Answer:
Conservation
Explanation:
She has observation conservation because If the temperature of the liquids stays constant and the container is insulated and not heat or cool the liquid much would not change the density of the liquid very much so that it's original volume could remain constant.
The interesting thing is not that the child assumes the taller glass holds more liquid but that they fail to understand conservation: the fact that the water from one glass is going to be the same amount after being poured into any other container. It's as if they did not realize the water came from the same glass.
The velocity of a sky diver t seconds after jumping is given by v(t) = 80(1 − e−0.2t). After how many seconds is the velocity 65 ft/s? (Round your answer to the nearest whole number.)
Answer:
8 seconds
Explanation:
Given:
The velocity of the sky diver 't' seconds after jumping is given as:
[tex]v(t)=80(1-e^{-0.2t})[/tex]
The velocity is given as, [tex]v=65\ ft/s[/tex]
So, in order to find the time required to reach the above given velocity, we plug in 65 for 'v' in the above equation and solve for time 't'. This gives,
[tex]65=80(1-e^{-0.2t})\\\\\frac{65}{80}=1-e^{-0.2t}\\\\0.8125=1-e^{-0.2t}\\\\e^{-0.2t}=1-0.8125\\\\\textrm{Taking natural log on both sides, we get:}\\\\-0.2t=\ln(0.1875)\\\\t=\frac{\ln(0.1875)}{-0.2}\\\\t=8.4\ s\approx 8\ s(Nearest\ whole\ number)[/tex]
Therefore, the time taken to reach a velocity of 65 ft/s is nearly 8 seconds.
The velocity of the skydiver is 65 ft/s after approximately 9 seconds. This is found by solving the provided velocity function for the given speed.
To determine after how many seconds the velocity of the skydiver is 65 ft/s, we need to solve the equation
[tex]v(t) = 80(1 - e^-^0^.^2^t).[/tex]
Given v(t) = 65, we set up the equation:
[tex]65 = 80(1 - e^-^0^.^2^t)[/tex]
First, isolate the exponential term:
[tex]65/80 = 1 - e^-^0^.^2^t\\0.8125 = 1 - e^-^0^.^2^t[/tex]
Subtract 1 from both sides:
[tex]-0.1875 = -e^-^0^.^2^t[/tex]
Divide by -1:
[tex]0.1875 = e^-^0^.^2^t[/tex]
Take the natural logarithm (ln) of both sides to solve for t:
[tex]ln(0.1875) = -0.2t[/tex]
Solve for t:
[tex]t = ln(0.1875) / -0.2[/tex]
Using a calculator, we get:
[tex]t = 8.6 seconds[/tex]
Rounding to the nearest whole number, the velocity is 65 ft/s after about 9 seconds.
Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to upper case or lower case, respectively, and then use the converted string in the comparison.1. True2. False
Answer:
True, check attachment for code
Explanation:
To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.
The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.
The third line sets of a second String variable called result.
The fourth line is where the conversion is done.
We can compare the string
We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.
Check attachment for the code
A(n) ? is a premises wiring system whose power is derived from a source of electric energy or equipment other than a service. Such systems have no direct connection from circuit conductors of one system to circuit conductors of another system, other than those established through bonding or grounding connections.
Answer: Separately derived system
Explanation: A separately derived system is used to describe a premise wiring system whose power is derived from a source of electrical energy such as transformer, solar photovoltaic cell or generator. A separately derived system has no direct connection to any conductor from another system or doesn't generate it's power from any direct connection to a conductor from another system or source except those from established from bonding or grounding connections. Separately derived systems usually generate it's power on it's own.
An isolated power system is isolated from other systems, using bonding or grounding for safety, and may include an isolation transformer. Electrical safety devices like circuit breakers or fuses prevent thermal hazards, and a three-wire system enhances both thermal and shock safety.
Explanation:The student's question refers to an "isolated power system" which is a type of electrical distribution system. An isolated power system is unique because it is separated from other power systems. It has no direct connection with the circuit conductors of another system, with the exception of connections through bonding or grounding. This premise of design is intended to enhance safety, and it usually involves the use of an isolation transformer to prevent shock. Moreover, it avoids a sudden increase in the voltage that could disrupt the power supply.
Electrical safety devices, such as circuit breakers and fuses, are critical in such systems to interrupt excessive currents and prevent thermal hazards. The three-wire system, which is essential for safety in modern household and industrial wiring, utilizes live/hot, neutral, and earth/ground wires. The neutral wire and the case of any connected appliance are both grounded, which means they are connected to the earth to ensure they exist at zero volts, providing an alternative return path for the current through the earth, thereby guarding against thermal and shock hazards.
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
If you weigh 685 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 kmkm ? Take the mass of the sun to be msmsm_s = 1.99×1030 kgkg , the gravitational constant to be GGG = 6.67×10−11 N⋅m2/kg2N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be ggg = 9.8 m/s2m/s2 .
Answer:
[tex] W = 5.94 \cdot 10^{15} N [/tex]
Explanation:
To calculate the weight on the surface of a neutron star we can use the following equation:
[tex] W = m*g [/tex]
Where:
W: is the weight of the person
m: is the mass of the person
g: is the gravity of the neutron star
Hence, first we need to find m and g. The mass is equal to:
[tex]m = \frac{W}{g} = \frac{685 N}{9.81 m/s^{2}} = 69.83 kg[/tex]
Now, the gravity of the neutron star can be found using the followig equation:
[tex]F = \frac{G*m*M}{r^{2}} = m*g \rightarrow g = \frac{G*M}{r^{2}}[/tex]
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ m³ kg⁻¹ s⁻²
M: is the mass of the neutron star = 1.99x10³⁰ kg
r : is the distance between the person and the surface of the neutron star = 25/2 = 12.5 km
[tex] g = \frac{6.67 \cdot 10^{-11} m^{3}kg^{-1}s^{-2}*1.99 \cdot 10^{30} kg}{(12.5 \cdot 10^{3} m)^{2}} = 8.50 \cdot 10^{13} m/s^{2} [/tex]
Now, we can find the weight on the surface of the neutron star:
[tex]W = m*g = 69.83 kg * 8.50 \cdot 10^{13} m/s^{2} = 5.94 \cdot 10^{15} N[/tex]
I hope it helps you!
Your weight on a neutron star with the same mass as the Sun and a diameter of 25.0 km would be approximately 5.95 × 10¹⁴ N. This is due to the extremely high gravitational acceleration on the neutron star's surface. Such high gravity results from the star's compactness and mass.
To determine your weight on the surface of a neutron star, we need to calculate the gravitational acceleration on its surface and then use this to find the weight force.
Step-by-Step Solution:
Calculate the gravitational acceleration, gns, at the surface of the neutron star using the formula for gravitational acceleration: g = G * M / R², where:G is the gravitational constant, 6.67 × 10⁻¹¹N⋅m²/kg²M is the mass of the neutron star, which is equal to the mass of the Sun, 1.99 × 10³⁰ kgR is the radius of the neutron star, which is half of its diameter, so R = 25.0 km / 2 = 12.5 km = 1.25 × 10⁴ mSubstitute these values into the formula:gns = (6.67 × 10⁻¹¹N⋅m²/kg²) * (1.99 × 10³⁰ kg) / (1.25 × 10⁴ m)² = 8.51 × 10¹² m/s²To find your weight, use the weight formula: Weight = mass × gravitational acceleration.Your mass (m) can be found from your weight on Earth: WeightEarth = m × g, so:m = WeightEarth / g = 685 N / 9.8 m/s² = 69.9 kgNow, calculate your weight on the neutron star:Weightns = m × gns = 69.9 kg × 8.51 × 1012 m/s² ≈ 5.95 × 10¹⁴NSummary: Your weight on a neutron star with the same mass as the Sun and a diameter of 25.0 km would be approximately 5.95 × 10¹⁴N, which is significantly more than your weight on Earth.
A 497−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?
Final answer:
To calculate the initial temperature of the copper piece, use the principle of energy conservation and the equations for heat gained and heat lost. The final temperature of the system is approximately 24.8 °C.
Explanation:
To calculate the initial temperature of the copper piece, we can use the principle of energy conservation. The heat gained by the water is equal to the heat lost by the copper. The heat gained by the water can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The heat lost by the copper can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat lost, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature. Since the final temperature is known, we can rearrange the formulas to solve for the initial temperature of the copper:
Initial temperature of copper = (heat gained by water / (m * c)) + final temperature
Substituting the given values into the formulas, we get:
Heat gained by water = (159 g) * (4.18 J/g·°C) * (39.9 °C - 22.8 °C)
Heat lost by copper = (497 g) * (0.387 J/g·°C) * (final temperature - 89.5 °C)
Setting the two equations equal, we can solve for the final temperature:
(159 g) * (4.18 J/g·°C) * (39.9 °C - 22.8 °C) = (497 g) * (0.387 J/g·°C) * (final temperature - 89.5 °C)
Solving the equation, the final temperature of the system is approximately 24.8 °C.
When two tuning forks are stuck simultaneously, 6 beats per second are heard. The frequency of one fork is 560 Hz. A piece of wax is placed on the 560 Hz fork to lower its frequency slightly. If the beat frequency is increased, what is the correct frequency for the second fork
Answer:
The correct frequenvy of the second fork is; ν2 = 566Hz
Explanation:
First of all, when two wave sources with slightly differing frequencies ν1 and ν2 generate waves at the same time and are superposed, then an interference effect will occur in time.
The intensity will be found to oscillate with time with a frequency (ν) called the beat frequency. It is depicted by:
ν = ± (ν1 −ν2).
In this question, there are two tuning forks, one with a frequency of let's say ν1=560Hz
The beat frequency is ν=6Hz
Therefore,
ν2=560 - 6 or ν2=560 + 6
i.e ν2=554 or ν2=566
For us to know the correct frequency, if the 560 Hz fork is loaded with wax, the increased inertia will lower its frequency.
Then it is found that the beat frequency increases. This can only mean that the other fork has a higher frequency. Hence the unkown frequency of the second fork must be,
ν2 = 566Hz
A current is established in a gas discharge tube when a sufficiently high potential difference is applied across the two electrodes in the tube. The gas ionizes; electrons move toward the positive terminal and singly charged positive ions move toward the negative terminal. What is the current in a hydrogen discharge tube in which 3.4 ✕ 1018 electrons and 1.4 ✕ 1018 protons move past a cross-sectional area of the tube each second? (Enter the magnitude.)
Explanation:
It is given that the number of electrons passing through the cross-sectional area in 1 s is [tex]3.4 \times 10^{18}[/tex]. Also, we know that charge on an electron is [tex]-1.60 \times 10^{-19} C[/tex], then negative charge crossing to the left per second is as follows.
I- = [tex]3.4 \times 10^{18} electrons \times -1.6 x 10^{-19} C/electrons[/tex]
I- = 0.544 A
As it is given that the number of protons crossing per second is [tex]1.4 \times 10^{18}[/tex], as the charge on the proton is [tex]+1.60 \times 10^{-19} C[/tex], then positive charge crossing to the right per second is calculated as follows.
I+ = [tex]1.4 \times 10^{18} electrons \times 1.6 \times 10^{-19} electrons/C[/tex]
I+ = 0.224 A
I = l I+ l + l I- l
So, I = 0.544 + 0.224
= 0.768 A
Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.
When charged particles are separated by an infinite dis- tance, the electric potential energy of the pair is zero. When 19. 7. the particles are brought close, the electric potential energy of a pair with the same sign is positive, whereas the electric potential energy of a pair with opposite signs is negative. Give a physical explanation of this statement.
Answer:
Explanation:
Potential energy Is given as.
U=kq1q2/r
Where k is a constant
q1 is charge
And q2 is also a charge
r is the distant between the two charges
When distance between the two charges is infinity then, the potential energy will b zero
U=kq1q2/∞
Therefore
U=0
Thus, the potential energy is zero when charged particles are separated by an infinite distance.
Let considered that both charges are positive.
Then U becomes
U=k(+q1)(+q2)/r
This shows that the energy is positive when the charges are positively charge. Also if the charges are also both negative charged the total energy will also be positive
U=k(-q1)(-q2)/r.
Both when their are opposite charges, the energy will be negative
U=k(-q1)(+q2)/r
U will be negative
A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 27.2. A potential difference of 5.0 V is maintained between the ends of the 2.0 m composite wire.
a.) What is the current in the copper section? b.) What is the current in the silver section?
b.) What is the magnitude of E in the copper section?
c.) What is the magnitude of E in the silver section?
d.) What is the potential difference between the ends of the silver section of the wire?
Using physics concepts of Ohm's law, resistance calculation, and electric field magnitude calculation, we can find the current in copper and silver sections, the electric field magnitudes, and the potential difference in silver section. The potential difference across the silver section will be the same as that applied across the entire wire due to copper's negligible resistance.
Explanation:The student's question is related to electricity and magnetism, a branch of physics. In the scenario described, the resistance (R) of each wire section can be found using the formula R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. Here, resistivity needs to be known from Table 27.2.
Once the resistance of each wire section is known, Ohm's law (V = IR) can be used to find the current (I) in each section. The same current will flow in both sections of the wire since it is a series circuit. The potential difference across each section can then be calculated using Ohm's law.
The magnitude of the electric field E in each section can be calculated from E = V/d, where V is the potential difference across the section and d is its length.
The potential difference across the silver section is the same as the potential difference applied across the entire wire because the copper wire has negligible resistance compared to silver.
Learn more about electricity and magnetism here:https://brainly.com/question/33523755
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A box with its contents has a total mass of 40 kg. It is dropped from a very high building. After reaching terminal speed, what is the magnitude of the air resistance force acting upward on the falling box
Answer:
The magnitude of air = 392N
Explanation:
We use Newton's 2nd law. The sum of the vertical forces must be equal to zero because at terminal speed , the acceleration is zero. Solving for the air resistance force,F(air ) gives:
EFvertical = mg - F(air)= ma
F(air) = mg = 40 × 9.8 = 392N
Answer: 392N
Explanation:
Newton's second law of motion states that "The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object."
the sum of vertical forces has to be equal to zero because by the time the terminal speed has been attained, the acceleration is zero. Now, we solve for air resistance force.
summation of F(vertical) = mg - F(air) = ma
a = 0 m/s²
thus, F(air) = mg
F(air) = 40kg*9.8m/s²
F(air) = 392N
A man on the 14 th floor of a building sees a bucket (dropped by a window washer) passing his window and notes that it hits the ground 1 second later. Assuming a floor is 4.9 meters high (and neglecting air friction), from what floor was the bucket dropped?
Answer:
The bucket was the dropped from 56 th floor.
Explanation:
Given that,
Height of floor = 4.9 m
Height of 14 floor = 68.6 m
Time taken = 1 sec
We need to calculate the speed of the bucket
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]68.6=v\times1+\dfrac{1}{2}\times9.8\times(1)^2[/tex]
[tex]v=68.6-\dfrac{1}{2}\times9.8\times(1)^2[/tex]
[tex]v=63.7\ m/s[/tex]
We need to calculate the time
Using equation of motion
[tex]v=u+gt[/tex]
[tex]t=\dfrac{v}{g}[/tex]
Put the value into the formula
[tex]t=\dfrac{63.7}{9.8}[/tex]
[tex]t=6.5\ sec[/tex]
We need to calculate the distance
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
[tex]s=0+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]s=\dfrac{1}{2}\times9.8\times(6.5)^2[/tex]
[tex]s=207.025\ m[/tex]
We need to calculate the number of floor
[tex]n=\dfrac{s}{h_{f}}[/tex]
Put the value into the formula
[tex]n=\dfrac{207.025}{4.9}[/tex]
[tex]n=42.25\approx42[/tex]
The bucket was the dropped from
[tex]f=14+42= 56[/tex]
Hence, The bucket was the dropped from 56 th floor.
energy conservation-problems 1. A slingshot fires a pebble from the top of a building at a speed of 14.0m/is. The building is 31.0m tall. Ignoring all frictional effects, find the speed with which the pebble strikes the ground
Answer:
Horizontal velocity is 14 m/s
Vertical velocity is 28.3 m/s
Explanation:
Hello dear friend, you have not mentioned the type of speed that you require for this problem.
In this case, there are three possibilities with which the slingshot can be fired i.e. Horizontally, Vertically straight up and vertically straight down. Below is the explanation / answer to all three possibilities
Fired horizontally:
initial conditions:
Vertical Velocity = 0 ; Horizontal Velocity = 14m/s
final conditions:
Vertical Velocity (v² = u² + 2gs) but initial vertical velocity is zero
v² = 2gs so v² = 2(9.8)(31) = 607
v = 24.6m/s
but Horizontal Velocity is still = 14m/s
Resultant velocity from these two velocity components (Pythagoras theorem)
V² = (v horizontal)² + (v vertical)² = 14² + 24.6²
V = 28.3m/s
angle = tan ⁻¹(24.3/14) = 60.1⁰
V = 28.3m/s at angle of 60.1⁰ to the horizontal
Fired Vertically Straight Up
distance before the pebble reaches maximum height from top of building
v² = u² + 2gs
where, v is zero at maximum height
g is minus for upward motion.
v² = u² + 2gs
0 = 14² - 2(9.8)s
s = 196/19.6 = 10.0m
totals distance from maximum height to the ground = 10.0 m + 31.0 m = 41.0m
v² = u² + 2gs
now u from maximum height is 0 and g is positive for downward motion
v² = 2gs
v² = 2(9.8)(41.0)
v = 28.3m/s
v = 28.3m/s vertically straight up
Fired Vertically Straight Down
v² = u² + 2gs
u = 14m/s, g = 9.8m/s², s = 31.0m
v² = 14² + 2(9.8)(31.0)
v = 28.3m/s
v = 28.3m/s vertically straight down
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.16 m/s. Suppose the collision is elastic.
Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you release the ball 1.4 mm above the ground? Solve this problem using energy.
Answer:
The initial velocity is 0.5114 m/s or 511.4 mm/s
Explanation:
Let the initial velocity be 'v'.
Given:
Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]
Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]
Final height of the ball (h₂) = 15 mm = 0.015 m
Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.
Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.
Change in kinetic energy is given as:
[tex]\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity[/tex]
As it just touches the 15 mm high roof, the final velocity will be zero. So,
[tex]v_f=0\ m/s[/tex].
Now, the change in kinetic energy is equal to:
[tex]\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2[/tex]
Change in gravitational potential energy = Final PE - Initial PE
So,
[tex]\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J[/tex] [ g = 9.8 m/s²]
Now, Change in KE = Change in PE
[tex]0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s[/tex]
Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s
A 25-kg iron block initially at 350oC is quenched in an insulated tank that contains 100 kg of water at 18oC. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process.
Answer:
The value of total entropy change during the process
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
Explanation:
mass of iron [tex]m_{iron}[/tex] = 25 kg
Initial temperature of iron [tex]T_{1}[/tex] = 350°c = 623 K
Mass of water [tex]m_{w}[/tex] = 100 kg
Initial temperature of water [tex]T_{2}[/tex] = 180°c = 453 K
When iron block is quenched inside the water the final temperature of both iron & water becomes equal. this is = [tex]T_{f}[/tex]
Thus heat lost by the iron block = heat gain by the water
⇒ [tex]m_{iron}[/tex] [tex]C_{iron}[/tex] ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = [tex]m_{w}[/tex] [tex]C_{w}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ 25 × 0.448 × ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = 100 × 4.2 × ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ [tex]( T_{1} - T_{f} ) = 37.5 ( T_{f} - T_{2} )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 ( T_{f} - 453 )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 T_{f} - 16987.5[/tex]
⇒ [tex]38.5 T_{f} = 17610.5[/tex]
⇒ [tex]T_{f} = 457.41 K[/tex]
This is the final temperature after quenching.
The total entropy change is given by,
[tex]dS = m_{iron}\ C_{iron} \ ln \frac{T_{f} }{T_{1} } + m_{w}\ C_{w} \ ln \frac{T_{f} }{T_{2} }[/tex]
Put all the values in above formula,
[tex]dS =[/tex] 25 × 0.448 × [tex]ln \frac{457.41}{623}[/tex] + 100 × 4.2 × [tex]ln \frac {457.41}{453}[/tex]
[tex]dS =[/tex] - 3.46 + 4.06
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
This is the value of total entropy change.
The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1, 2, 2) is 160°. (a) Find the rate of change of T at (1, 2, 2) in the direction toward the point (4, 1, 3). Incorrect: Your answer is incorrect. (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that
Answer:
The answers to the questions are as follows;
(a) The rate of change of T at (1, 2, 2) in the direction toward the point (4, 1, 3) is [tex]\frac{160\sqrt{11} }{33}[/tex]
(b) The direction of the gradient is in the direction of greatest increase and it is towards the origin.
Explanation:
To solve the question, we note that the shape of the ball is that of a sphere.
Therefore the distance of a point from the center is given by
f(x, y, z) = [tex]\sqrt{x^2+y^2+z^2}[/tex]
The temperature T in a metal ball is inversely proportional to the distance from the center of the ball
Therefore T ∝ [tex]\frac{1}{\sqrt{x^2+y^2+z^2}}[/tex] or T = [tex]\frac{C}{\sqrt{x^2+y^2+z^2}}[/tex]
Where
C = Constant of proportionality
x, y, and z are the x, y and z coordinates values
To find C, we note that at point (1, 2, 2), T = 160 °C.
Therefore 160 °C = [tex]\frac{C}{\sqrt{1^2+2^2+2^2}}[/tex] = [tex]\frac{C}{\sqrt{9}}[/tex] = [tex]\frac{C}{3}}[/tex]
Therefore C = 160 × 3 = 480 °C·(Unit length)
We therefore have the general equation as
T = [tex]\frac{480}{\sqrt{x^2+y^2+z^2}}[/tex]
The vector from points (1, 2, 2) to point (4, 1, 3) is given by
1·i + 2·j +2·k - (4·i + 1·j +3·k) = -3·i + j -k
From which we find the unit vector given by
u = [tex]\frac{1}{\sqrt{(-3)^2+1^2+(-)^2} } (-3, 1, -1)= \frac{1}{\sqrt{11} } (-3, 1, -1)[/tex]
From which we have the gradient equal to
∇T(x, y, z) = -480×(x²+y²+z²)[tex]^-{\frac{3}{2}}[/tex] in (x, y, z)
This gives D[tex]_u[/tex] = ∇T·u
= -480×(x²+y²+z²)[tex]^-{\frac{3}{2}}[/tex] in (x, y, z)·[tex]\frac{1}{\sqrt{11} } (-3, 1, -1)[/tex]
That is
[tex]-\frac{480}{\sqrt{11} }[/tex](x²+y²+z²)[tex]^-{\frac{3}{2}}[/tex] (-3·x + y - z)
From where D[tex]_u[/tex]Tat point (1, 2, 2) is = [tex]\frac{160\sqrt{11} }{33}[/tex]
(b) The direction of greatest increase in temperature is in the direction of the gradient and the direction of the gradient is opposite to the direction of {x, y, z}, which is away from the origin.
Hence the direction of the greatest increase in temperature is towards the origin.
What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.20 km on the moon some 384000 km away? Assume an average wavelength of 550 nm for the light received.
cm
Answer:
= 4.96cm
Explanation:
distance between objects of moon = 5.20km = 5.2 × 10³ m
Wavelength of light, λ = 550nm = 5.50 × 10⁻⁷m
distance of moon, L = 384000 km = 3.84 × 10⁸m
formula for resolving power of two objects
d = (1.22 × λ ×L) / D
D = (1.22 × λ ×L) / d
D = (1.22 × 5.50 × 10⁻⁷ ×3.84 × 10⁸) / 5.2 × 10³
D = 4.96cm
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is 4.00 atm, is opened, what will be the final pressure in the two bulbs
Answer:
[tex]P_{C} = 3.2\, atm[/tex]
Explanation:
Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:
Bulb A (2 L, 2 atm) - Before opening:
[tex]P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T[/tex]
Bulb B (3 L, 4 atm) - Before opening:
[tex]P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T[/tex]
Bulbs A & B (5 L) - After opening:
[tex]P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T[/tex]
After some algebraic manipulation, a formula for final pressure is derived:
[tex]P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}[/tex]
And final pressure is obtained:
[tex]P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}[/tex]
[tex]P_{C} = 3.2\, atm[/tex]
The astronomer who discovered the dwarf planet Eris suggests there might be another object far beyond the Kupier belt. If this Planet X exists, it would be about 10 times the mass of Earth and 2-3 times the size of Earth, putting it in the ice giant category, and have an orbit with a semi-major axis of 700 AU. You can read more about this object on NASA's page. If this object exists, what would we classify it as?
These objects would be classified as extreme trans Neptunian object (ETNO).
Explanation:
ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.
Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)
The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.
Planet X, if it exists, would likely be classified as a planet rather than a dwarf planet because it is much larger and more massive, suggesting it could clear its orbital neighborhood, which is one of the criteria that differentiate planets from dwarf planets.
The hypothetical object proposed by the astronomer beyond the Kuiper Belt, nicknamed Planet X, would be classified differently from Eris and other dwarf planets like Pluto, Makemake, and Haumea. Since it is speculated to be about 10 times the mass of Earth and 2-3 times its size, placing it within the ice giant category, it would resemble Uranus or Neptune rather than the smaller dwarf planets in the solar system.
Dwarf planets are generally smaller bodies that, while orbiting the Sun and having sufficient mass for their self-gravity to overcome rigid body forces, have not cleared their neighboring region of other objects. In contrast, Planet X, being significantly larger and more massive, would likely be considered a full-fledged planet if its existence were confirmed, primarily because of its mass, size, and potential to clear its orbit, aligning closer with the current criteria for a planet.
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar system. have the Sun at their exact center. are highly inclined to the ecliptic. are almost circular, with low eccentricities.
Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
[tex]T^{2} = a^{3}[/tex]
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
. A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall
Explanation:
Below is an attachment containing the solution.
An airplane needs to reach a velocity of 199.0 km/h to take off. On a 2000-m runway, what is the minimum acceleration necessary for the plane to take flight? Assume the plane begins at rest at one end of the runway.
Answer:
[tex]0.76m/s^2[/tex]
Explanation:
We are given that
Final velocity, v=199 km/h=[tex]199\times \frac{5}{18}=55.3m/s[/tex]
1km/h=[tex]\frac{5}{18}m/s[/tex]
Initial velocity, u=0
S=2000 m
We know that
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex](55.3)^2=2a(2000)[/tex]
[tex]a=\frac{(55.3)^2}{2\times 2000}[/tex]
[tex]a=0.76m/s^2[/tex]
Hence, the minimum acceleration necessary for the plane to take flight=[tex]0.76m/s^2[/tex]
A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the ball (about 0.15 s). If the football has a mass of 0.44 kg, what average force does the punter exert on the ball?
Answer:
Force, F = 44 N
Explanation:
Given that,
Initial speed of the football, u = 0
Final speed, v = 15 m/s
The time of contact of the ball, t = 0.15 s
The mass of football, m = 0.44 kg
We need to find the average force exerted on the ball. It is given by the formula as :
[tex]F=ma\\\\F=\dfrac{mv}{t}\\\\F=\dfrac{0.44\times 15}{0.15}\\\\F=44\ N[/tex]
So, the average force exerted on the ball is 44 N. Hence, this is the required solution.
An object of mass 0.77 kg is initially at rest. When a force acts on it for 2.9 ms it acquires a speed of 16.2 m/s. Find the magnitude (in N) of the average force acting on the object during the 2.9 ms time interval. (Enter a number.)
Answer: 4.3KN
Explanation:f=m(v-u)/t
m=0.77kg
t=0.0029s
s=16.2m/s
F= 0.77*16.2/0.0029
F=12.474/0.0029
F= 4301.38N
F=4.3KN
To find the magnitude of the average force acting on an object, one can use the derived form of the second law of motion, F = mΔv/Δt. Placing the given values into the equation, we calculate the force to be approximately 4307 N.
Explanation:To solve this question, we can use the equation F = mΔv/Δt which is derived from the second law of motion (Force = mass × acceleration), where F is the average force, m is the mass of the object, Δv is the change in velocity, and Δt is the time interval.
Substituting the given values, m = 0.77 kg, Δv = 16.2 m/s (the final velocity) - 0 m/s (the initial velocity) = 16.2 m/s and Δt = 2.9 ms = 2.9 × 10⁻³ s (as time should be converted to seconds).
Therefore, F = (0.77 kg × 16.2 m/s) / 2.9 × 10⁻³ s = 4306.9 N. Therefore, the magnitude of the average force acting on the object during the 2.9 ms time interval is approximately 4307 N.
Learn more about Force Calculation here:https://brainly.com/question/33703918
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Frequency of electromagnetic waves that a radio station is assigned
Answer:
Carrier Wave
Explanation:
A carrier wave is described or known as the continuous electromagnetic radiation, of constant amplitude and frequency, which is being given out or released by a transmitter. It is modulated in direct proportion to the signal,that is voice or music, which is meant to be transmitted or broadcasted.
It is mostly used for the transmission of information such as speech and music which can be seen in radio communication.
what will be the speed of a solid sphere of mass 2.0 kilograms and radius 15.0 centimeters when it reaches the bottom of a incline of length 5.0 meters. Assume the sphere starts from rest and rolls without slipping.
Explanation:
Below is an attachment containing the solution.
What happens to the direction of the magnetic field about an electric current when the direction of the current is reversed?
Answer:
The direction of the magnetic field is also reversed.
Explanation:
The direction of the magnetic field is also reversed when viewed form the same side if the direction of current is reversed. The direction of the magnetic field with respect to the direction of electric current is determined by the Maxwell's right-hand thumb rule.According to this rule we place our palm with the thumb pointing the direction of current flow and curling our finger in an action of gripping the wire. This position the the direction of curled fingers represents the direction of magnetic field.during a crash, an airbag inflates to stop a dummys forward motion. the dummy mass is 75 kg. if the net force is on the dummy is 825 N toward the rear of the car, what is the dumys deceleration
Answer:
11 m/s²
Explanation:
Deceleration: This can be defined as the rate of decrease of velocity. The S.I unit of deceleration is m/s².
From the question,
F = md ..................... Equation 1
Where F = Force acting on the dummy towards the rear of the car, m = mass of the dummy, d = deceleration of the dummy.
make d the subject of the equation
d = F/m............... Equation 2
Given: F = 825 N, m = 75 kg.
substitute into equation 2
d = 825/75
d = 11 m/s²
Answer:
11 m/s^2.
Explanation:
Given:
Mass = 75 kg
Force = 825 N
F = m × a
a = 825 ÷ 75
= 11 m/s^2.
Ball bearings can be made by letting spherical drops of molten metal fall inside a tall tower - called a shot tower - and solidify as they fall. If a bearing needs 4.0 s to solidify enough for impact, how high must the tower be?
Answer:
height of tower= 78.48 meters
Explanation:
So here are your givens:
time(t)= 4 s.
initial velocity(u) = 0 m/s
acceleration due to gravity (g)= 9.81 m/s^2
distance(s)=s meters
using one of Newton's equation of motion ;
[tex]s=ut+\frac{1}{2} gt^{2} \\s=0(4)+\frac{1}{2}(9.81)(4)^{2} \\s=\frac{1}{2} (9.81)(16)\\s=\frac{156.96}{2} \\s=78.48[/tex]
height of the tower from the ground=total distance covered by bearing
height of tower= 78.48 meters
Final answer:
The shot tower needs to be at least 78.4 meters tall for a ball bearing to solidify in 4.0 seconds given the acceleration due to gravity.
Explanation:
Calculating the Height of the Shot Tower
To determine how high the tower must be for a ball bearing to solidify in 4.0 seconds, we can use the kinematic equation for free-fall motion without initial velocity, which is h = 1/2gt², where h is the height, g is the acceleration due to gravity (9.8 m/s² on Earth), and t is the time in seconds. For t = 4.0 seconds, we get:
h = 1/2 * 9.8 m/s² * (4.0 s)²
h = 1/2 * 9.8 * 16
h = 4.9 * 16
h = 78.4 meters
Therefore, the shot tower must be at least 78.4 meters tall to allow a ball bearing to solidify in mid-air within 4.0 seconds before impact.