3. Substituting for α and r we get: 4*(0.01 rad/s2)2*(2.0 m)3 = 0.0016 m/s2.
The total linear acceleration is the vector sum of the radial and tangential accelerations. Since these vectors are perpendicular, we calculate their resultant by Pythagoras' theorem: √(at2 + ar2) = √(0.022 m/s2 + 0.00162 m/s2) = 0.02016 m/s2. Hence, the choice that's closest is (a) 0.066 m/s2.
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A heavy steel ball is hung from a cord to make a pendulum. The ball is pulled to the side so that the cord makes a 5 ∘ angle with the vertical. Holding the ball in place takes a force of 40 N . If the ball is pulled farther to the side so that the cord makes a 8 ∘ angle, what force is required to hold the ball?
Answer:
f = 63.8 N
Explanation:
initial angle to the vertical = 5 degrees
initial holding force = 40 N
final angle to the vertical = 8 degrees
final holding force = ?
find the final holding force
force = mgSinθm = mass and g = acceleration due to gravity
for the initial holding force:40 = mgSin5
mg = [tex]\frac{40}{sin5}[/tex] ....equation 1
for the final holding force:f = mgSin8 ......equation 2
substituting the value of mg from equation 1 (where mg = [tex]\frac{40}{sin5}[/tex] ) into equation 2
f = mgSin8 = [tex]\frac{40}{sin5}[/tex] x Sin8
f = 63.8 N
Answer:
64 N
Explanation:
Given that the angles are very small, the following approximation can be made:
F ≈ m*g*α
where F is the force needed to hold the ball, m is the ball mass, g is the acceleration of gravity and α is the angle between the cord and the vertical.
Let's call F1 the force needed to hold the ball at 5° (α1) and F2 the force needed to hold the ball at 8° (α2).
F1 ≈ m*g*α1
F2 ≈ m*g*α2
Dividing the equations:
F1/F2 = α1/α2
F2 = (α2/α1)*F1
F2 = (8/5)*40
F2 = 64 N
A new gel is being developed to use inside padding and helmets to cushion the body from impacts. The gel is stored in a 4.1 m^3 cylindrical tank with a diameter of 2 m. The tank is pressurized to 1.3 atm of surface pressure to prevent evaporation. A total pressure probe located at the bottom of the tank reads 60 ft of water. What is the specific gravity of the gel contained in the tank?
Answer:
[tex]SE_{gel} = 3.75[/tex]
Explanation:
First, we have to calculate the gel's column height using the cylinder's volume, as follows:
[tex]V=\pi\times r \times h\\h=\frac{V}{\pi \times r}\\h=\frac{4.1m^3}{\pi \times 1m}= 1.30 m[/tex]
Then, as the pressure given at the bottom of the tank is the sum of the surface pressure and the gel's column pressure, we need to calculate only the gel's column pressure:
ft of water is a unit of pressure, but we need to convert it to atm and then to Pa, in order to calculate our results in the correct units. Therefore, the conversion factor is:
1 ft of water (4°C) = 0.0295 atm
[tex]60 ft water \times \frac{0.0295 atm}{1 ft water}= 1.77 atm\\P_{bottom}=P_{surface}+P_{gel}\\P_{gel}=P_{bottom}-P_{surface}=1.77 atm - 1.3 atm\\P_{gel}= 0.47 atm\times \frac{101325Pa}{1 atm}=47622.75 Pa[/tex]
Now, to calculate the specific gravity, we need to find first the gel's density:
[tex]P_{gel} = \rho gh\\\rho = \frac{P_{gel}}{gh}=\frac{47622.75 Pa}{9.8 m/s^2 \times 1.30m}= 3738.04 \frac{kg}{m^3}[/tex]
[tex]SE_{gel} = \frac{\rho_{gel}}{\rho_{water}}= \frac{3738.04 kg/m^3}{997 kg/m^3} = 3.75\\SE_{gel} =3.75[/tex]
The specific gravity of the gel is 3.75.
Final answer:
To calculate the specific gravity of a gel, convert the pressure reading from feet of water to pascals, add the tank's pressurized atmosphere, then use the relation between pressure, density, gravity, and height to solve for the density of the gel, which is then compared to the density of water.
Explanation:
The question is asking to calculate the specific gravity of a gel based on the pressure reading from a probe at the bottom of a cylindrical tank. We are given the pressure as 60 ft of water. To convert this to a pressure that we can use to find density, we need to convert feet of water to pascals:
1 ft H2O = 2989.07 Pa
60 ft H2O = 60 * 2989.07 Pa = 179344.2 Pa
Since the tank is pressurized to 1.3 atmospheres at the surface, we must also consider this in our pressure calculation:
1 atm = 101325 Pa
1.3 atm = 1.3 * 101325 Pa = 131722.5 Pa
The total pressure at the bottom of the tank is the sum of the pressure due to the gel and the pressurized air:
Total pressure = 179344.2 Pa + 131722.5 Pa
Total pressure = 311066.7 Pa
To find the specific gravity, we use the following relation where the specific gravity is the ratio of the density of the gel (ρgel) to the density of water (ρH2O):
Specific gravity = ρgel / ρH2O
We know that pressure is also the product of density (ρ), gravity (g=9.81 m/s²), and height (h=60 ft * 0.3048 m/ft) for the fluid:
Pressure = ρgel * g * h
ρgel = Pressure / (g * h)
ρgel = 311066.7 Pa / (9.81 m/s² * 60 ft * 0.3048 m/ft)
After calculating ρgel, we divide that by the density of water (1000 kg/m³) to find the specific gravity.
A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pressure and one that reads gauge pressure. The tank is then brought to the top of a mountain.Which statement is true for the gauge that reads gauge pressure?
Select the correct answer
Both gauges will change by the same amount.
The pressure reading increases.
The pressure reading stays the same.
The pressure reading decreases.
Final answer:
The gauge pressure reading stays the same when a steel cylinder is moved to the top of a mountain because it measures pressure relative to the external atmospheric pressure, which does not influence the pressure inside the cylinder.
Explanation:
The correct statement for the gauge that reads gauge pressure when a steel cylinder is brought to the top of a mountain is that the pressure reading stays the same. Gauge pressure is the pressure relative to atmospheric pressure, which means it measures the excess pressure in the tank over the external atmospheric pressure. Since the pressure inside the steel cylinder is not dependent on external atmospheric pressure changes, the reading on the gauge pressure meter would remain constant, even if the cylinder is taken to a different altitude.
Two balls are thrown horizontally. Ball C is thrown with a force of 20 N, and ball D is thrown with a force of 40 N. Assuming all other factors are equal, ball D will fall toward the ground
Answer:
ball D will fall toward the ground at the same time as ball C
Explanation:
both balls experience the same downward (vertical) force of gravity as such they will both fall down at the same time, given that all other factors are equal.
although the ball were through with different forces,
those forces where in the horizontal direction but the force of gravity (downward force) will act on them equally to bring them down at the same time
What is evaporation? Condensation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. ResetHelp liquid gaseous solid Evaporation is the physical change in which a substance converts from its state to its state. Condensation is the physical change in which a substance converts from its state to its state.
Answer:
Evaporation is the physical change in which a substance converts from its liquid state to its gaseous state. Condensation is the physical change in which a substance converts from its gaseous state to its liquid state.
Explanation:
Evaporation and condensation are opposite processes to each other. Evaporation changes a liquid to a gas and condensation is the reverse.
Standing at a crosswalk, you hear a frequency of 540 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 446 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)
Answer:
Speed of the ambulance is 32.7 metres per second.
Explanation:
Let the actual frequency of the siren be [tex]f_{0}[/tex].
Frequency observed by me when ambulance is approaching = 540 Hz
Frequency observed by me when ambulance is moving away = 446 Hz
Let [tex]v_{s}[/tex] be the speed of sound and [tex]v_{a}[/tex] be the speed of ambulance.Then according to Doppler effect:
When source is moving towards observer,frequency observed is given as
[tex]f_{0} \times \frac{v_{s} }{v_{s} - v_{a} }[/tex] = 540 Hz
When source is moving away from observer,frequency observed is given as
[tex]f_{0} \times \frac{v_{s} }{v_{s} + v_{a} }[/tex] = 446 Hz
Taking [tex]v_{s} = 343 \frac{m}{s}[/tex] and solving the above two equations by eliminating [tex]f_{0}[/tex],
we get [tex]v_{a} = 32.7 \frac{m}{s}[/tex]
To calculate current flow through any branch of a circuit by substituting the values of IX and RX for the branch
values when total circuit current and the resistance are known, use the _______________ formula.
A. reciprocal
B. current divider
Answer:
Current divider
Explanation:
To calculate current flow through any branch of a circuit by substituting the values of [tex]I_X\ and\ R_X[/tex] for the branch values when total circuit current and the resistance are known, use the current divider formula.
A current divider is a circuit that produces output current as a function of input current. It is a rule to find the splitting of the current in all branches of the circuit. Hence, the correct option is (B).
The formula to calculate the current flow through any branch of a circuit when the total circuit current and resistance are known is the current divider formula.
Explanation:To calculate the current flow through any branch of a circuit by substituting the values of IX and RX for the branch values when total circuit current and the resistance are known, use the current divider formula. The current divider rule is particularly useful in parallel resistor circuits and it allows for the easy calculation of the current flowing through a resistor in parallel.
In a parallel circuit, the voltage across each branch is the same, but the current through each branch can be different, depending on the resistance of that branch. The current divider rule states that the current through a branch is the ratio of the total parallel resistance to the branch resistance, times the total current entering the parallel combination.
For example, if we want the current through resistor R1 (I1), and we know the total current (Itot) and the total parallel resistance (Rp), as well as the resistance of R1 (R1), we can use the formula:
I1 = Itot * (Rp / R1)
By knowing the total resistance and the total current, one can deduce the individual branch currents using Ohm's Law and the principles of parallel circuits.
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You have been hired to check the technical correctness of an upcoming made-for-TV murder mystery that takes place in a space shuttle. In one scene, an astronaut's safety line is cut while on a spacewalk. The astronaut, who is 200 meters from the shuttle and not moving with respect to it, finds that the suit's thruster pack has also been damaged and no longer works and that only 4 minutes of air remains. To get back to the shuttle, the astronaut unstraps a 10-kg tool kit and throws it away with a speed of 8 m/s. In the script, the astronaut, who has a mass of 80 kg without the toolkit, survives, but is this correct?
Answer:
The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air
Explanation:
Due the astronaut throws the 10-kg tool kit away with a speed of 8 m/s, it gives a momentum equivalent but in the other direction, so [tex]I=mv=(10Kg)(8m/s)=80kg*m/s[/tex], then we can find the speed that the astronaut reaches due to its weight we get, [tex]v=\frac{I}{m} =\frac{80kg*m/s}{80Kg} =1m/s[/tex].
Finally, as the distance to the space shuttle is 200m, the time taken to the astronaut to reach it at the given speed will be [tex]t=\frac{d}{v}=\frac{200m}{1m/s}=200s[/tex], as the remaining air time is 4 min or 240 seconds, The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air.
Approximately how long does it take the sun to orbit the milky way galaxy?
Answer:
it takes the sun about 230 million years to orbit the milky way.
Explanation:
we're moving at an average velocity of 828,000 km/hr but it still takes us a long time to orbit the milky way.
Which of the following would constitute evidence against Turcotte's model?
A) the success of gradualist models explaining the surface of Mars
B) an even more detailed map of the surface of Venus
C) an even longer river of lava on Io, a moon of Jupiter
D) a few active volcanoes on Ishtar Terra, a continent on Venus
E) a volcano on Earth releasing a massive burst of thermal energy all at once
Answer: answer is option A
Explanation: gradually model ascertained the features of the surface of mass is explaining it as gradual incremental changes continuously over a long period of time which negates the catastrophic model nature of turtles model
Tech A says the first external component to install when assembling an engine is the intake manifold. Tech B says the exhaust manifolds should be installed after the intake manifold. Who is correct?
Answer:
both technicians are correct
Explanation:
both technicians are correct because technician A says the first external component to be installed is the intake manifold and technician B says the exhaust manifold should be installed after the intake manifold has been installed, this implies the same thing because installing the exhaust manifold after installing the intake manifold means you are installing the intake manifold first.
A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.020 kg m2. A string is wound around the cylinder and pulled with a force of 1.0 N. The angular acceleration of the cylinder is:
Answer:[tex]5 rad/s^2[/tex]
Explanation:
Given
Radius of cylinder r=0.1 m
Length L=0.2 in.
Moment of inertia I=0.020 kg-m^2
Force F=1 N
We Know Torque is given by
[tex]Torque =I\alpha =F\cdot r[/tex]
where [tex]\alpha =angular\ acceleration[/tex]
[tex]I\alpha =F\cdot r[/tex]
[tex]0.02\cdot \alpha =1\cdot 0.1[/tex]
[tex]\alpha =5 rad/s^2[/tex]
The angular acceleration of the cylinder is 5 rad/s².
To calculate the angular acceleration of the cylinder, we use the formula of torque below.
What is torque?Torque is the force that causes a body to rotate about an axis.
Formula:
Iα = Fr........... Equation 1Where:
I = rotational initialα = Angular accelerationr = Radius of the cylinderF = Forceθ = Angle.
Make α the subject of the equation
α = RFsinθ/I............. Equation 2From the question,
Given:
r = 0.1 mF = 1.0 NI = 0.020 kgm²θ = 90° ( Since the rope is tangental to the side of the cylinder).Substitute these values into equation 2
α = (1.0×0.1×sin90°)/(0.02)α = 5 rad/s²Hence, The angular acceleration of the cylinder is 5 rad/s².
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The Earth and the moon are attracted to each other by gravitational force. The more massive Earth attract the less massive moon with a force that is (greater than, less than, the same as) the force with which the moon attracts the Earth.
Answer:
Earth attract the less massive moon with a force that is the same as the force with which the moon attracts the Earth.
Explanation:
This can be explained by Newton's third law:
[tex]F_{12}=-F_{21}[/tex]
The force exerted by body 1 on body 2 is the same as that exerted by body 2 on body 1, only with the opposite sign.
In this case that force is the gravitational force, but the law still applies.
So the moon and the earth are attracted with the same magnitude of force.
Based on discoveries to date, which of the following conclusions is justified?a) Most stars have one or more terrestrial planets orbiting within their habitable zones.b) Planets are common, but planets as small as Earth are extremely rare.c) Planetary systems are common and planets similar in size to Earth are also common.d) Although planetary systems are common, few resemble ours with terrestrial planets near the Sun and jovian planets far from the Sun.
Based on discoveries to date, the conclusion as “Planetary systems are common and planets similar in size to Earth are also common” is justified.
Answer: Option C
Explanation:
Some studies show that on average, each star has at least single planet. This means that most stars, such as the Solar System, possess planets (otherwise exoplanets). It is known that small planets (more or less Earthly or slightly larger) are more common than giant planets. The mediocrity principles state that planet like Earth should be universal in the universe, while the rare earth hypothesis says they are extremely rare.
Size is often considered an important factor, because planets the size of the Earth are probably more terrestrial and can hold the earth's atmosphere. The planetary system is a series of gravitational celestial objects orbiting a star or galaxy. Generally, planetary systems describe systems with one or more planets, although such systems may also consist of bodies such as dwarf planets, asteroids and the like.
Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N. What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?
Answer:
The tension in the string connecting block 50 to block 51 is 50 N.
Explanation:
Given that,
Number of block = 100
Force = 100 N
let m be the mass of each block.
We need to calculate the net force acting on the 100th block
Using second law of newton
[tex]F=ma[/tex]
[tex]100=100m\times a[/tex]
[tex]ma=1\ N[/tex]
We need to calculate the tension in the string between blocks 99 and 100
Using formula of force
[tex]F_{100-99}=ma[/tex]
[tex]F_{100-99}=1[/tex]
We need to calculate the total number of masses attached to the string
Using formula for mass
[tex]m'=(100-50)m[/tex]
[tex]m'=50m[/tex]
We need to calculate the tension in the string connecting block 50 to block 51
Using formula of tension
[tex]F_{50}=m'a[/tex]
Put the value into the formula
[tex]F_{50}=50m\times a[/tex]
[tex]F_{50}=50\times1[/tex]
[tex]F_{50}=50\ N[/tex]
Hence, The tension in the string connecting block 50 to block 51 is 50 N.
We have that for the Question "What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?"
it can be said that
The tension in the string connecting block 100 to block 99 = [tex]1N[/tex]The tension in the string connecting block 50 to block 51 = [tex]50N[/tex]
From the question we are told
Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N.
Assuming mass of each block is 1 kg
The equation for the force is given as
[tex]F = ma\\\\a = \frac{F}{m}\\\\ = \frac{100}{100*1}\\\\ = 1 m/s^2[/tex]
Now, between block 100 and 99,
[tex]F = ma\\\\F = 1*1 \\\\= 1 N[/tex]
Now between block 50 and 51. There are 50 blocks behind 51 st block, so,
[tex]m = 50 Kg\\\\a = 1 m/s2 (assuming all blocks accelerate at same rate)\\\\F = 50 * 1\\\\F= 50 N[/tex]
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A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the equilibrium point (y= 0) with a negative velocity at t= 0.Assume that downward is the positive direction of motion. What equation describes this motion as a function of time?
Answer:
y = -19.2 sin (23.15t) cm
Explanation:
The spring mass system is an oscillatory movement that is described by the equation
y = yo cos (wt + φ)
Let's look for the terms of this equation the amplitude I
y₀ = 19.2 cm
Angular velocity is
w = √ (k / m)
w = √ (245 / 0.457
w = 23.15 rad / s
The φ phase is determined for the initial condition t = 0 s , the velocity is negative v (0) = -vo
The speed of the equation is obtained by the derivative with respect to time
v = dy / dt
v = - y₀ w sin (wt + φ)
For t = 0
-vo = -yo w sin φ
The angular and linear velocity are related v = w r
v₀ = w r₀
v₀ = v₀ sinφ
sinφ = 1
φ = sin⁻¹ 1
φ = π / 4 rad
Let's build the equation
y = 19.2 cos (23.15 t + π/ 4)
Let's use the trigonometric ratio π/ 4 = 90º
Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a
y = -19.2 sin (23.15t) cm
What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?
Answer:
Explanation:
Facts about magnetic flux;
(1). Change in magnetic flux is equal to the current generated.
(2). The magnetic field moves from the north to the south.
(3). Magnetic flux is equal to how much magnetic field goes through a direction.
(4). Magnetlc direction can be found found by using Lenz's law.
Faraday's law= emf= N(BA/ ∆t.
Where N= number of loop
∆B/∆T = rate at which flux changes.
The magnitude φ(initial) of the magnetic flux through one turn of the coil before it is rotated is
φ(initial) = BA while φ(final)= zero(0).
The initial magnetic flux through one turn of the coil, with the coil initially perpendicular to Earth's magnetic field, is 6.10 × 10⁻⁹ T×m².
To calculate the magnitude of the initial magnetic flux (φinitial) through one turn of the coil, we use the formula:
φ = B × A × cos(θ)
where:
B is the magnetic field (5.00 × 10⁻⁵ T)A is the area the coil encloses (12.2 cm²)θ is the angle between the magnetic field and the normal to the plane of the coilSince initially, the plane of the coil is perpendicular to Earth's magnetic field, θ = 0°, and hence cos(θ) = 1. First, we must convert the area from cm² to m²:
A = 12.2 cm² × (1 m² / 10,000 cm²) = 12.2 × 10⁻⁴ m²
Now, we can calculate the magnetic flux:
φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m² × cos(0°)
φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m²
φinitial = 6.10 × 10⁻⁹ T×m²
This is the magnitude of the initial magnetic flux through one turn of the coil before it is rotated.
Complete Question:
A coil has 230 turns enclosing an area of 12.2 cm³. In a physics laboratory experiment, the coil is rotated during the time interval 3.00 × 10⁻² s from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00 × 10⁻⁵ T.
What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?
In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exerts an upward force of 400 N. (a) What is the mass of the car in kilograms? (b) What is its weight in pounds?
Answer:
Explanation:
Given
Each student exert a force of [tex]F=400 N[/tex]
Let mass of car be m
there are 18 students who lifts the car
Total force by 18 students [tex]F=18\times 400=7200 N[/tex]
therefore weight of car [tex]W=7200[/tex]
mass of car [tex]m=\frac{W}{g}[/tex]
[tex]m=\frac{7200}{9.8}=734.69 kg[/tex]
(b)[tex]7200 N \approx 1618.624\ Pound-force[/tex]
[tex]734.69 kg\approx 1619.71 Pounds[/tex]
The mass of the car lifted by eighteen students is 734.69 kg, and its weight is approximately 1620.29 pounds.
To determine the mass of the car that eighteen students lift in a college homecoming competition, given that each student exerts an upward force of 400 N, first calculate the total force exerted by the students: 18 students × 400 N per student = 7200 N.
Next, we use the equation for weight (W = mg) to find the mass (m), where W is the weight, m is the mass, and g is the acceleration due to gravity (9.8 m/s²). Thus, the mass of the car is m = W/g = 7200 N / 9.8 m/s² ≈ 734.69 kg.
To find the weight in pounds, we can use the conversion factor 1 kg ≈ 2.20462 lbs. Therefore, the weight of the car in pounds is approximately 734.69 kg × 2.20462 lbs/kg ≈ 1620.29 lbs.
Waves on a swimming pool propagate at 0.720 m/s. You splash the water at one end of the pool and observe the wave go to the opposite end, reflect, and return in 31.0 s. How far away is the other end of the pool?
Answer:
11.16 m
Explanation:
total distance traveled by the wave = speed * time
= 0.720 m/s * 31.0 s
= 22.32 m
Since this is an echo back to the starting point then the length of the swimming pool must be half of the distance traveled.
length of swimming pool = 22.32 m÷2 = 11.16 m
A cabin has a 0.159-m thick wooden floor [k = 0.141 W/(m · C°)] with an area of 13.4 m2. A roaring fire keeps the interior of the cabin at a comfortable 18.0 °C while the air temperature in the crawl space below the cabin is –16.4°C. What is the rate of heat conduction through the wooden floor?
Answer:
[/tex] 408.8[/tex]
Explanation:
[tex]t[/tex] = thickness of the floor = 0.159 m
[tex]k[/tex] = thermal conductivity of wooden floor = 0.141 Wm⁻¹ ⁰C⁻¹
[tex]T_{i}[/tex] = Temperature of the interior of the cabin = 18.0 ⁰C
[tex]T_{o}[/tex] = Temperature of the air below the cabin = - 16.4 ⁰C
Difference in temperature is given as
[tex]\Delta T[/tex] = Difference in temperature = [tex]T_{i} - T_{o} = 18 - (- 16.4) = 34.4[/tex]⁰C
[tex]A[/tex] = Area of the floor = 13.4 m²
[tex]Q[/tex] = Rate of heat conduction
Rate of heat conduction is given as
[tex]Q = \frac{kA \Delta T}{t}[/tex]
[tex]Q = \frac{(0.141) (13.4) (34.4)}{0.159}\\Q = 408.8[/tex] W
A stone is launched vertically upward from a cliff 384 ft above the ground at a speed of 80 ft divided by s. Its height above the ground t seconds after the launch is given by s equals negative 16 t squared plus 80 t plus 384 for 0 less than or equals t less than or equals 8. When does the stone reach its maximum height?
Answer:
2.5 seconds
Explanation:
s(t) = -16t^2 + 80t + 384
for
0≤t≤8
First we differentiate s(t) to get s'(t)
s'(t) = -32t + 80
Let us then find the critical point; thus we will equate s'(t) to zero and then search for values where s'(t) is undefined
s'(t) = -32t + 80 = 0
t = 80/32
t = 2.5 sec
Let us evaluate s at the critical points and end points
s(0) = -16(0)^2 + 80(0) + 384 = 384
s(2.5) = -16(2.5)^2 + 80(2.5) + 384 = 684
s(8) = -16(8)^2 + 80(8) + 384 = 0
Thus, the stone attains it maximum height of 684ft at at t=2.5s
What can turn a natural hazard into a natural disaster?
Answer: A hazard is a condition that has the potential to cause harm.
A natural hazard is a potentially harmful situation, where a person places himself in a naturally unsafe zone.
A natural disaster is a large scale destruction of life and properties by the forces of nature.
A natural hazard can become a natural disaster in some cases where the natural harmful situation a person places himself in, is acted upon by the forces of nature on a large scale.
1. The ___________ was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.2. Discovering planets through the __________ requires obtaining and studying many spectra of the same star.3. The________________ successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.4. The ____________ is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.5. Compared to the planets of our solar system, the composition of a _________________ most resembles the compositions of Uranus and Neptune.6. Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call______.7. An extrasolar planet that is rocky and larger than Earth is often called a_____.
Answer:
(1) The ____astrometric method_______ was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.
(2) Discovering planets through the ____doppler method ______ requires obtaining and studying many spectra of the same star.
(3) The____kepler mission ____________ successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.
(4) The ___transit method_________ is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.
(5) Compared to the planets of our solar system, the composition of a ____water world _____________ most resembles the compositions of Uranus and Neptune.
(6) Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call__migration ____.
(7) An extrasolar planet that is rocky and larger than Earth is often called a__super-Earth ___.
The Doppler technique and transit technique are used to discover extrasolar planets such as Jupiter-sized ones and those larger than Earth respectively. The Kepler mission has been successful in identifying thousands of these. The composition of extrasolar planets can differ, some resembling Uranus and Neptune, while the concept of 'planet migration' explains the closeness of large planets to their stars.
Explanation:1. The Doppler technique was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.
2. Discovering planets through the spectroscopy requires obtaining and studying many spectra of the same star.
3. The Kepler mission successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.
4. The transit technique is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.
5. Compared to the planets of our solar system, the composition of a mini-Neptune most resembles the compositions of Uranus and Neptune.
6. Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call planet migration.
7. An extrasolar planet that is rocky and larger than Earth is often called a super-Earth.
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A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light in the material?
Answer:
80% (Eighty percent)
Explanation:
The material has a refractive index (n) of 1.25
Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s
We can find the speed of light in the material (v) using the relationship
n = c/v, similarly
v = c/n
therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s
v = 239 833 966 m/s
Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as
(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%
Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)
Answer:
The percentage of speed of light in vacuum to the speed of light in the said material is 80%
Explanation:
The common values of refractive index are between 1 and 2 since nothing can travel faster than the speed of light, therefore, no material has a refractive index lower than 1.
According to the formula [tex]n=\frac{c}{v}[/tex]
where [tex]n[/tex] is the index of refraction
[tex]c[/tex] is the speed of light in vacuum
and [tex]v[/tex] is the speed of light in the material, it can be seen that n and v are inversely proportional which means greater the refractive index lower is the speed of light.
Since we know that speed of light in vacuum is 300,000 km/s using the formula we get,
[tex]v=\frac{c}n}[/tex]
[tex]v=\frac{300000}{1.25}=240,000 km/s[/tex]
for finding percentage,
[tex]=\frac{240000}{300000}*100 = 80[/tex] %
(a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.20 kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.
Answer:
a= 0.063 m
b = 0.116 m
Explanation:
First of all, we need the spring constant in order to solve this problem. You are not giving that data, but I will tell you how to solve this assuming a value of k, In this case, let's assume the value of k is 1600 N/m. (I solved an exercise like this before, using this value).
Now, we need to use the expressions to calculate the distance of the spring.
The elastic potential energy (Uel) is given with the following formula:
Uel = 1/2 kx²
Solving for x:
x = √2*Uel/k
Replacing the data in the above formula (And using the value of k os 1,600):
x = √2 * 3.2 / 1600
x = 0.063 m
b) For this part, we need to apply the work energy theorem which is:
K1 + Ugrav1 + Uel1 + Uo = K2 + Ugrav2 + Uel2
Since in this part, the exercise states that the book is dropped, we can say that the innitial and the end is 0, therefore, K1 = K2 = 0.
The spring at first is not compressed, so Uel1 = 0, and Uo which is the potential energy of other factors, is also 0, because there are no other force or factor here. Therefore, our theorem is resumed like this:
Ugrav1 = Uel2
The potential energy from gravity is given by:
Ug = mgy
And as the spring is placed vertically, we know the height which the book is dropped, so the distance y is:
y = x + h
And this value of x, is the one we need to solve. Replacing this in the theorem we have:
mg(h+x) = 1/2kx²
g would be 9.8 m/s²
Now, replacing the data:
1.2*9.8(0.8 + x) = 1/2*1600x²
Rearranging and solving for x we have:
1.2*9.8*2(0.8 + x) = 1600x²
18.82 + 23.52x = 1600x²
1600x² - 23.52x - 18.82 = 0
Now we need to solve for x, using the general formula:
x = - (-23.52) ± √(-23.52)² - 4 * 1600 * (-18.82) / 2*1600
x = 23.52 ± √553.19 + 120,448 / 3200
x = 23.52 ± 347.85 / 3200
x1 = 23.52 + 347.85 / 3200 = 0.116 m
x2 = 23.52 - 347.85 / 3200 = -0.101 m
Using the positive value, we have that the distance is 0.116 m.
Termination of translation requires a termination signal, RNA polymerase, and a release factor. a release factor, initiator tRNA, and ribosomes. initiation factors, the small subunit of the ribosome, and mRNA. elongation factors and charged tRNAs.
Termination of translation requires a termination signal, and a release factor
Explanation:
There are 3 stops codons of the 64 possible codons. These are UAA, UAG, or UGA. These do not code for amino acids and are therefore not recognized by any anticodons for any of the ‘charged’ tRNA. These codons, are recognized by release factors that ‘knock off’ the newly synthesized peptide from the ribosome through peptidyl-tRNA hydrolysis. There are several release factors (RF1 and RF2) in bacteria but in eukaryotes only one RF has been discovered to recognize the 3 stop codons.
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Translation termination is signaled by a stop codon and facilitated by release factors and ribosomes, which lead to the release of the newly synthesized protein and the disassembly of the ribosomal complex.
Explanation:Termination of translation requires a termination signal, which is a stop codon (UAA, UAG, or UGA) that doesn't code for an amino acid but rather signals the end of the translation process. During this stage, release factors recognize the stop codon and prompt the addition of a water molecule to the carboxyl end of the peptidyl-tRNA in the P site. This action leads to the release of the newly synthesized polypeptide chain. Ribosomes, which consist of a large and a small subunit, dissociate from the mRNA and from each other upon the completion of translation.
Ribosomes, release factors, and the mRNA template are the key participants in the termination process. Initiator tRNA and elongation factors with charged tRNAs are involved in the initiation and elongation stages of translation, but not in termination.