Answer:
Option C is the correct answer.
Explanation:
Heat required to melt solid in to liquid is calculated using the formula
H = mL, where m is the mass and L is the latent heat of fusion.
Latent heat of fusion for water = 333.55 J/g
Mass of ice = 0.3 kg = 300 g
Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature
H = mL = 300 x 333.55 = 100,375 J
Option C is the correct answer.
A mass of 100 kg is pulled by a 392 N force in the +X direction along a rough surface (uk=0.4) with uniform velocity v=20 m/s. What is the TOTAL work done by ALL forces on the object after s=10 m?
Answer:
The total work done will be zero.
Explanation:
Given that,
Mass = 100 kg
Force = 392 N
Velocity = 20 m/s
Distance s= 10 m
We need to calculate the work done
Using balance equation
The net force will be
[tex]F'=F-\mu mg[/tex]
[tex]F'=392-0.4\times100\times9.8[/tex]
[tex]F'=0[/tex]
The net force is zero.
Hence, The total work done will be zero by all forces on the object.
The sound intensity at the ear of passenger in a car with a damaged muffler is 8.0 × 10-3 W/m2. What is the intensity level of this sound in decibels? Use the threshold of hearing (1.0 × 10-12 W/m2) as the refere
Answer:
99 dB
Explanation:
We have given that sound intensity with a damaged muffler =8× [tex]10^{-3}[/tex]
we have to find the intensity level of the this sound in decibels
for calculating in decibels we have to use the formula
β=[tex]10log\frac{I}{I_0}[/tex]
=[tex]10log\frac{.008}{10^{-12}}[/tex]
=10 log8+10 log[tex]10^{9}[/tex]
=10 log8+90 log10
=10×0.9030+90
=99 dB
Final answer:
To calculate the intensity level of a sound in decibels, use the formula β = 10 log10(I/I0). Given the intensity of [tex]8.0 \times 10^{-3[/tex] threshold of hearing of [tex]10^{-12[/tex] the sound intensity level is approximately 99 decibels.
Explanation:
The student asks about calculating the intensity level of sound in decibels, given the sound intensity at the ear of a passenger in a car with a damaged muffler (8.0 × 10-3 W/m2). To find the intensity level in decibels (dB), we use the formula:
β = 10 log10(I/I0)
where:
β is the sound intensity level in decibels (dB)I is the sound intensity in watts per meter squared (W/m2)I0 = 10-12 W/m2 is the reference intensity, which is the threshold of hearingPlugging in the values:
β = 10 log10(8.0 × 10-3 / 10-12)
β = 10 log10(8.0 × 109)
β = 10 (log10(8) + log10(109))
β = 10 (0.903 + 9)
β = 10 × 9.903
β = 99.03 dB
The intensity level of the sound in the car with the damaged muffler is approximately 99 decibels.
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in parallel with the same battery, the current is 1.60 A. Part A What are the values of the two resistors?
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
[tex]\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2[/tex]
In parallel
[tex]\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}[/tex]
[tex]\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0[/tex]
Solving the above quadratic equation
[tex]\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}[/tex]
[tex]\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega[/tex]
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
The two resistors in question, when wired in a series, have a combined resistance of about 36.36 Ohms. When wired in parallel, they share a resistance of roughly 7.5 Ohms. The two resistor values would then most likely be around 28.86 Ohms and 7.5 Ohms.
Explanation:The nature of your question indicates a focus on the properties and calculations associated with electrical resistors in a circuit. These resistors can either be configured in a series or parallel connection, which will drastically change their behavior and derived readings.
From the context of your question, it is evident that a 12V battery is being used in conjunction with two resistors. In a series connection, the sum of the voltages across each resistor will equal the total voltage, effectively segmenting the 12V battery's output. In a parallel connection, each resistor would experience the full 12V impact, leading to larger current readings. Your values indicate a 0.33A current in a series scenario and a 1.60A current in a parallel situation.
Knowing this, we can apply Ohm's Law, which states Voltage (V) equals Current (I) times Resistance (R). In the series connection, the total resistance can be calculated as 12V divided by 0.33A equals approximately 36.36 Ohms. For the parallel connection, the total resistance would be 12V divided by 1.60A equals approximately 7.5 Ohms. Subtracting these two values, we can find that one of the resistors is around 28.86 Ohms while the other is roughly 7.5 Ohms. This would satisfy both the series and parallel conditions outlined in your question.
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the pressure and absolute temperature of an ideal gas are both tripled, the volume is __________. not changed
increased nine times
decreased to one ninth
decreased to one third
Answer:
Not Changed
Explanation:
To know what happened with the volume you need to know the Ideal gas Law
[tex]\frac{P_{1}V_{1}}{T1} =\frac{P_{2}V_{2} }{T2}[/tex]
This law is a combination of the other four laws: Boyles's, Charles's, Avogadro's, and Guy-Lussac's.
The initial state is represented by P1, V1, T1 and the final by P2, V2, T2.
In this case:
[tex]T_{2} =3T_{1} \\P_{2} =3P_{1}[/tex]
Replacing on the equation
[tex]\frac{P_{1}V_{1}}{T1} =\frac{3P_{1}V_{2}}{3T_{1} }[/tex]
If we clear from the equation V2
[tex]\frac{P_{1}V_{1}3T_{1}}{T_{1} 3P_{1}} ={V_{2}}[/tex]
Then cancel both P1 and T1
[tex]\frac{3V_{1}}{3} =V_{2}[/tex]
You will found that
[tex]V_{1} =V_{2}[/tex]
A large in-falling fragment could be tracked using radar. Explain how distance, speed, and the direction of motion, of the fragment could be determined. (15 points)
Answer:
speed, distance and direction of motion of the object can be determined by analyzing the radio wave.
Explanation:
We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.
Distance can be assessed by taking following consideration, the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.
Therefore distance can be determine as D= v*t/2,
here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)
Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by
[tex]v= (\Delta f/f)(c/2),[/tex]
where[tex]\Delta f[/tex]is the change in frequency and
c is the speed of light (the wave).
Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:
When the frequency transition is very low, the object moves away from the radar and vice verse.
Be sure to answer all parts. Convert: (a) 29.8°C (the melting point of gallium) to degrees Fahrenheit °F (b) 172.9°F (the boiling point of ethanol) to degrees Celsius °C (c) 4.2 K ,the boiling point of helium, to degrees Celsius °C
Answer:
(a) 85.64°F
(b) 78.28°C
(c) - 268.8°C
Explanation:
(a) 29.8°C into °F
To convert °C into °F, we use the formula given below
C / 100 = (F - 32) / 180
C / 5 = (F - 32) / 9
9 x 29.8 / 5 = F - 32
F = 85.64°F
(b) 172.9°F into °C
C / 5 = (F - 32) / 9
C = (172.9 - 32) x 5 / 9
C = 78.28°C
(c) 4.2 K into °C
C = K - 273
C = 4.2 - 273 = - 268.8°C
In a head-on collision, an alpha particle (Z = 2) of energy 8.80 MeV bounces straight back from a nucleus of charge 82.0 e. How close were the centers of the objects at closest approach?
The close were the centers of the objects at closest approach 2.7 x 10^-14 m
E =8.8 MeV = 8.8 x 1.6 x 10^-13 J
q = 2 e = 2 x 1.6 x 10^-19 C
Q = 82 e = 82 x 1.6 x 10^-19 C
Let d be the distance of closest approach
E = k Q q / d
Where, K = 9 x 10^9 Nm^2 / C^2
d = k Q q / E
d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)
d = 2.7 x 10^-14 m
A ball having a mass of 200 g is released from rest at a height of 400 mm above a very large fixed metal surface. If the ball rebounds to a height of 325 mm above the surface, determine the coefficient of restitution between the ball and the surface.
Answer:
0.9
Explanation:
h = 400 mm, h' = 325 mm
Let the coefficient of restitution be e.
h' = e^2 x h
325 = e^2 x 400
e^2 = 0.8125
e = 0.9
The coefficient of restitution between the ball and the surface is 0.9.
The coefficient of restitution (e) between the ball and the surface can be determined using the formula:
[tex]\[ e = \sqrt{\frac{\text{height after collision}}{\text{height before collision}}} \][/tex]
First, we need to convert the heights from millimeters to meters for consistency, since the standard units for height in physics are meters.
The initial height (h_i) before the collision is 400 mm, which is equivalent to 0.4 m (since 1 m = 1000 mm).
The final height (h_f) after the collision is 325 mm, which is equivalent to 0.325 m.
Now, we can plug these values into the formula for the coefficient of restitution:
[tex]\[ e = \sqrt{\frac{0.325 \text{ m}}{0.4 \text{ m}}} \][/tex]
[tex]\[ e = \sqrt{\frac{325}{400}} \][/tex]
[tex]\[ e = \sqrt{0.8125} \][/tex]
[tex]\[ e = 0.9 \][/tex]
Therefore, the coefficient of restitution between the ball and the surface is 0.9. This means that the collision is relatively elastic, with the ball retaining a significant portion of its initial kinetic energy after the rebound.
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of (c) a 2300 kg rhinoceros; (d) a 22 g song sparrow.
Explanation:
1. Force applied on an object is given by :
F = W = mg
(a) A 160 lb human being, F = 160 lb
g = acceleration due to gravity, g = 32 ft/s²
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{160\ lb}{32\ ft/s^2}[/tex]
m = 5 kg
(b) A 1.9 lb cockatoo, F = 1.9 lb
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{1.9\ lb}{32\ ft/s^2}[/tex]
m = 0.059 kg
2. (a) A 2300 kg rhinoceros, m = 2300 kg
[tex]W=2300\ kg\times 32\ ft/s^2=73600\ lb[/tex]
(b) A 22 g song sparrow, m = 22 g = 0.022 kg
[tex]W=0.022\ kg\times 32\ ft/s^2=0.704\ lb[/tex]
Hence, this is the required solution.
Final answer:
The mass of a 160 lb human being is 72.576 kg, and the mass of a 1.9 lb cockatoo is 0.86184 kg. A 2300 kg rhinoceros weighs 5060 lbs in English units, while a 22 g song sparrow weighs 0.0484 lbs.
Explanation:
To calculate the mass (in SI units) of a 160 lb human being, we use the conversion factor 1 lb = 0.4536 kg. So, the calculation is:160 lb x 0.4536 kg/lb = 72.576 kg
(a) The mass of a 160 lb human being in SI units is 72.576 kg.
For a 1.9 lb cockatoo, the conversion to kilograms is:1.9 lb x 0.4536 kg/lb
= 0.86184 kg
(b) The mass of a 1.9 lb cockatoo in SI units is 0.86184 kg
To convert the mass of a 2300 kg rhinoceros to weight in English units, knowing that 1 kg = 2.2 lbs (where weight in pounds is considered the gravitational force on the mass), the weight is calculated as:
2300 kg x 2.2 lbs/kg= 5060 lbs
(c) The weight of a 2300 kg rhinoceros in English units is 5060 lbs.
Finally, for converting a 22 g song sparrow to weight in English units:
22 g x (1 kg/1000 g) x (2.2 lbs/kg)= 0.0484 lbs
(d) The weight of a 22 g song sparrow in English units is 0.0484 lbs.
A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cross section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.
Answer:
The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
(c) is correct option.
Explanation:
Given that,
Length of steel wire = 4.7 m
Cross section[tex]A = 3\times10^{-3}\ m^2[/tex]
Length of copper wire = 3.5 m
Cross section[tex]A = 4\times10^{-5}\ m^2[/tex]
We need to calculate the ratio of young's modulus of steel and copper
Using formula of young's modulus for steel wire
[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]
[tex]Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}[/tex]....(I)
The young's modulus for copper wire
[tex]Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}[/tex]....(II)
From equation (I) and (II)
The ratio of the young's modulus of steel and copper
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}[/tex]
Hence, The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)
Answer:
Frequency, [tex]f=1.2\times 10^{10}\ Hz[/tex]
Explanation:
It is given that,
Magnetic field, B = 0.43 T
We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
[tex]f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}[/tex]
[tex]f=1.2\times 10^{10}\ Hz[/tex]
Hence, this is the required solution.
A vertical spring (ignore its mass), whose spring constant is 594-N/m, is attached to a table and is compressed down by 0.196-m. What upward speed (in m/s) can it give to a 0.477-kg ball when released?
Answer:
Speed, v = 6.91 m/s
Explanation:
Given that,
Spring constant, k = 594 N/m
It is attached to a table and is compressed down by 0.196 m, x = 0.196 m
We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]
[tex]v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}[/tex]
v = 6.91 m/s
So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 4.5 x 10 -3 Wb. What is the flux that passes through the circular loop?
Assume that you can heat water with perfect insulation (all the heat from combustion of ethanol is transferred to water). What is the volume of ethanol required to heat 200 mL of water by 10 °C? The density of ethanol is 0.78 g/mL, the specific heat capacity of water is 4.184 J∙g−1∙°C−1and the heat of combustion of ethanol is -1368 kJ/mol. Show your work.
Answer:
volumme =0.36 ml
Explanation:
total heat required can be obtained by using following formula
[tex]q= mC \Delta T[/tex].......(1)
where,
m - mass of water,
C - specific heat capacity of water and = 4.184 j g^{-1} °C
[tex]\Delta T[/tex] - total change in temperature. = 10°C
The density of water is 1 g/cc. hence, 200 mL of water is equal to 200 g
putting all value in the above equation (1)
q = 200*4.184* 10 ° = 8368 J.
Therefore total number of moles of ethanol required to supply 8368 J of heat is
[tex]\frac {8368}{1368000} = 0.006117 moles.[/tex]
The molar mass of ethanol is 46 g/mol.
The mass of ethanol required is 46* 0.006117 = 0.28138 g
The density of ethanol is 0.78 g/ml.
The volume of ethanol required is
[tex]\frac {0.28138}{0.78} = 0.36 ml[/tex]
To heat 200 mL of water by 10 °C, you would need approximately 0.00785 mL of ethanol. This is calculated by converting the volume of water to its mass, calculating the heat energy using the specific heat capacity of water, and then converting that energy to the amount of ethanol required using the heat of combustion of ethanol.
Explanation:To calculate the amount of ethanol required to heat 200 mL of water by 10 °C, we need to use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to convert the volume of water to its mass. Since the density of water is 1 g/mL, the mass of 200 mL of water is 200 g.
Next, we can calculate the heat energy using the formula: Q = mcΔT. The specific heat capacity of water is 4.184 J/g °C, the mass of water is 200 g, and the change in temperature is 10 °C. Plugging in these values, we get: Q = (200 g)(4.184 J/g °C)(10 °C) = 8376 J.
Finally, we can convert this heat energy to the amount of ethanol required using the heat of combustion of ethanol. The heat of combustion of ethanol is -1368 kJ/mol. To convert from J to kJ, we divide the heat energy by 1000 to get: 8376 J / 1000 = 8.376 kJ.
Now we can calculate the volume of ethanol. Since the density of ethanol is 0.78 g/mL, we can use the formula: V = m/ρ, where V is the volume, m is the mass, and ρ is the density. The mass of ethanol can be calculated by rearranging the formula: m = Q/ΔH, where Q is the heat energy and ΔH is the heat of combustion. Plugging in the values, we get: m = 8.376 kJ / -1368 kJ/mol = -0.00611 mol.
Now we can calculate the volume of ethanol using the formula: V = m/ρ. The mass of ethanol is -0.00611 mol and the density of ethanol is 0.78 g/mL. Plugging in these values, we get: V = -0.00611 mol / (0.78 g/mL) = -0.00785 mL.
Since volume cannot be negative, the volume of ethanol required to heat 200 mL of water by 10 °C is 0.00785 mL.
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A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:
Explanation:
Charge on proton, q₁ = e
Charge on alpha particles, q₂ = 2e
The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90=sin(90) = 1[/tex]
For proton, [tex]F_p=ev_pB[/tex]..........(1)
For alpha particle, [tex]F_a=2ev_aB[/tex]..........(2)
Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,
[tex]ev_pB=2ev_aB[/tex]
[tex]\dfrac{v_p}{v_a}=\dfrac{2}{1}[/tex]
So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.
If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.
Explanation:The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.
The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.
This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.
Therefore, the proton is moving twice as fast as the alpha particle.
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A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms value of the current in this circuit is 0.680 A, what is the inductance of the inductor?
Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor [tex]C= 2.00\mu F[/tex]
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current
[tex]I = \dfrac{V}{Z}[/tex]
[tex]Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}[/tex]
Put the value of Z into the formula
[tex]I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}[/tex]
Put the value into the formula
[tex]0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}[/tex]
[tex]L=35.8\ mH[/tex]
Hence, The inductance of the inductor is 35.8 mH
Answer:
Inductance,L:
"The property of the conductor or the solenoid to generate the electromotive force,emf due to the flow of current,I."
Unit: henry,H as it is equivalent to, kg.m².sec⁻².A⁻².
Explanation:
Data:
Voltage,v=120 v-rms,Frequency,f=1000 Hz,Capacitor, C=2.00 μF,Current,I=0.680 A,Solution:
We need to calculate the inductance, L of the solenoid inside a circuit,
I=v/z,Z=√R²+(Lω-1/Cω)²,putting the values I=V/√R²+(Lω-1/Cω)²,0.680=120/√(100)²+(L×2π×1000-1/2×10⁻⁶×2π×1000)²,L=35.8×10⁻³H, or L=35.8 mH.⇒Answer9.96 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed. Use data from the steam tables.
By using steam tables or similar thermodynamic property charts, you can determine the initial and final temperature and enthalpy of R-134a under specified pressure conditions in a rigid container. As the pressure changes with heating, you can find the corresponding changes in temperature and enthalpy.
Explanation:The subject matter of this question falls under the domain of thermodynamics in physics. It involves an element of heating, changes in pressure and the resultant alterations in temperature. Given the specifics, we can't directly provide a numerical solution in this setting as it requires the usage of refprop or other similar thermodynamic property database or the proper specific enthalpy and temperature charts for R-134a. However, conceptually, it can be handled via using steam tables (or more specifically refrigerant tables) for the substance R-134a. Initially, you lookup the properties using the given pressure and known mass and volume to calculate the initial temperature and enthalpy. Similarly, when the container is heated and the pressure raises to 600 kPa, you can again refer to the steam tables with this new pressure (and the volume which remains unchanged as the container is rigid) to determine the new temperature and enthalpy.
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What If? Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the speed of light, and 1 MeV = 1.6 ✕ 10−13 kg · m2/s2. Based on this, what are the units of momentum for a high-speed subatomic particle in terms of fundamental SI units?
Answer:
(kg⋅m/s)
Explanation:
The unit of momentum is the product of the units of mass and velocity. In SI units, if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s)
This question is dealing with fundamental SI units. Thus, let's list the seven basic SI Units upon which other units are expressed;
Mass - kilogram (kg)
Length - meter (m)
Time - second (s)
Amount of substance - mole (mol)
Electric current - ampere (A)
Thermodynamic temperature - kelvin (K)
Luminous intensity - candela (cd)
Fundamental SI unit of momentum is Kg.m/s
Now, we want to write the SI Unit of momentum.
From the question, we are told that Physicists often measure the momentum of subatomic particles using the formula;
MeV/c, where c is the speed of light, and 1 MeV = 1.6 ✕ 10−13 kg.m²/s²
Now, we know that unit of speed is in m/s.
Thus, in units, momentum = MeV/c = (kg.m²/s²)/(m/s)
Simplifying this gives; Kg.m/s
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A 61 kg skier starts from rest at the top of a 1200 m long trail which drops a total of 227 m from top to bottom. At the bottom, the skier is moving 11 m/s. How much energy was dissipated by friction?
Answer:
energy dissipated = 132 kJ
Explanation:
mass = 61 kg
height drop = 227 m
velocity = 11 m/s
potential energy due to height drop from top to bottom
P.E. = m g h
P.E. = 61× 9.8× 227
P.E. = 135,700 J
kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
= [tex]\frac{1}{2}\times 61 \times 11^2[/tex]
= 3690.5 J
energy dissipated = P.E - K.E.
= 135,700 J -3690.5 J
=132,009.5 J = 132 kJ
Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liquid that has a mass of 0.155 g and a volume of 0.000235 L. B. Calculate the volume in milliliters of a 4.71-g sample of a solid with a density of 3.63 g/mL. C. Calculate the mass of a 0.293-mL sample of a liquid with a density of 0.930 g/mL.
Answer:
A) 0.660 g/ml
B) 1.297 ml
C) 0.272 g
Explanation:
Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density [tex]D[/tex] appears as a physical characteristic property of matter that establishes a relationship between the mass [tex]m[/tex] of a body or substance and the volume [tex]V[/tex] it occupies:
[tex]D=\frac{m}{V}[/tex] (1)
Knowing this, let's begin with the answers:
Answer A:Here the mass is [tex]m=0.155g[/tex] and th volume [tex]V=0.000235L=0.235mL[/tex]
Solving (1) with these values:
[tex]D=\frac{0.155g}{0.235mL}[/tex] (2)
[tex]D=0.660g/mL[/tex] (3)
Answer B:In this case the mass of a sample is [tex]m=4.71g[/tex] and its density is [tex]D=3.63g/mL[/tex].
Isolating [tex]V[/tex] from (1):
[tex]V=\frac{m}{D}[/tex] (4)
[tex]V=\frac{4.71g}{3.63g/mL}[/tex] (5)
[tex]V=1.297mL[/tex] (5)
Answer C:In this case the volume of a sample is [tex]V=0.293mL[/tex] and its density is [tex]D=0.930g/mL[/tex].
Isolating [tex]m[/tex] from (1):
[tex]m=D.V[/tex] (6)
[tex]m=(0.930g/mL)(0.293mL)[/tex] (7)
[tex]m=0.272g[/tex] (8)
To compute density, volume or mass, utilize the density formula (Density = Mass / Volume). For problem A, the liquid's density is 0.660 g/mL. For B, the volume of the sample is 1.30 mL, and for C, the mass of the sample is 0.272 g.
Explanation:The formula to calculate density is Density = Mass / Volume. A. To calculate the density in g/mL of a liquid with a mass of 0.155 g and a volume of 0.000235 L, we first need to convert the volume from liters to milliliters: 0.000235 L is equal to 0.235 mL. Then, divide the mass by the volume to get the density: Density = 0.155 g / 0.235 mL = 0.660 g/mL.
B. To calculate the volume in milliliters of a 4.71-g sample of a solid with a density of 3.63 g/mL, divide the mass by the density: Volume = 4.71 g / 3.63 g/mL = 1.30 mL.
C. To calculate the mass of a 0.293-mL sample of a liquid with a density of 0.930 g/mL, multiply the volume by the density: Mass = 0.293 mL * 0.930 g/mL = 0.272 g.
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A capacitor has a charge of 4.6 μC. An E-field of 1.8 kV/mm is desired between the plates. There's no dielectric. What must be the area of each plate?
Answer:
[tex]A = 0.2875 m^2[/tex]
Explanation:
As we know that
[tex]Q = 4.6 \mu C[/tex]
E = 1.8 kV/mm
now we know that electric field between the plated of capacitor is given as
[tex]E = \frac{\sigma}{\epsilon_0}[/tex]
now we will have
[tex]1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}[/tex]
[tex]\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})[/tex]
[tex]\sigma = 1.6 \times 10^{-5} C/m^2[/tex]
now we have
[tex]\sigma = \frac{Q}{A}[/tex]
now we have area of the plates of capacitor
[tex]A = \frac{Q}{\sigma}[/tex]
[tex]A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}[/tex]
[tex]A = 0.2875 m^2[/tex]
At a carnival, you can try to ring a bell by striking a target with a 8.91-kg hammer. In response, a 0.411-kg metal piece is sent upward toward the bell, which is 3.88 m above. Suppose that 21.9 percent of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?
Answer:
4 m/s
Explanation:
M = mass of the hammer = 8.91 kg
m = mass of the metal piece = 0.411 kg
h = height gained by the metal piece = 3.88 m
Potential energy gained by the metal piece is given as
PE = mgh
PE = (0.411) (9.8) (3.88)
PE = 15.6 J
KE = Kinetic energy of the hammer
Given that :
Potential energy of metal piece = (0.219) Kinetic energy of the hammer
PE = (0.219) KE
15.6 = (0.219) KE
KE = 71.2 J
v = speed of hammer
Kinetic energy of hammer is given as
KE = (0.5) M v²
71.2 = (0.5) (8.91) v²
v = 4 m/s
What is the wavelength of an electron that has a kinetic energy of 0.50 MeV (relativistic)?
Answer:
The wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]
Explanation:
The kinetic energy of the electron is, [tex]E=0.5\ MeV=0.5\times 10^6\ eV[/tex]
We need to find the wavelength of this electron. It can be calculated using the concept of DE-broglie wavelength as :
[tex]\lambda=\dfrac{h}{\sqrt{2mE} }[/tex]
h is Plank's constant
m is the mass of electron
[tex]\lambda=\dfrac{6.67\times 10^{-34}\ J-s}{\sqrt{2\times 9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ eV} }[/tex]
[tex]\lambda=6.99\times 10^{-22}\ m[/tex]
So, the wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]. Hence, this is the required solution.
The wavelength of an electron with a kinetic energy of 0.50 MeV (relativistic) is approximately 7.28 x 10^-12 m.
Explanation:The wavelength of an electron with a kinetic energy of 0.50 MeV can be calculated using the relativistic de Broglie equation:
λ = h/(m*c)
Where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 Js), m is the mass of the electron (9.11 x 10^-31 kg), and c is the speed of light (3.00 x 10^8 m/s).
Substituting the values:
λ = (6.63 x 10^-34 Js)/((9.11 x 10^-31 kg)*(3.00 x 10^8 m/s))
λ ≈ 7.28 x 10^-12 m
Therefore, the wavelength of the electron is approximately 7.28 x 10^-12 m.
The heater element of a particular 120-V toaster is a 8.9-m length of nichrome wire, whose diameter is 0.86 mm. The resistivity of nichrome at the operating temperature of the toaster is 1.3 × 10-6 Ω ∙ m. If the toaster is operated at a voltage of 120 V, how much power does it draw
Answer:
Power, P = 722.96 watts
Explanation:
It is given that,
Voltage, V = 120 V
Length of nichrome wire, l = 8.9 m
Diameter of wire, d = 0.86 mm
Radius of wire, r = 0.43 mm = 0.00043 m
Resistivity of wire, [tex]\rho=1.3\times 10^{-6}\ \Omega-m[/tex]
We need to find the power drawn by this heater. Power is given by :
[tex]P=\dfrac{V^2}{R}[/tex]
And, [tex]R=\rho\dfrac{l}{A}[/tex]
[tex]P=\dfrac{V^2\times A}{\rho\times l}[/tex]
[tex]P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}[/tex]
P = 722.96 watts
So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.
Velocity of a wave is the. (a) Wavelength x frequency (b) Wave number x frequency (c) Time period x phase (d) None
Answer:
option (a)
Explanation:
Wavelength is defined as the distance traveled by the wave in one complete oscillation.
The number of oscillations completed in one second is called frequency.
The relation for the wave velocity is given by
wave velocity = frequency x wavelength
A car driving at an initial speed of 10.0 m/s accelerates on a straight road at 3.00 m/s^2. a) what is the speed of the car after one quarter of a mile? (1 mile =1.609km). B) the driver in part A slams on the brakes after reaching the quarter mile. If the car can decelerate at a rate of 4.50 m/s^2, what is the stopping distance of the car?
Answer:
The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.
Explanation:
Given that,
Initial speed = 10.0 m/s
Acceleration = 3.00 m/s^2
Distance [tex]d = \dfrac{1}{4}\times1.609[/tex]
[tex]d = 0.40225\ km=0.40225\times10^{3}\ m[/tex]
We need to calculate the speed of the car,
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Where, u = initial velocity
v = final velocity
a = acceleration
Put the value in the equation
[tex]v^2=(10.0)^2+2\times3.00\times0.40225\times10^{3}[/tex]
[tex]v^2=2513.5[/tex]
[tex]v=50.13\ m/s[/tex]
(B). We need to calculate the stopping distance of the car,
Using equation of motion again
[tex]v^2=u^2+2as[/tex]
Here,initial velocity = 50.13 m/s
Final velocity = 0
Acceleration = -4.50 m/s²
Put the value in the equation
[tex]0=(50.13)^2-2\times4.50\times s[/tex]
[tex]s=\dfrac{(50.13)^2}{2\times4.50}[/tex]
[tex]s=279.22\ m[/tex]
Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.
Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them?
Explanation:
Force between two point changes, F₁ = 1 N
Distance between them, r₁ = 2 cm = 0.02 m
We know that the electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
i.e.
[tex]F\propto \dfrac{1}{r^2}[/tex]
i.e
[tex]\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2[/tex]
Let F₂ is the force when the distance between the charges is 8 cm, r₂ = 0.08 m
[tex]F_2=\dfrac{F_1\times r_1^2}{r_2^2}[/tex]
[tex]F_2=\dfrac{1\ N\times (0.02\ m)^2}{(0.08\ m)^2}[/tex]
F₂ = 0.0625 N
So, the distance between the sphere is 8 N, the new force is equal to 0.0625 N. Hence, this is the required solution.
Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectric constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.29 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?
Answer:
current = 1.51 A
Explanation:
Initially the capacitor without any dielectric is connected across AC source
so the capacitive reactance of that capacitor is given as
[tex]x_c = \frac{1}{\omega c}[/tex]
now we have
[tex]i = \frac{V_{rms}}{x_c}[/tex]
here we know that
[tex]i = 0.29 A[/tex]
now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor
so here net capacitance is given as
[tex]c_{eq} = 4.2c + c = 5.2c[/tex]
now the equivalent capacitive reactance is given as
[tex]x_c' = \frac{1}{\omega(5.2c)}[/tex]
[tex]x_c' = \frac{x_c}{5.2}[/tex]
so here we have new current in that circuit is given as
[tex]i' = \frac{V_{rms}}{x_c'}[/tex]
[tex]i' = 5.2 (i) = 5.2(0.29)[/tex]
[tex]i' = 1.51 A[/tex]
A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 17.117.1 m/s at a distance of 0.4700.470 m from the joint and the moment of inertia of the forearm is 0.5500.550 kg·m2, what is the rotational kinetic energy of the forearm?
Answer:
364.4 J
Explanation:
I = Moment of inertia of the forearm = 0.550 kgm²
v = linear velocity of the ball relative to elbow joint = 17.1 m/s
r = distance from the joint = 0.470 m
w = angular velocity
Using the equation
v = r w
17.1 = (0.470) w
w = 36.4 rad/s
Rotational kinetic energy of the forearm is given as
RKE = (0.5) I w²
RKE = (0.5) (0.550) (36.4)²
RKE = 364.4 J
Answer:
363.96 J
Explanation:
v = 17.1 m/s
r = 0.47 m
I = 0.55 kgm^2
Let ω be the angular velocity
ω = v / r = 17.1 / 0.47 = 36.38 rad/s
The kinetic energy of rotation is
K = 1/2 I ω^2 = 0.5 x 0.55 x 36.38 x 36.38 = 363.96 J
A motorcycle rider executes a loop-to-loop stunt in a loop with a diameter of 8.4 m At the top of the loop, the rider has a speed of 7.5 m/s. The combined mass of the rider and motorcycle is 175 kg. Determine the normal force between the motorcycle and track at the top of the loop
Answer:
The normal force between the motorcycle and track at the top of the loop = 2343.75 N
Explanation:
At the top of the loop the track is directed tangential to the motion, the normal to the tangent is along radius of circle. that is here we need to find centripetal force.
We have expression for centripetal force
[tex]F=\frac{mv^2}{r}[/tex]
Here mass, m = 175 kg
Velocity, v = 7.5 m/s
Diameter, d = 8.4 m
Radius, r = 0.5d = 0.5 x 8.4 = 4.2 m
Substituting
[tex]F=\frac{mv^2}{r}=\frac{175\times 7.5^2}{4.2}=2343.75N[/tex]
The normal force between the motorcycle and track at the top of the loop = 2343.75 N