Answer:
n=5, l=1, m(l) = -1, m(s)= + 1/2
Explanation:
Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.
(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5
(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).
Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)
(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.
Here, our value is -1 that is m(l)= -1
(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.
A set of quantum numbers for an electron in a 5p orbital could be: principal quantum number (n) = 5, angular momentum quantum number (l) = 1, magnetic quantum number (m) = 0, spin quantum number (m_s) = -1/2.
Explanation:
An electron in a 5p orbital can be described using four quantum numbers. Knowing that p orbitals correspond to the angular momentum quantum number l=1, and the principal quantum number n for the fifth energy level is 5, we can deduce the following. The magnetic quantum number, m, ranges from -l to +l, and thus it can have the values -1, 0, or +1 representing different p orbitals in the same shell. The spin quantum number, m_s, can either be -1/2 or +1/2, indicating the orientation of the electron's spin. Thus, an example of a set of four quantum numbers that could be assigned to an electron occupying a 5p orbital could be: (5, 1, 0, -1/2).
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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
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.
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.
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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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 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
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.
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
A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is 9.8 m/s2.
What is the gravitational potential energyassociated with the ball relative to the ceiling?Answer in units of J.
Answer:-81.29 J
Explanation:
Given
mass of ball [tex]m=2.35 kg[/tex]
Length of string [tex]L=3.53 m[/tex]
height of Room [tex]h=5.03 m[/tex]
Gravitational Potential Energy is given by
[tex]P.E.=mgh[/tex]
where h=distance between datum and object
here Reference is ceiling
therefore [tex]h=-3.53 m[/tex]
Potential Energy of ball w.r.t ceiling
[tex]P.E.=2.35\times 9.8\times (-3.53)=-81.29 J[/tex]
i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling
A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the spring has a height of 82 cm and a constant of 24 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
Answer:
45.6 cm
Explanation:
Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation
[tex]E_g = E_p[/tex]
[tex]mgh = 0.5kx^2[/tex]
where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.
[tex]4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2[/tex]
[tex]47.088x + 227.906 = 1200x^2[/tex]
[tex]1200x^2 - 47.088x - 227.906 = 0[/tex]
[tex]x = 0.456m[/tex] or 45.6 cm
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
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.
A football player kicks a 0.41-kg football initially at rest; and the ball flies through the air. If the kicker's foot was in contact with the ball for 0.051 s and the ball's initial speed after the collision is 21 m/s, what was the magnitude of the average force on the football?
Answer:
Average force on the football = 168.82 N
Explanation:
Force = Mass x Acceleration
F = ma
Mass, m = 0.41 kg
We have equation of motion, v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 21 m/s
Time, t = 0.051 s
Substituting
21 = 0 + a x 0.051
a = 411.76 m/s²
Substituting in force equation,
F = ma = 0.41 x 411.76 = 168.82 N
Average force on the football = 168.82 N
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the balloon has a radius of 0.498 m. If the filled balloon is fastened to a vertical line, what is the tension in the line? The density of air is 1.29 kg/m3.
Answer:
5.5 NExplanation:
mass of balloon (m) = 12.5 g = 0.0125 kg
density of helium = 0.181 kg/m^{3}
radius of the baloon (r) = 0.498 m
density of air = 1.29 kg/m^{3}
acceleration due to gravity (g) = 1.29 m/s^{2}
find the tension in the line
the tension in the line is the sum of all forces acting on the line
Tension =buoyant force + force by helium + force of weight of rubber
force = mass x acceleration
from density = \frac{mass}{volume} , mass = density x volume
buoyant force = density x volume x accelerationwhere density is the density of air for the buoyant force
buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N
force by helium = density x volume x accelerationforce by helium = 0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N
force of its weight = mass of rubber x accelerationforce of its weight = 0.0125 x 9.8 = 0.1225 N
Tension = buoyant force + force by helium + force of weight of rubberthe force of weight of rubber and of helium act downwards, so they
carry a negative sign.
Tension = 6.54 - 0.917 - 0.1225 = 5.5 NCalculating the tension on a string tied to a helium-filled balloon involves principles of buoyancy and weight. The difference between the buoyant force exerted by the displaced air and the sum of the weight of the empty balloon and weight of the helium gives the tension on the line.
Explanation:The tension on the vertical line tied to the helium-filled balloon can be found using Archimedes' Principle and Buoyancy. First, the volume (V) of the helium-filled balloon can be calculated using the formula for the volume of a sphere; V = 4/3 x π x (0.498 m)³ = ~0.516 m³. The mass (m) of the helium in the balloon is calculated by multiplying this volume by the density of helium, which is m = 0.516 m³ x 0.181 kg/m³ = ~0.093 kg.
Next, the weight (w) of air displaced by the balloon is calculated. This weight is equal to the product of the volume of the balloon, the density of the air, and gravitational acceleration; w = 0.516 m³ x 1.29 kg/m³ x 9.8 m/s² = ~6.5 N. This is the buoyant force that lifts the balloon.
The weight (W) of the filled balloon is the sum of the weight of the empty balloon and the helium; W = (0.0125 kg + 0.093 kg) x 9.8 m/s² = ~1.03 N. The tension (T) in the line is then the difference between the buoyant force and this weight (i.e., T = 6.5 N - 1.03 N = ~5.47 N).
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A wood-burning stove (emissivity = 0.900 and surface area = 3.75 m2) is being used to heat a room. The fire keeps the stove surface at a constant 182 °C (455 K) and the room at a constant 22 °C (295 K). Determine the net radiant power generated by the stove.
Answer:
Radiated power will be [tex]675241.7175watt[/tex]
Explanation:
We have given emissivity [tex]\epsilon =0.9[/tex]
Stove temperature T = 455 K
Room temperature [tex]T_C=295K[/tex]
We know the Stephan's constant [tex]\sigma =5.67\times 10^{-6}w/m^2K^4[/tex]
We know that radiated power is given by [tex]P=\epsilon \sigma A(T^4-T_C^4)=0.9\times5.67\times 10^{-6}\times 3.75\times (455^4-295^4)=675241.7175watt[/tex]
The net radiant power generated by the wood-burning stove can be determined by using the Stefan-Boltzmann law. This law combines the factors of Stefan-Boltzmann constant, emissivity of the body, surface area, and the difference of the fourth powers of the absolute temperatures.
Explanation:The rate of radiant power generated by the stove can be found using the Stefan-Boltzmann law, which describes the rate of heat transfer by emitted radiation. This law can be stated as P = σeAT⁴, where P is the radiant power, σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ J/s.m².K⁴), e is the emissivity of the body, A is the surface area, and T is the absolute temperature in Kelvins.
For the net rate of heat transfer by radiation (or the net radiant power) for the wood-burning stove, you would use the Stefan-Boltzmann law as Q-net = σeA(T₂⁴ - T₁⁴), where T₂ is the absolute temperature of the stove and T₁ the absolute temperature of the room.
Plugging in the stove's values from the problem: emissivity = 0.900, surface area = 3.75 m², T₂ = 455 K (stove temperature), and T₁ = 295 K (room temperature), we can calculate the net radiant power allocated by the stove.
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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 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
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]
A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. What is the height and orientation with respect to the original object of the final image?
Answer:
The object for the converging lens is upright and 0.429 cm tall, the image of this converging lens is inverted and 1.375 cm high
Explanation:
Let
[tex]d_{o}=distance of object[tex]\\f=focal length\\d_{i}=distance of image\\I_{h}=Image height[/tex]
For diverging lens:
[tex]d_{o} = 0.50\\f = -0.20\\\frac{1}{d_{o}}+\frac{1}{-0.20}\\\frac{1}{d_{i}}=\frac{1}{-0.20}-\frac{1}{0.50}=-7\\d_{o}=-\frac{1}{7}[/tex]
Magnification = [tex]\frac{d_{i}}{d_{o}}= -\frac{1}{7}÷ 0.5 = -0.286[/tex]
Image height [tex]= -0.286 * 1.5 = -0.429 cm[/tex] (negative sign means the image is virtual, inverted.
This image is [tex]\frac{1}{7}[/tex] meter to left of the center of the diverging lens.
The converging lens is located 0.08 m to the right of the diverging lens
The distance between the image of the diverging lens and center of the converging lens = [tex]\frac{1}{7} + 0.08 = 0.229 m[/tex]
The image of the diverging lens becomes the object of the converging lens.
[tex]d_{o} = 0.223\\f = 0.17\\\frac{1}{d_{i}}=\frac{1}{0.17}-\frac{1}{0.223}=0.715\\d_{i}=0.715m to the right of the converging lens[/tex]
[tex]Magnification =\frac{d_{i}}{d_{o}} = \frac{0.715}{0.223}=3.206\\image height=3.206 * 0.429 = 1.375 cm.[/tex]
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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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.
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
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.
(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.
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.
A uniform rectangular plate of length B = 15 cm and height A = 24 cm has a rectangular corner cut out of it of length D = 6 cm and height C = 8 cm. The plate is made of a material of area mass density σ. For this problem we set the origin at the lower left corner of the plate with the x-axis horizontal pointing right and the y-axis vertical pointing up.
a. Calculate the value of the y-coordinate, in centimeters, for the center of mass of the plate.
b. Calculate the value of the x-coordinate, in centimeters, for the center of mass of the plate.
The center of mass for a rectangle with a corner cut out can be calculated using the centroid formula by considering the plate and the cut-out as separate figures and then determining the average weighted by area.
Explanation:This problem concerns the calculation of the center of mass for 2-dimensional figures. The center of mass represents the point at which we could balance the object in a uniform gravitational field and is calculated by integrating the mass distributed in the object.
First, consider that area mass density σ is constant. Thus, this is equivalent to finding the centroid of the shape. The location of the centroid C (x,y) is given by:
x = (A1x1 + A2x2)/(A1 + A2)
and
y = (A1y1 + A2y2)/(A1 + A2)
where Ax and Ay represent the area and centroid coordinate of the individual figure, respectively. For this problem, A1 is the whole rectangular plate and A2 is the cutout. The centroid of a rectangle is at the midpoint of its diagonals, so the x and y coordinates for A1 are (B/2, A/2) and for A2 are (D, D/2).
Then, the x-coordinate for the center of mass becomes, x = [(AB*(B/2) - CD*(D/2)]/(AB - CD) and y-coordinate for center of mass becomes, y = [(AB*(A/2) - CD*(C/2)]/(AB - CD) Calculate them to get numerical value.
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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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A playground merry-go-round has a radius of 3.0 m and arotational inertia of 600 kg ⋅ m2. It
is initially spinning at 0.80 rad/s when a 20-kg child crawlsfrom the center to the rim. When
the child reaches the rim the angular velocity of themerry-go-round is:_________
A) 0.61 rad/s
B) 0.73 rad/s
C) 0.80 rad/s
D) 0.89 rad/s
E) 1.1rad/s
Answer:
Final angular velocity, [tex]\omega_f=0.61\ rad/s[/tex]
Explanation:
Given that,
Radius of merry- go- round, r = 3 m
Inertia of the merry- go- round, [tex]I=600\ kg-m^2[/tex]
Angular speed, [tex]\omega=0.8\ rad/s[/tex]
Mass, m = 20 kg
Let I' is the new rotational inertia of the merry- go- round. Here, the angular momentum of the system remains conserved. So,
[tex]L_f=L_o[/tex]
[tex]I_f\omega_f=I_o\omega_o[/tex]
[tex]\omega_f=(\dfrac{I}{I+mr^2})\omega_o[/tex]
[tex]\omega_f=(\dfrac{600}{600+20\times 3^2})\times 0.8[/tex]
[tex]\omega_f=0.61\ rad/s[/tex]
So, the angular velocity of the merry-go-round is 0.61 rad/s. So, the correct option is (A). Hence, this is the required solution.
1. A pair of oppositely charged parallel plates is separated by 5.51 mm. A potential difference of 614 V exists between the plates. What is the strength of the electric field between the plates? The fundamental charge is 1.602 × 10?19 . Answer in units of V/m.
2. What is the magnitude of the force on an electron between the plates? Answer in units of N.
3. How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.7 mm from the positive plate? Answer in units of J.
Answer:
Part a)
[tex]E = 1.11 \times 10^5 N/C[/tex]
Part 2)
[tex]F = 1.78 \times 10^{-14} N[/tex]
Part 3)
[tex]W = 5 \times 10^{-17} J[/tex]
Explanation:
Part 1)
As we know that electric field and potential difference related to each other as
[tex]E = \frac{\Delta V}{x}[/tex]
so we will have
[tex]\Delta V = 614 V[/tex]
[tex]x = 5.51 mm[/tex]
so we have
[tex]E = \frac{614}{5.51 \times 10^{-3}}[/tex]
[tex]E = 1.11 \times 10^5 N/C[/tex]
Part 2)
Charge of an electron
[tex]e = 1.6 \times 10^{-19} C[/tex]
now force is given as
[tex]F = qE[/tex]
[tex]F = (1.6 \times 10^{-19})(1.11 \times 10^5)[/tex]
[tex]F = 1.78 \times 10^{-14} N[/tex]
Part 3)
Work done to move the electron
[tex]W = F.d[/tex]
[tex]W = (1.78 \times 10^{-14})(5.51 - 2.7) \times 10^{-3}[/tex]
[tex]W = 5 \times 10^{-17} J[/tex]
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.