Answer:
Technicians A and B are both correct
Explanation:
A quick take-up valve in a quick take-up master cylinder assembly which reduces fluid flow noise and restriction caused by fluid flowing in one mode of operation through an incipiently or substantially closed valve spring coil stack. The check valve arrangement has a coil compression limiting device and also has the outlet orifice so positioned as not to require fluid flow through the check valve spring coils while the check valve is open.
If the brake warning light is NOT on and there are no visible brake fluid leaks, your master cylinder may be worn or leaking internally allowing the brake pedal to slowly sink when pressure is applied to it. This type of condition will be most noticeable when you are holding constant pressure against the brake pedal at a stop light. If the pedal sinks or requires pumping to keep your car from creeping ahead, the master cylinder needs to be replaced.
High energy environments are most likely to contain which one of the following?
a.large particles such as gravel
b.silt-sized particles
c.manganese nodules
d.cosmogenous sediments
e.clay-sized particles
Answer:
a.large particles such as gravel
Explanation:
A high energy environment is composed of the Aquas sedimentary environment and features the high energy of turbulent waves and currents in motion and contains the large parts of gravel that is an aggregation of rocks fragments and are formed as result of the weathering and erosion of rocks.High energy environments are typically associated with larger particles such as gravel due to the ability of these environments to move larger matter. Smaller particles are generally found in lower energy environments.
Explanation:High energy environments, such as fast-flowing rivers or coastal areas with strong waves, are most capable of moving larger particles. So, among these options, the environments with high energy are most likely to contain large particles such as gravel (a). The other particles listed (silt, manganese nodules, cosmogenous sediments, clay) require less energy to be moved, so they are likely to be found in lower energy environments.
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An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 8,000 g and an area of 5.00 cm2 and is free to slide up and down, keeping the pressure of the gas constant.
a. How much work is done on the gas as the temperature of 0.280 mol of the gas is raised from 15.0 degrees C to 27.0 degrees C?
b. What does the sign of your answer to part (a) indicate?
A. The surroundings do positive work on the gas.
B. The gas does positive work on its surroundings.
C. There is no work done by the gas of the surroundings.
Answer:
Explanation:
Given
mass of gas [tex]m=8000 gm=8 kg[/tex]
Area of Piston [tex]A=5 cm^2[/tex]
[tex]T_1=15^{\circ}C[/tex]
[tex]T_2=27^{\circ}C[/tex]
no of moles [tex]n=0.28[/tex]
Work done on the gas is given by
[tex]W=-P\Delta V[/tex]
[tex]P\Delta V[/tex] can also be written as [tex]nR\Delta T[/tex]
as PV=nRT
[tex]W=-nR\Delta T[/tex]
[tex]W=-0.28\times 8.314\times (27-15)[/tex]
[tex]W=-27.93 J[/tex]
negative sign indicates that work is done on the system i.e. surrounding done a positive work on the gas
A) A motorist traveling at 12 m/s encounters a deer in the road 41 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? B) If his or her reaction time is 1.1333s, how fast will (s) he be traveling when(s) he reaches the deer?
Answer:
A) Reaction Time = 2.73 seconds.B) He will not reach the deer.Explanation:
A) Initial speed, u = 12 m/s
Acceleration, a = -7 m/s²
Final velocity, v = 0 m/s
We have equation of motion v² = u² + 2as
0² = 12² + 2 x -7 x s
s = 10.29 m
He need 10.29 m to stop.
Remaining distance = 43 - 10.29 = 32.71 m
We have
Remaining distance = Reaction Time x Initial velocity
32.71 = Reaction Time x 12
Reaction Time = 2.73 seconds.
B) Distance traveled in 1.1333 s = 1.1333 x 12 = 13.60 m
Remaining distance = 41 - 13.6 = 27.4 m
Initial speed, u = 12 m/s
Acceleration, a = -7 m/s²
Displacement, s = 27.4 m
v² = 12² + 2 x -7 x 27.4
v² = -239.6
Not possible.
Motorist will not hit deer.
He will not reach the deer.
A soccer ball is kicked from point Pi at an angle above a horizontal field. The ball follows an ideal path before landing on the field at point Pf . Which of the following statements is true when the ball is at point X?
A. The horizontal velocity vector points to the right & equals v cos θ.
B. The vertical velocity points up & equals v sin θ.
C. The acceleration vector points up.
D. The total velocity equals the initial velocity v.
Answer:
A. The horizontal velocity vector points to the right & equals v cos θ.
Explanation:
The motion describes a parabolic path, where the horizontal speed is constant and the horizontal velocity vector always points to the right and equals v*cos θ.
Examine the five words and/or phrases and determine the relationship among the majority of words/phrases. Choose the one option that does not fit the pattern. A. abyssal clay B. calcareous ooze C. coarse lithogenic sediment D. manganese nodule E. siliceous ooze
Answer:
The Correct Option is C (coarse lithogenic sediment)
Explanation:
This is because all other options are found on sea floors, except for coarse lithogenic sediments that are as a result of erosion on land. Both calcareous and siliceous ooze are common sea sediments. Abyssal clay and manganese nodule are red clay and rock concretion respectively found in the bottom of the sea.
2. In the winter, weather reporters often day "It will be a very cold night because there are no clouds." a. Use the sim to see if you can understand why this could be true. b. Describe your observations. c. Would there be a difference between daytime and nighttime cloud effects
Answer:
a. Clouds at night prevent the heat radiation of the earth from escaping back into the space by continuously reflecting them back.
b. If during the day the earth has been heated well by the sun and then if the night is cloudy then that night is not so cold.
c. Day time clouds prevent the heating of the earth by reflecting back the radiations of the sun back into the space.
Explanation:
During the day if there is sunshine and then if the night is cloudy then that night will not be so cold because during the day the earth has absorbed the heat of sun's radiation which it radiates at night but when there is cloud at night it acts as a reflecting screen and reflects the heat radiations back to the earth.
But if there is cloud during the day time it prevents the radiations from the sun to fall on the earth by reflecting them back into the space preventing the rise in temperature of the earth.
how do the wavelengths of electromagnetic energy absorbed by materials on earth compare?
Explanation:
The shorter wavelength electromagnetic waves from sun are absorbed by earth material in form of short wavelength and the radiated wavelength are longer ones. Also higher energy waves are of shorter wavelength and lower energy waves have longer wavelength. So, they are absorbed as short wavelength and radiated back as long wavelength.
A ball is dropped from rest and falls to the floor. The initial gravitational potential energy of the ball-Earth-floor system is 10 J. The ball then bounces back up to a height where the gravitational potential energy is 7 J. What was the mechanical energy of the ball-Earth-floor system the instant the ball left the floor?
The mechanical energy of the ball-Earth-floor system at the instant when the ball left the floor is of 7 J.
What is Mechanical Energy?At any instant, the energy of an object by virtue of its motion or position is known as the mechanical energy of the object. In other words, mechanical energy is either the potential energy or the kinetic energy at any time instant.
Given data -
The initial gravitational potential energy is, U = 10 J.
The final gravitational potential energy is, U' = 7 J.
Now, considering the given case when the ball loses energy in the rebound due to ball deformation, heat loss from the bounce. The total mechanical energy is not conserved so mechanical energy as it bounces off the floor is equal to transformed potential energy at a maximum height of second bounce which is 7 J.
Thus, we can conclude that the mechanical energy of the ball-Earth-floor system at the instant when the ball left the floor is of 7 J.
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A stubborn 130 kg pig sits down and refuses to move. To drag the pig to the barn, the exasperated farmer ties a rope around the pig and pulls with his maximum force of 800 NN. The coefficients of friction between the pig and the ground are μs=0.80μs=0.80 and μk=0.50μk=0.50.
Answer:
Explanation:
Given
mass [tex]m=130 kg[/tex]
Force [tex]F=800 N[/tex]
coefficient of static friction [tex]\mu _s=0.8[/tex]
coefficient of static friction [tex]\mu _k=0.5[/tex]
maximum Static Friction will be given by
[tex]F_s=\mu _smg[/tex]
[tex]F_s=0.8\times 130\times 10=1040 N[/tex]
initially we need to provide a force of 1040 N to move the Pig . As soon Pig starts moving kinetic friction will come into play.
But maximum force is less than Maximum static friction force so it is impossible to move pig.
Answer:
C
Explanation:
The distance that the pig moves up will be less than the distance that the strongman pulls down.
Explanation: The strongman is further from the fulcrum than the pig is, so the distance that he pulls down will be greater than the distance that the pig moves up.
If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is A. 600 m/s2 B. 60 m/s2 C. 3 m/s2 D. 6 m/s2
Answer:
D. 6 m/s²
Explanation:
Acceleration is change in velocity over time.
a = Δv / Δt
a = (60 m/s − 0 m/s) / 10 s
a = 6 m/s²
The acceleration of the car is 6m/s².
To find the correct option among all the options, we need to know about the acceleration.
What is acceleration?Acceleration is the rate of change of velocity with respect to time. It's given as a= ∆V/∆t.What is the acceleration for ∆V and and ∆t are 60 m/s and 10s respectively?Acceleration = ∆V/∆t = 60/10= 6 m/s².
Thus, we can conclude that the option (D) is correct.
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Archaeologists find an object that is known to have been created 18,000 years ago. Measurements indicate that 1000 atoms of 14C are present in the object. How many atoms of 14C were present when the object was made?
Final answer:
About 8,960 atoms of carbon-14 were present in the object when it was made 18,000 years ago, based on its current carbon-14 content and the known half-life of carbon-14.
Explanation:
To calculate how many atoms of carbon-14 (14C) were present in an object that was created 18,000 years ago, we use the half-life of 14C, which is 5,730 years. Given that there are 1,000 atoms of 14C present now, we need to find out how many half-lives have passed to work backwards and determine the initial quantity of 14C atoms.
First, we divide the age of the object by the half-life of 14C:
18,000 years / 5,730 years per half-life ≈ 3.14 half-lives
Now, we apply the formula for exponential decay, taking into account the number of half-lives:
Initial Quantity = Present Quantity × (2⁰ of half-lives)
Initial Quantity = 1000 atoms × (2≈ 3.14)
Initial Quantity ≈ 1000 atoms × 8.96
Initial Quantity ≈ 8960 atoms of 14C were present when the object was made.
You must dim your high beams for oncoming vehicles by the time they are within 500 feet of your vehicle.
Answer:
True
Explanation:
when oncoming vehicle driver is at a distance of 500 feet from you, then it is advisable to dim your high beam light so that it cannot blind oncoming vehicle driver. However, of the oncoming driver is at 200 to 300 ft then you must use low beam light so to pass the oncoming driver safely from your side. Hence the statement is true.
the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous athanol is -168.6 kj/mol. calculate the vapour pressure of ehtanol at
Answer:
P=133.71mmHg
Explanation:
the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous ethanol is -168.6 kj/mol. calculate the vapour pressure of ethanol at
assumption:
is that temperature is at 25C, at standard pressure of 1bar(750mmHg)
ethanol is an ideal gas
The free energy of ethanol liquid does not vary with pressure,
C2H5OH(l)⟶C2H5OH(g)
free energy of formation on the reactant side is -174.9 kj/mol
fro the product side is -168.6 kj
∅Gvap-∅G(l)=-168.6kj/mol-(-174.9kj/mol)
+6.3kj/mol
∅G=∅Gvap+RTlnK-∅Gliq
∅G=0
0=+6.3kj/mol+8.314Jk/mol/k(298)InK
-6.3/(RT)=Lnk
taking the exponential of both sides
[tex]e^{-6300/(8.314*298)} =K[/tex]
0.178=k
k=p/[tex]p^{0}[/tex]
P^0=refers to the pressure of ethanol vapour at its standard state
partial pressure , which is 750 mmHg
P=0.178*750
P=133.71mmHg
Final answer:
The vapor pressure of ethanol at a given temperature can be calculated using the Clausius-Clapeyron equation.
Explanation:
The vapor pressure of a substance is related to its standard free energy of formation and temperature. To calculate the vapor pressure of ethanol at a given temperature, we can use the Clausius-Clapeyron equation:
ln(P₂/P₁) = △Hvap/R × (1/T₁ - 1/T₂)
Where P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively, △Hvap is the enthalpy of vaporization, R is the gas constant, and ln denotes the natural logarithm.
By substituting the given values for △Hvap = -174/9 kJ/mol, T₁ = 20.0 °C (293 K), and T₂ = desired temperature, we can solve for P₂.
When the following equation is balanced, the sum of all the coefficients is ____________. CO + NO → CO2 + N2.
Answer:
7
Explanation:
The balanced reaction is as follows:
[tex]2CO+2NO \rightarrow 2CO_2 + N_2[/tex]
Coefficient of CO = 2
Coefficient of NO = 2
Coefficient of CO2 = 2
Coefficient of N2 = 1
Sum of all coefficient = 2 + 2+ 2+ 1
= 7
You are in an airplane going down a runway currently going 24 m/s and accelerating at 8m/s for 800 meters until take off. How long does it take the plane to go those 800 meters?
Answer:
Time taken to cover the distance 11.45 s.
Explanation:
The given parameters:
Initial Velocity(u)=24 m/s
Acceleration (a)=8 [tex]m/s^{2}[/tex]
Distance or Displacement(s)=800 m
Displacement or Distance is equal in the above because it travels in a straight line.
We have to apply Second Equation of Motion,
s=ut+[tex]\frac{1}{2} at^{2}[/tex]
800=24t + 4[tex]t^{2}[/tex]
200=6t + [tex]t^{2}[/tex]
[tex]t^{2}[/tex] + 6t - 200=0
Solving, the quadratic equation to find out the roots, we get that the possible values of t will be 11.45 s and a negative value.
The negative value will be neglected as time cannot be negative.
Hence, the time taken is 11.45 s.
A large building has an inclined roof. The length of the roof is 54.0 m and the angle of the roof is 17.0° below horizontal. A worker on the roof goes of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 2.87 m/s2. After leaving the edge of the roof, it falls a vertical distance of 46.5 m before hitting the ground.
a. How much time does it take the hammer to fall from the edge of the roof to the ground?
b. How far horizontally does the hammer travel from the edge of the roof until it hits the ground?
Answer:
a) [tex]t=2.6\ s[/tex]
b) [tex]s=43.7747\ m[/tex]
Explanation:
Given:
length of inclined roof, [tex]l=54\ m[/tex]
inclination of roof below horizontal, [tex]\theta=17^{\circ}C[/tex]
acceleration of hammer on the roof, [tex]a_r=2.87\ m.s^{-2}[/tex]
height from the lower edge of the roof, [tex]h=46.5\ m[/tex]
Now, we find the final velocity when leaving the edge of the roof:
Using the equation of motion:
[tex]v^2=u^2+2.a_r.l[/tex]
[tex]v^2=0^2+2\times 2.87\times 54[/tex]
[tex]v=17.6057\ m.s^{-1}[/tex]
The direction of this velocity is 17° below the horizontal.
∴Vertical component of velocity:
[tex]v_y=v.sin\ \theta[/tex]
[tex]v_y=17.6057\times sin\ 17^{\circ}[/tex]
[tex]v_y=5.1474\ m.s^{-1}[/tex]
a.
So, the time taken to fall on the ground:
[tex]h=ut+\frac{1}{2} g.t^2[/tex]
here:
initial velocity, [tex]u=v_y=5.1474\ m.s^{-1}[/tex]
putting respective values
[tex]46.5=5.1474\times t+0.5\times 9.8\times t^2[/tex]
[tex]t=2.6\ s[/tex]
b.
Horizontal component of velocity, [tex]v_x=v.cos\ \theta=17.6057\ cos\ 17^{\circ}=16.8364\ m.s^{-1}[/tex]
Since there is no air resistance so the horizontal velocity component remains constant.
∴Horizontal distance from the edge of the roof where the hammer falls is given by:
[tex]s=v_x.t[/tex]
[tex]s=16.8364\times 2.6[/tex]
[tex]s=43.7747\ m[/tex]
Using equations of motion, it's calculated that it takes about 3.07 seconds for the hammer to fall to the ground from the roof. The hammer also travels approximately 89.2 meters horizontally from the roof to the ground.
Explanation:The question is related to physics concepts of motion under gravity and kinematics. For simplicity, let's ignore air resistance.
Part a:The time it takes for the hammer to fall to the ground from the edge of the roof can be calculated using the equation of motion: h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s²), and t is the time. Solving the equation for time (t): t = sqrt(2h/g). Substituting the given values, we get t = sqrt((2*46.5)/9.8) ≈ 3.07 s.
Part b:The horizontal distance travelled by the hammer can be calculated using the formula: distance = speed × time. The horizontal speed of the hammer when it falls off the roof will be the same speed it had just as it left the roof due to the roof slope acting on it with constant acceleration. This can be gotten from the equation v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (2.87 m/s²) and t is the time. The time here is the time it takes for the hammer to slide down the roof, gotten by time = distance/speed = 54.0/v. Solving all these gives a distance of approximately 89.2 m.
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With each bounce off the floor, a tennis ball loses 11% of its mechanical energy due to friction. When the ball is released from a height of 3.4 m above the floor, what height will it reach after the third bounce?
240 mm
240 cm
24 cm
270 cm
Answer:
240 cm
Explanation:
Gradpoint
"According to the Navigation Rules, what factor should be considered in determining a safe speed?
Answer:
Visibility conditions.
Explanation:
Safe speed is a speed at which the operator of the boat can take effective action to avoid stops within the distance. To calculate the safe speed visibility conditions (fog, rain, mist, and darkness) should be included. We should also include some other factors given below:
i) Traffic density
ii) Type of vessels in the area
iii) Wind
iv) Water conditions
The Navigation Rules dictate that a safe speed is determined by factors such as visibility, traffic density, vessel maneuverability, weather conditions, and proximity to navigational hazards.
Explanation:According to the Navigation Rules, several factors should be taken into account in determining a safe speed. These include the visibility, the traffic density, the maneuverability of the vessel in immediate circumstances, the state of wind, sea, and current, and the proximity of navigational hazards. For example, on a clear day with sparse traffic and calm seas, higher speeds might be safe. However, in heavy traffic or poor visibility, the navigation rules would call for lower speeds to ensure safety.
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Laser light of wavelength lambda passes through a thin slit of width a and produces its first dark fringes at angles of +/- 30 degree with the original direction of the beam. The slit is then reduced in size to a circle of diameter a. When the same laser light is passed through the circle, the first dark fringe occurs at
A) +/- 66.9 degree.
B) +/- 45.0 degree.
C) +/- 37.6 degree.
D) +/- 36.6 degree.
E) +/- 15.0 degree.
Answer:
C- ±37.6°
Explanation:
This is the case of a single slit diffraction. In this case, the dark fringe occurs at
sin θ = ± mλ/a
where m = 1
θ = ± 30°
Hence we have sin ± 30° = ± λ/a
±0.5 = ± λ/a
Hence, we have a = 2λ
For a circular aperture, the condition for the first dark fringe is
D = diameter of circle = a = 2λ
so we have sin θ1 = 1.22λ/a
sin θ1 = 1.22λ/2λ
hence sin θ1 = 0.61
θ1 = sin⁻¹0.61 = ±37.6°
According to the second law of thermodynamics, the disorder of a system always decreases. 1. True 2. False
Answer:
2. False
Explanation:
Entropy can be interpreted as a measure of the random distribution of a system. According to the second law of thermodynamics a system in an unlikely condition will have a natural tendency to reorganize itself to a more probable condition, this reorganization will result in an increase in entropy (disorder).
The following is a limitation of a bar chart schedule. a. Scheduling information is presented at a summary level b. activity interrelations are not considered c. ease of preparation d. ease of interpretation e. None of the above
Answer:
b. activity interrelations are not considered
Explanation:
A bar chart schedule or Gantt chart is used to visualize task that scheduled over time.
Bird bones have air pockets in them to reduce their weight–this also gives them an average density significantly less than that of the bones of other animals. suppose an ornithologist weighs a bird bone in air and in water and finds its mass is 43.0 g and its apparent mass when submerged is 3.60 g (the bone is watertight).a. what mass of water is displaced? b. what is the volume of the bone? c. what is its average density?
Answer:
39.4 g
39.4 cm³
1.09137 g/cm³
Explanation:
[tex]\rho[/tex] = Density of water = 1 g/cm³
Mass of water displaced will be the difference of the
[tex]m=43-3.6\\\Rightarrow m=39.4\ g[/tex]
Mass of water displaced is 39.4 g
Density is given by
[tex]\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{39.4}{1}\\\Rightarrow v=39.4\ cm^3[/tex]
So, volume of bone is 39.4 cm³
Average density of the bird is given by
[tex]\rho=\dfrac{43}{39.4}\\\Rightarrow \rho=1.09137\ g/cm^3[/tex]
The average density is 1.09137 g/cm³
A 2.0-g piece of Mg reacts with HNO3. Which conditions would produce the GREATEST reaction rate
0.5 M HNO3 at 20 C
0.5 M HNO3 at 40
1.0 M HNO3 at 20
1.0 M HNO3 at 40
Answer:
1.0 M HNO3 at 40°C
Explanation:
Rate of chemical reaction: This can be defined as the number of moles of reactant, converted or product formed per unit time.
Factors that affect rate of chemical reaction:
(a) Temperature: Generally, an increase in temperature increase the rate of chemical reaction by (1) increasing the number of particles with energy equal to or greater than the activation energy, (2) Increasing the average speed of all the reactant particles, due to greater kinetic energy, leading to higher frequency of collision.
(b) Concentration: An increase or decrease in the concentration of the reactant will result to a corresponding increase or decrease in the effective collision of the reactant and hence in the reaction rate.
other factors that affect the rate of chemical reaction are
(i) Nature of the reactant
(ii) Surface area of reactant
(iii) presence of light
(iv) presence of catalyst.
From the question above,
The condition with the highest temperature and concentration will produce the GREATEST reaction rate.
And that is 1.0 M HNO3 at 40 °C
What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.A) This low value is possible because the density of Saturn is so low.B) This low value is possible because the magnetic field of Saturn is so weak.C) This low value is possible because the magnetic field of Saturn is so strong.D) This low value is possible because the density of Saturn is so high.
Answer:
Approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex] at that distance from the center of the planet.
Option A) The low value of [tex]g[/tex] near the cloud top of Saturn is possible because of the low density of the planet.
Explanation:
The value of [tex]g[/tex] on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:
[tex]\displaystyle \frac{G \cdot M \cdot m}{R^2}[/tex],
where
[tex]G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2[/tex].[tex]M[/tex] is the mass of the planet, and[tex]m[/tex] is the mass of the object.To find an equation for [tex]g[/tex], divide the equation for gravity by the mass of the object:
[tex]\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}[/tex].
In this case,
[tex]M = 5.68\times 10^{26}\; \rm kg[/tex], and[tex]R = 5.82 \times 10^7\; \rm m[/tex].Calculate [tex]g[/tex] based on these values:
[tex]\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}[/tex].
Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.
Refer to the equation for [tex]g[/tex]:
[tex]\displaystyle g = \frac{G \cdot M}{R^2}[/tex].
The mass of the planet is in the numerator. If two planets are of the same size, [tex]g[/tex] would be greater at the surface of the more massive planet.
On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since [tex]R[/tex] is in the denominator of [tex]g[/tex], increasing the value of [tex]R[/tex] while keeping [tex]M[/tex] constant would reduce the value of [tex]g[/tex]. That explains why the value of [tex]g[/tex] near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately [tex]9.81\; \rm N \cdot kg^{-1}[/tex].
As a side note, [tex]5.82\times 10^7\rm \; m[/tex] likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of [tex]g[/tex] near the cloud tops of Saturn is approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex].
concept is gravitational force. Saturn's low value of g on its surface is possible because of its low average density, which is less than the density of water.
The value of g on the surface of Saturn is determined by its mass and radius. The low value of g on Saturn is possible because of its low average density. Saturn's mass is much larger than Earth's, but its density is much lower, resulting in a lower value of g on its surface.
The low density of Saturn, which is only 0.7 g/cm³, is less than the density of water. This means that Saturn would be light enough to float if placed in water. Therefore, despite its large mass, the low density of Saturn allows for a low value of g on its surface.
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A network engineer is subnetting the 10.0.240.0/20 network into smaller subnets. Each new subnet will contain between a minimum of 20 hosts and a maximum of 30 hosts. Which subnet mask will meet these requirements?
a. 255.255.224.0
b. 255.255.240.0
c. 255.255.255.224
d. 255.255.255.240
Answer:255.255.255.224
Explanation:
/20 has a subnet mask of 255.255.240.0
We have 20 on bits for the network and 16 off bits for the host. Since the designer wants 20 to 30 hosts in each subnet we will be having 5 off bits for the host. Therefore we will be having a /27..
Which will be 255.255.255.224
Final answer:
The suitable subnet mask for creating subnets with 20-30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27, which allows for 32 IP addresses, 30 of which can be used for hosts.
Explanation:
The correct subnet mask to meet the requirement for new subnets that support between a minimum of 20 hosts and a maximum of 30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27. This subnet mask allows for 32 IP addresses per subnet, with 30 usable addresses for hosts when excluding the network and broadcast addresses. Subnet masks 255.255.224.0 and 255.255.240.0 provide too many hosts per subnet, and subnet mask 255.255.255.240 or /28 provides only 16 IP addresses, which would be insufficient for the required minimum of 20 hosts.
A certain radioactive isotope placed near a Geiger counter registers 120 counts per minute. If the half-life of the isotope is one day, what will the count rate be at the end of four days?
Answer:
The count rate at the end of four days will be 7.5 counts per minute.
Explanation:
First it is important to know that half-life is the time to a piece of radioactive material to decay 50%. So if we know we start with 120 counts per minute and we already know the half-life of the isotope is 1 hour we expect that past 1 hour the material decays 50% (it's halved) so we will count 60 counts per minute, now if we wait another hour 60 counts will decay in to 30 counts per minute and so on. That should be translate to a math equation as:
final counts = initial counts * [tex](\frac{1}{2})^{\#half\,life\,periods}[/tex]
After 4 days we have 4 half-life periods passed so:
final counts= 120 counts per minute * [tex](\frac{1}{2})^{4}[/tex]
final material = 7.5 counts per minute
True or False. The spin quantum number (ms) describes the orientation of the spin of the electron. The magnetic quantum number (ml) describes the the size and energy associated with an orbital. An orbital is the path that an electron follows during its movement in an atom. The angular momentum quantum number (l) describes the orientation of the orbital. The principal quantum number (n) describes the shape of an orbital.
The spin quantum number (ms) describes the orientation of the spin of the electron: TRUE
The magnetic quantum number (ml) describes the size and energy associated with an orbital. An orbital is the path that an electron follows during its movement in an atom: FALSE
The angular momentum quantum number (l) describes the orientation of the orbital: FALSE
The principal quantum number (n) describes the shape of an orbital: FALSE
Explanation:
The magnetic quantum number (ml) - The number of orbitals and the orientation within a subshell is determined. The orbital angular momentum quantum number (l) - The shape of an orbital is determined.The principal quantum number (n) - The energy of an electron and the distance of the electron from the nucleus is described.The statement contains mixed truths and falsehoods. The spin quantum number describes the orientation of electron spin, the magnetic quantum number pertains to orbital orientation, and the angular momentum quantum number concerns orbital shape, while the principal quantum number describes orbital size and energy.
Explanation:The statement is false. The spin quantum number (ms) does describe the orientation of the spin of the electron, either up or down. The magnetic quantum number (ml) describes the orientation of the orbital in space. An orbital is defined as a region in space where there's a high probability of finding an electron, not their path. The angular momentum quantum number (l) determines the shape of the orbital. The principal quantum number (n) describes the size and energy level of an orbital.
Learn more about Quantum Numbers here:https://brainly.com/question/35385397
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With each bounce off the floor, a tennis ball loses 23% of its mechanical energy due to friction. When the ball is released from a height of 2.5 m above the floor, what height will it reach after the third bounce?
110 mm
110 cm
11 cm
150 cm
Answer:
110 cm
Explanation:
Gradpoint
A sample of radiosodium () has a half-life of 15 hr. If the sampleâs activity is 100 millicuries after 24 hr, approximately what must its original activity have been?
Answer : The original activity will be, 303 millicuries.
Explanation :
Half-life = 15 hr
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{15hr}[/tex]
[tex]k=4.62\times 10^{-2}\text{ hr}^{-1}[/tex]
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]4.62\times 10^{-2}\text{ hr}^{-1}[/tex]
t = time passed by the sample = 24 hr
a = initial amount of the reactant = ?
a - x = amount left after decay process = 100 millicuries
Now put all the given values in above equation, we get
[tex]24=\frac{2.303}{4.62\times 10^{-2}}\log\frac{a}{100}[/tex]
[tex]a=302.97\text{ millicuries}\approx 303\text{ millicuries}[/tex]
Therefore, the original activity will be, 303 millicuries.
Old Isaac took a little nosedive from his perch on top of the building, 25 feet above ground. Given that he fell as a result of a gentle tap to his noggin (initial velocity was zero), how fast is Isaac traveling when he hits the ground? (In case you forgot, acceleration due to gravity is -32 ft/sec2 .) Make sure your answer is in feet/sec. Leave your reply under the brick beside your car.
Answer:
40 ft/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = -32 ft/s² = a (downward is taken a negative)
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times -25+0^2}\\\Rightarrow v=40\ ft/s[/tex]
The speed of Isaac as he reached the ground is 40 ft/s
"The final velocity of Isaac when he hits the ground can be calculated using the kinematic equation for an object in free fall under the influence of gravity with an initial velocity of zero:
[tex]\[ v_f^2 = v_i^2 + 2a\Delta y \][/tex]
Plugging in the known values:
[tex]\[ v_f^2 = 0^2 + 2(-32 \text{ ft/sec}^2)(25 \text{ ft}) \] \[ v_f^2 = 0 + (-2)(32)(25) \] \[ v_f^2 = -1600 \][/tex]
Since velocity cannot be negative, we take the positive square root of the absolute value of [tex]\( v_f^2 \)[/tex] to find [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \sqrt{1600} \] \[ v_f = 40 \text{ ft/sec} \][/tex]
Therefore, Isaac is traveling at 40 feet per second when he hits the ground.
The answer is: [tex]40 \text{ ft/sec}.[/tex]