c. The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object.
Explanation:The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object. This is because energy transfers from the object with the higher temperature to the object with the lower temperature until they reach thermal equilibrium.
Mass and specific heat do not directly affect the temperature change. The mass only affects the amount of energy transferred, while the specific heat determines how much energy is needed to raise the temperature.
Therefore, the correct answer is c. The one with the higher initial temperature.
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A chamber fitted with a piston can be controlled to keep the pressure in the chamber constant as the piston moves up and down to increase or decrease the chamber volume. The chamber contains an ideal gas at 296 K and 1.00 atm.What is the work done on the gas as the piston compresses it from 1.00 L to 0.633 L ?
Express your answer with the appropriate units.
W =
J
Answer:
W = 37.2J
Explanation:
See attachment below.
Two identical wires A and B are subject to tension. The tension in wire A is 3 times larger than that in wire B. Find the ratio of the frequencies of the first harmonic in these two wires, fA1 / fB1.
Answer:
1.732
Explanation:
Let
Tension in wire B=T
Tension in wire A=3 T
We have to find the ratio of the frequencies of the first harmonic in these two wires.
When two wires are identical then the length of both wires are same.
Suppose, the length of each wire=l
Frequency=[tex]\frac{1}{2l}\sqrt{\frac{T}{\mu}}[/tex]
Where [tex]\mu=[/tex]Mass per unit length
Mass per unit length of both wires are same because the two wires are identical.
[tex]\mu_A=\mu_B[/tex]
[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_B}}}[/tex]
[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_A}}}=\sqrt{\frac{T_A}{T_B}}=\sqrt{\frac{3T}{T}}[/tex]
[tex]\frac{f_A}{f_B}=\sqrt 3=1.732[/tex]
The ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].
The given parameters;
tension in wire A = Ttension in wire B = 3TThe frequency of first harmonic of each wire is calculated as follows;
[tex]F_ A = \frac{1}{2l} \sqrt{\frac{3T}{\mu} } \\\\F_B = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
l is the length of the wiresT is the tension on the wireμ is the mass per unit lengthThe ratio of the two frequencies is calculated as follows;
[tex]\frac{F_A}{F_B} = \frac{\frac{1}{2l} \sqrt{\frac{3T}{\mu} } }{\frac{1}{2l} \sqrt{\frac{T}{\mu} } } \\\\\frac{F_A}{F_B} = \sqrt{\frac{3T}{T} } \\\\\frac{F_A}{F_B} = \sqrt{3}[/tex]
Thus, the ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].
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Ball B is suspended from a cord of length l attached to cart A, which can roll freely on a frictionless, horizontal track. The ball and the cart have the same mass m. If the cart is given an initial horizontal velocity v0 while the ball is at rest, describe the subsequent motion of the system, specifying the velocities of A and B for the following successive values of the angle θ (assume positive counterclockwise) that the cord will form with the vertical:
(a) θ = θmax
(b) θ = 0
(c) θ = θmin
The motion involves energy conversion between kinetic and potential energy, with ball B reaching maximum potential energy at θmax and θmin, and maximum kinetic energy at θ = 0. Cart A's velocity will be adjusted according to the change in motion of ball B, based on the conservation of momentum.
Explanation:The question relates to the physics concept known as conservation of momentum and energy in the context of pendular motion and collisions within an isolated system. When cart A is given an initial velocity and ball B is at rest, the force that accelerates ball B will be the component of the tension in the string that acts horizontally, as there's no friction resistance on the track. During the motion, the total energy of ball B will be conserved, converting between potential energy when at its highest at θmax and θmin, and kinetic energy when passing through the lowest point at θ = 0. As the system consists of pendular motion, for a given displacement, the velocity of the pendulum and cart at various angles can be determined by using energy conservation.
Given that cart A and ball B have the same mass, the velocities can be tracked by considering the movement as a combination of the cart rolling and the pendulum swinging, respecting the conservation of angular momentum around the axis from which the ball B is suspended.
A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to this axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to 0.30 of its maximum value?
Answer:
The angle that the door must rotate is 70.5 graus
Explanation:
When the door rotates through and angel X, the magnetic flux that passes through the door decreases from its maximum value to 0.3 of its maximum value
This way,
X = 0.3 Xmax
X = B A cosX = 0.3 B A
or
cosX = 0.3
or
X = [tex]cos^{-1}[/tex] (0.3) = 70.5
X = 70.5 graus
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?
Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by
[tex]V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }[/tex]
Taking natural log on both sides,
[tex]e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })[/tex]
[tex]t = -RC ln (1-\frac{V}{V_{0} })[/tex] (1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
= [tex]\frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }[/tex]
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)
[tex]t = -RC ln (1-\frac{V}{V_{0} })[/tex]
= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
How much work does it take for an external agent to move a 45.0-nC charge from a point on the +x-axis, 3.40 cm from the origin to a point halfway between the 41.0-nC and 52.0-nC charges?
Answer:
The work done on the 45.0nC charge is 2.24×10^-3J. The detailed solution can be found in the attachment below.
Explanation:
The Problem solution makes use of the potential energy relationship between the charges. This solution assumes the distance as shown in the diagram. For a different distance arrangement adjustments should be made appropriately.
The question in Physics pertains to the work needed by an external agent to move a charge within an electric field, and it involves applying the concept of electric potential energy. The work done is equal to the change in electric potential energy, which requires knowledge of the charges' positions and potentials.
Explanation:The question asks: How much work does it take for an external agent to move a 45.0-nC charge from a point on the +x-axis, 3.40 cm from the origin to a point halfway between the 41.0-nC and 52.0-nC charges? To answer this question, we would use the concept of electric potential energy in the electric field created by point charges. Work is required to move a charge within an electric field against electric forces. The work done by an external force to move a charge from one point to another is equal to the change in electric potential energy, which can be calculated from the initial and final electric potentials at the points in question.
To find the required work, we need to know the initial and final positions relative to other charges, and then we apply the electric potential energy formula. However, a complete solution would require additional specifics about the configuration and distances between the charges. Without these details, we cannot provide an exact numerical answer.
A 2.40 kg snowball is fired from a cliff 7.69 m high. The snowball's initial velocity is 13.0 m/s, directed 49.0° above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?
Answer:
a) W = 180.87 J , b) ΔU = -180.87 J , c) ΔU = -180.87 J
Explanation:
a) Work is defined as
W = F .ds
Where bold indicates vectors, we can write the scalar product
W = F s cos θ
Where the angle is between force and displacement.
The force of gravity is the weight of the body, which is directed downwards and the displacement thickens the tip of the cliff at the bottom, so that it is directed downwards, therefore the angle is zero degrees
W = [tex]F_{g}[/tex] y
W = m g y
For this problem we must fix a reference system, from the statement it is established that the system is placed at the base of the cliff, so that final height is zero and the initial height (y₀ = 7.69m)
W = 2.40 9.8 (7.69-0)
W = 180.87 J
b) The potential energy is
U = mg y
The change in potential energy,
ΔU = [tex]U_{f}[/tex]- U₀
ΔU = mg ([tex]y_{f}[/tex]- y₀)
ΔU = 2.4 9.8 (0 -7.69)
ΔU = -180.87 J
c) in this case we change the reference system to the height of the cliffs, for this configuration
y₀ = 0
[tex]y_{f}[/tex] = -7.69 m
ΔU = 2.4 9.8 (-7.69 -0)
ΔU = -180.87 J
A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational potential energy ber ground level the ball leves 50% .(a) Give the total mechanical energy of the ball E in terms of maximum height Am it reaches, the mass m, and the gravitational acceleration g.(b) What is the height, hm in meters?
Complete Question:
A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level
a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.
b) What is the height, hn in meters?
Answer:
a) Energy = mghₙ
b) Height, hₙ = 5.02 m
Explanation:
a) Total energy in terms of maximum height
Let maximum height be hₙ
At maximum height, velocity, V=0
Total mechanical energy , E = mgh + 1/2 mV^2
Since V=0 at maximum height, the total energy in terms of maximum height becomes
Energy = mghₙ
b) Height, hₙ in meters
mghₙ = mgh + 1/2 mV^2
mghₙ = m(gh + 1/2 V^2)
Divide both sides by mg
hₙ = h + 0.5 (V^2)/g
h = 2.15m
g = 9.8 m/s^2
V = 7.5 m/s
hₙ = 2.15 + 0.5(7.5^2)/9.8
hₙ = 2.15 + 2.87
hₙ = 5.02 m
A Carnot engine receives 250 kJ·s−1 of heat from a heat-source reservoir at 525°C and rejects heat to a heat-sink reservoir at 50°C. What are the power developed and the heat rejected?
a. The quantity of heat rejected by the Carnot engine is equal to -101.2 kJ/s.
b. The power developed by the Carnot engine is equal to 148.8 kJ/s.
Given the following data:
Quantity of heat received = 250 kJ/sTemperature of heat-source = 525°CTemperature of heat rejected = 50°CConversion:
Temperature of heat-source = 525°C to Kelvin = 525 + 273 = 798K
Temperature of heat rejected = 50°C to Kelvin = 50 + 273 = 323K
To find the power developed and the heat rejected, we would use Carnot's equation:
[tex]-\frac{Q_R}{T_R} = \frac{Q_S}{T_S}[/tex]
Where:
[tex]Q_R[/tex] is the quantity of heat rejected.[tex]T_R[/tex] is the heat-sink temperature.[tex]T_S[/tex] is the heat-source temperature.[tex]Q_S[/tex] is the quantity of heat received.Making [tex]Q_R[/tex] the subject of formula, we have:
[tex]Q_R = -( \frac{Q_S}{T_S})T_R[/tex]
Substituting the given parameters into the formula, we have;
[tex]Q_R = -( \frac{250}{798}) \times 323\\\\Q_R = -( \frac{80750}{798})[/tex]
Quantity of heat rejected = -101.2 kJ/s
Now, we can determine the power developed by the Carnot engine:
[tex]P = -Q_S - Q_R\\\\P = -250 - (-101.2)\\\\P = -250 + 101.2[/tex]
Power, P = 148.8 kJ/s.
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The power developed by the Carnot engine is approximately 148.8 kilowatts. The heat rejected by the Carnot engine is approximately 101.2 kilowatts.
The Carnot efficiency (η) of a heat engine is given by the formula:
η = 1 - (T₁ ÷ T₂)
Heat received (Q(in)) = 250,000 J/s
Temperature of the heat-source reservoir (T(Hot)) = 798.15 K
Temperature of the heat-sink reservoir (T(Cold) = 323.15 K
The Carnot efficiency:
η = 1 - (T₁ ÷ T₂)
η = 1 - (323.15 ÷ 798.15 )
η = 0.5952
Therefore, The Carnot efficiency is 0.5952, which means the engine converts 59.52% of the heat received into useful work.
Power Developed:
P = η × Q(in)
P = 0.5952 × 250,000
P = 148,800 = 148.8 kW
Therefore, The power developed by the Carnot engine is approximately 148.8 kilowatts.
Heat Rejected:
Q(out) = Q(in) - P
Q(out) = 250,000- 148,800 = 101,200 J/s
Q(out) = 101.2 kW
Therefore, The heat rejected by the Carnot engine is approximately 101.2 kilowatts.
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The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x) = (2.0 {\rm J/m}^{3})x^{3}- (15 {\rm J/m}^{2})x^{2}+ (36 {\rm J/m})x - 23 {\rm J}. Find the location(s) where the force on the mass is zero.
Answer:x=2 and x=3
Explanation:
Given
Potential Energy for a certain mass is
[tex]U(x)=2x^3-15x^2+36x-23[/tex]
and we know force is given by
[tex]F=-\frac{\mathrm{d} U}{\mathrm{d} x}[/tex]
[tex]F=-(2\times 3x^2-15\times 2x+36)[/tex]
For Force to be zero F=0
[tex]\Rightarrow 6x^2-30x+36=0[/tex]
[tex]\Rightarrow x^2-5x+6=0[/tex]
[tex]\Rightarrow x^2-2x-3x+6=0[/tex]
[tex]\Rightarrow (x-2)(x-3)=0[/tex]
Therefore at x=2 and x=3 Force on particle is zero.
college physics 1, final exam 2015 a closed 2.0l container holds 3.0 mol of an idea gas. if 200j of heat is added, and no work is done, what is the chanhe in internal enegery of the system
Answer: The change in internal energy of the system is 200 Joules
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex]=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done by the system=[tex]-P\Delta V=0[/tex]
q = +200J {Heat absorbed by the system is positive}
[tex]\Delta E=+200+0=+200J[/tex]
Thus the change in internal energy of the system is 200 Joules
Atmospheric pressure is reported in a variety of units depending on local meteorological preferences. In many European countries the unit millibar (mbar) is preferred, in other countries the unit hectopascal (1 hPa = 1 mbar) is used, and in the United States inches of mercury (in Hg) is the commonly used unit. In most chemistry textbooks the units most commonly used are torr, mmHg, and atmospheres (atm). The unit atm is defined at sea level to be 1 atm = 760 mm Hg exactly. The density of mercury is 13.534 times that of water, if atmospheric pressure will support 769.6 mm Hg, what height of a water column would that same pressure support in mm?
Answer:
[tex]h_w=10415.7664\ mm[/tex] of water column.
Explanation:
Given:
density of mercury, [tex]S_m=13.534[/tex]
height of the mercury column supported by the atmosphere, [tex]h_m=769.6\ mm[/tex]
As we know that the equivalent pressure in terms of liquid column is given as:
[tex]P=\rho.g.h[/tex]
so,
[tex]S_m\times 1000\times g.h_m=\rho_w.g.h_w[/tex]
where:
[tex]g=[/tex] gravity
[tex]h_w=[/tex] height of water column
[tex]\rho_w=[/tex] density of water
[tex]13534\times 9.8\times 769.6=1000\times 9.8\times h_w[/tex]
[tex]h_w=10415.7664\ mm[/tex] of water column.
Final answer:
The atmospheric pressure that supports a 769.6 mm column of mercury will support a water column that is 13.6 times taller due to the density difference, resulting in a water column approximately 10,466.56 mm, or about 10.47 meters, tall.
Explanation:
When we talk about atmospheric pressure, we often use various units like millimeters of mercury (mmHg) or torr, which are equivalent to 1 atm of pressure, defined as exactly 760 mmHg. Since mercury is about 13.6 times denser than water, a column of mercury under atmospheric pressure is much shorter than a column of water under the same pressure. To find the height of a water column supported by a pressure of 769.6 mm Hg, we can set up a proportion using the densities of mercury and water. Since 1 atm supports a 760 mm column of mercury, and the density of mercury is 13.6 times that of water, the same pressure will support a water column that is 13.6 times taller.
The calculation is straightforward:
Height of mercury column (Hg): 769.6 mmDensity ratio (Hg:water): 13.6Height of water column = Height of mercury column × Density ratioHeight of water column = 769.6 mm × 13.6Height of water column = 10466.56 mmTherefore, the atmospheric pressure that supports a 769.6 mm Hg column will support a water column of approximately 10,466.56 mm, or about 10.47 meters tall.
Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 7.03 m below the point of release and she hears the splash 0.81 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)
Explanation:
The formula to calculate total time taken is as follows.
Total time = time to fall + time for sound
So, time for sound = [tex]\frac{distance}{velocity}[/tex]
= [tex]\frac{7.03}{343}[/tex]
= 0.0204 sec
Hence, time to fall is as follows.
(0.81 - 0.0204) sec
= 0.7896 sec
Now, we will calculate the time to fall as follows.
y = [tex]y_{o} + v_{o}yt + \frac{1}{2}at^{2}[/tex]
0 = [tex]h + v \times t - \frac{1}{2}gt^{2}[/tex]
0 = [tex]7.03 + v \times (0.81 - 0.0204) - 0.5 \times 9.81 \times(0.81 - 0.0204)^{2}[/tex]
= [tex]7.8196 - 0.5 \times 9.81 \times 0.623[/tex]
= 7.8196 - 3.058
= 4.7616 m/s
Therefore, she threw the coin at 4.76 m/s in the upward direction.
Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen
Answer:
Hydrogen takes 0.391s to get to distance x
Explanation:
From the chromatography table:
[tex]D_H_2=6.4\times10^{-5}m^2/s\\D_O_2=1.8\times10^{-5}m^2/s\\[/tex]
Using the equation[tex]x_m_s=\sqrt(2Dt)[/tex]. This equation relates time to distance during diffusion
[tex]x_m_s[/tex],[tex]_O_2[/tex]=[tex]\sqrt[/tex][tex]2D_o_2[/tex][tex]t_o_2[/tex] and [tex]x_m_s[/tex],[tex]__H_2[/tex]=[tex]\sqrt[/tex][tex]2D_H__2[/tex][tex]t_H__2[/tex]
Let the distance traveled be denoted by x(same distance traveled by both gases).
Distance is same when difference between[tex]t_H__2[/tex] and [tex]t_O__2[/tex] is 1.0 seconds.
[tex]t_O__2=t_H___2[/tex][tex]+1.0s[/tex]
At equal distance=>
[tex]2D_O__2[/tex][tex]t_O__2[/tex]=[tex]2D_H__2[/tex][tex]t_H__2[/tex]
[tex]D_O_2[/tex][tex](t_H__2[/tex][tex]+1.0s)=D_H__2[/tex][tex]t_H__2[/tex]
Solving for hydrogen time:
[tex]t_H__2=(D_0__2)\div[/tex][tex](D_H__2[/tex]-[tex]D_O__2)[/tex][tex]\times1.0[/tex]
=[tex](1.8\times10^{-5}m^2/s)\div(6.4\times10^{-5}m^2/s-1.8\times10^{-5}m^2/s)\times1.0s[/tex]
=0.391s
Final answer:
To find the time before hydrogen is 1.00 s ahead of oxygen when both are diffusing through air, we use Graham's law of effusion. Hydrogen effuses four times faster than oxygen, so with calculations, we find that this time is approximately 2.33 seconds.
Explanation:
When hydrogen and oxygen are diffusing through air and are released simultaneously, we want to determine the time that passes before the hydrogen gas (H2) is 1.00 s ahead of the oxygen gas (O2). This concept involves diffusion rates in gases, which can be described using Graham's law of effusion. According to Graham's law, the rate of effusion for a gas is inversely proportional to the square root of its molar mass (M).
Given that hydrogen effuses four times as rapidly as oxygen, we can write the relationship between their rates as RH2 / RO2 = sqrt(MO2 / MH2). Knowing that the molar mass of hydrogen (MH2) is 2 g/mol and the molar mass of oxygen (MO2) is 32 g/mol, we can substitute these values into the equation to find the rate ratio.
Since hydrogen is four times faster, and we want hydrogen to be 1.00 s ahead of the oxygen, we can use the ratio to calculate how long it would take for oxygen to travel the same distance. This time can then be added to 1.00 s to find the total elapsed time. Here's a step-by-step approach to calculate this:
Calculate the square root of the ratio of the molar masses: sqrt(32/2) = sqrt(16) = 4.
The rate of hydrogen is therefore four times the rate of oxygen: RH2 = 4 × RO2.
If we let t be the time for oxygen to effuse, hydrogen will effuse in t/4 time.
To be 1.00 s ahead, we want t/4 = t - 1.00.
Solving for t gives us t = 4/3 seconds (approximately 1.33 s).
This is the time it would take for oxygen to cover the same distance that hydrogen covers in 1/3 second.
Thus, the total time before hydrogen is 1.00 s ahead of the oxygen is approximately 1.33 s (the time for oxygen) + 1.00 s = 2.33 seconds.
This concept is particularly useful in analytical chemistry, specifically in techniques like gas chromatography, where differences in the rate of diffusion of gases are used to separate and identify components in a mixture.
A white billiard ball with mass mw = 1.43 kg is moving directly to the right with a speed of v = 3.39 m/s and collides elastically with a black billiard ball with the same mass mb = 1.43 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 38° and the black ball ends up moving at an angle below the horizontal of θb = 52°. 1)What is the final speed of the white ball? m/s 2)What is the final speed of the black ball? m/s 3)What is the magnitude of the final total momentum of the system? kg-m/s 4)What is the final total energy of the system?
Answer: a) VW = 1.28m/s
b) Vb = 3.86m/s
c) p = 5.82kgm/s
d) E = 11.84J
Explanation: To solve this question, we make use of explosion formula in linear momentum concept.
Please find the attached file for the solution
A 1000 kg automobile is traveling at an initial speed of 20 m/s. It is brought to a complete stop in 5 s over a distance of 50 m. What is the work done in stopping the automobile
Answer:
W = -2*10⁵ J
Explanation:
Assuming no friction present, we can find the work done by an external force stopping the car, applying the work-energy theorem.This theorem says that the total work done on one object by an external net force, is equal to the change in the kinetic energy of the object.If the automobile is brought to a complete stop, we can find the change of the kinetic energy as follows:[tex]\Delta K = K_{f} - K_{0} = 0 - \frac{1}{2} * m* v_{0} ^{2} \\\\ \Delta K = -\frac{1}{2}*1000kg*(20 m/s)^{2} = -200000 J= -2e5 J[/tex]
So, the total work done in stopping the automobile, is -2*10⁵ J. The minus sign stems from the fact that the force and the displacement have opposite directions.Given values:
Mass, m = 1000 kgInitial speed, v = 20 m/sDistance, d = 50 mTime, t = 5 sAs we know,
→ [tex]\Delta K = K_f - K_0[/tex]
[tex]= 0-\frac{1}{2}mv_2^2[/tex]
By substituting the values,
[tex]= - \frac{1}{2}\times 1000\times (20)^2[/tex]
[tex]= - 500\times 400[/tex]
[tex]= -200000 \ J[/tex]
[tex]= -2\times 10^5 \ J[/tex]
Thus the above answer is right.
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A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. the lid of a pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. a separate metal piece, the petcock, sits on top of this opening and prevents steam from escaping until the pressure force overcomes the weight of the petcock. the periodic escape of the steam in this manner prevents any potentially dangerous pressure buildup and keeps the pressure inside at a constant value. determine the mass of the petcock of a pressure cooker whose operation pressure is 100 kpa gauge and has an opening cross-sectional area of 4 mm2. assume an atmospheric pressure of 101 kpa.
A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. The mass of the petcock of the pressure cooker is approximate [tex]40.8\ gm[/tex].
The pressure force on the petcock is the difference between the pressure inside the cooker and the atmospheric pressure, multiplied by the cross-sectional area of the opening:
Pressure force = (Pressure inside - Atmospheric pressure) * Opening area
Given that the operating pressure is 100 kPa gauge and the atmospheric pressure is 101 kPa,
Pressure inside[tex]= (100+ 101) \times 1000 = 201000\ Pa[/tex]
Atmospheric pressure [tex]= 101 \times 1000 = 101000\ Pa[/tex]
Now, calculate the pressure force:
Pressure force[tex]= (201000\ Pa - 101000\ Pa) \times 4\ mm^2[/tex]
Pressure force [tex]= 100000\ Pa \times 4 \times 10^{-6} m^2[/tex]
Pressure force [tex]= 0.4\ N[/tex]
The weight of the petcock is equal to the force of gravity acting on it:
Weight = mass × gravitational acceleration
[tex]Weight = mass \times 9.8 m/s^2\\m \times 9.8 = 0.4 \\m = 0.4 / 9.8 \\m = 0.0408\ kg[/tex]
So, the mass of the petcock of the pressure cooker is approximate [tex]40.8\ gm[/tex].
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To find the mass of the petcock, we calculate the force exerted by the steam using the given pressure and area, then equate it to the weight of the petcock. This gives us a mass of approximately 0.082 kg.
To determine the mass of the petcock in a pressure cooker, we need to understand the pressure dynamics and forces at play. The pressure cooker operates at a gauge pressure of [tex]100 kPa,[/tex] and the cross-sectional area of the opening through which steam escapes is given as [tex]4 mm^2[/tex].
First, let's convert the cross-sectional area from [tex]mm^2[/tex] to [tex]m^2[/tex]:
[tex]4 mm^2 = 4 \times 10^{-6} m^2[/tex]
The force exerted by the steam on the petcock can be calculated using the pressure-force relationship: [tex]F = PA[/tex]
Here, P is the absolute pressure inside the cooker, which is the sum of the gauge pressure and the atmospheric pressure:
[tex]P = 100 kPa (gauge) + 101 kPa (atmospheric) = 201 kPa[/tex]
The force is then:
[tex]F = P \times A = 201,000 Pa \times 4 \times 10^{-6} m^2 = 0.804 N[/tex]
The force exerted by the petcock due to its weight is [tex]F = mg[/tex], where m is the mass and g is the acceleration due to gravity ([tex]9.81 m/s^2[/tex]).
Setting the force exerted by the steam equal to the weight of the petcock gives us:
[tex]mg = 0.804 N[/tex]
Therefore, the mass of the petcock is:
[tex]m = F/g = 0.804 N / 9.81 m/s^2 \approx 0.082 kg[/tex]
Thus, the mass of the petcock is approximately [tex]0.082 kg[/tex].
The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P is 10.0 cm from the rotation axis, and point Q is 20.0 cm from the rotation axis. Find the speed of point P on the spinning wheel.
Answer:
0.938 m/s.
Explanation:
Given:
ω = 1 rev in 0.67 s
In rad/s,
1 rev = 2pi rad
ω = 2pi ÷ 0.67
= 9.38 rad/s
Rp = 10 cm
= 0.1 m
V = ω × r
= 9.38 × 0.1
= 0.938 m/s.
A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 0.9 µF capacitor then drops to 2 V. What is the capacitance of the second capacitor?
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
An object with total mass mtotal = 15.8 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.5 kg moves up and to the left at an angle of θ1 = 18° above the –x axis with a speed of v1 = 27.5 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of θ2 = 23° to the right of the -y axis at a speed of v2 = 21.4 m/s. 1)What is the magnitude of the final momentum of the system (all three pieces)?
Answer:
Explanation:
total mass, M = 15.8 kg
initial velocity, u = 0 m/s
m1 = 4.5 kg
m2 = 5.4 kg
v1 = 27.5 m/s at an angle 18°
v2 = 21.4 m/s at an angle 23°
As there is no external force acts on the system, so the moemntum of the system is conserved.
Momentum before collision = momentum after collision
as the momentum before collision is zero, so the momentum after collision is also zero.
A running back with a mass of 86 kg and a velocity of 9 m/s (toward the right) collides with, and is held by, a 129-kg defensive tackle going in the opposite direction (toward the left). What is the velocity of the tackle before the collision for their velocity afterward to be zero
Final answer:
The velocity of the defensive tackle before the collision is -6 m/s (toward the left).
Explanation:
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the two players stick together after the collision, their velocities will be equal but opposite in direction.
Let's assume that the velocity of the defensive tackle before the collision is v. The total momentum before the collision is given by:
m1 * v1 + m2 * v2 = (m1 + m2) * v
Where:
m1 = mass of the running back = 86 kg
v1 = velocity of the running back = 9 m/s (toward the right)
m2 = mass of the defensive tackle = 129 kg
v2 = velocity of the defensive tackle (to be determined)
Using the above equation, we can solve for v:
(86 kg * 9 m/s) + (129 kg * v2) = (86 kg + 129 kg) * 0
774 kg*m/s + 129 kg * v2 = 0 kg*m/s
v2 = -6 m/s
Therefore, the velocity of the defensive tackle before the collision is -6 m/s (toward the left).
The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Express your answer in terms of any or all of m, F, and t.
Answer:
Ft
Explanation:
We are given that
Initial velocity=u=0
We have to find the magnitude of p of the momentum of the particle at time t.
Let mass of particle=m
Applied force=F
Acceleration, [tex]a=\frac{F}{m}[/tex]
Final velocity , [tex]v=a+ut[/tex]
Substitute the values
[tex]v=0+\frac{F}{m}t=\frac{F}{m}t[/tex]
We know that
Momentum, p=mv
Using the formula
[tex]p=m\times \frac{F}{m}t=Ft[/tex]
The magnitude of the momentum of the particle after time, t is Ft = mv.
The given parameters:
initial velocity of the particle, u = 0The magnitude of the momentum of the particle after time, t is calculated as follows;
[tex]P =mv[/tex]
The force applied to the particle is calculated as;
[tex]F = ma = \frac{mv}{t} \\\\Ft = mv[/tex]
Thus, the magnitude of the momentum of the particle after time, t is Ft = mv.
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A charge Q is transfered from an initially uncharged plastic ball to an identical ball 12 cm away. The force of attraction is then 17 mN. How many electrons were transfered from one ball to the other
Answer:
Explanation:
The ball from which electrons are transferred will acquire Q charge and the ball receiving electrons will acquire - Q charge . force of attraction between them
= k Q² / d²
9 x 10⁹ x Q² / .12² = 17 X 10⁻³
Q² = .0272 X 10⁻¹²
Q = .1650 X 10⁻⁶ C
No of electrons = .1650 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .103125 x 10¹³
1.03125 x 10¹²
A conducting sphere of radius R1 carries a charge Q. Another conducting sphere has a radius R2 = 3 4 R1, but carries the same charge. The spheres are far apart. What is the ratio E2 E1 of the electric field near the surface of the sphere with radius R2 to the field near the surface of the sphere with radius R1?
Answer:
The ratio of electric field is 16:9.
Explanation:
Given that,
Radius [tex]R_{2}=\dfrac{3}{4}R_{1}[/tex]
Charge = Q
We know that,
The electric field is directly proportional to the charge and inversely proportional to the square of the distance.
In mathematically term,
[tex]E=\dfrac{kQ}{R^2}[/tex]
Here, [tex]E\propto\dfrac{1}{R^2}[/tex]
We need to calculate the ratio of electric field
Using formula of electric field
[tex]\dfrac{E_{2}}{E_{1}}=\dfrac{R_{1}^2}{R_{2}^2}[/tex]
Put the value into the formula
[tex]\dfrac{E_{2}}{E_{1}}=\dfrac{(4R_{1})^2}{(3R_{1})^2}[/tex]
[tex]\dfrac{E_{2}}{E_{1}}=\dfrac{16}{9}[/tex]
Hence, The ratio of electric field is 16:9.
Final answer:
The ratio of the electric fields E2/E1 at the surface of two conducting spheres with radii R2 = 3/4R1 carrying the same charge is 16/9.
Explanation:
The question is asking for the ratio of the electric fields near the surface of two conducting spheres carrying the same charge Q but having different radii, R1 and R2 where R2 is three-fourths of R1. To find the electric field E near the surface of a conducting sphere, we use the formula:
E = kQ/[tex]r^{2}[/tex]
where k is Coulomb's constant, Q is the charge, and R is the radius of the sphere. Since the charge Q is the same on both spheres, we can calculate the ratio of the electric fields by plugging in the radii:
E2/E1 = (kQ/[tex]r2^{2}[/tex]) / (kQ/[tex]r1^{2}[/tex])
Simplifying further, since k is a constant it cancels out, along with Q, which is the same for both spheres, we get:
E2/E1 = ([tex]r1^{2}[/tex]) / ([tex]r2^{2}[/tex])
Substituting R2 = 3/4R1:
E2/E1 = [tex]r1^{2}[/tex] / (3/4 R1)[tex]{2}[/tex]
E2/E1 = 1 / (3/4)^2
E2/E1 = 1 / (9/16) = 16/9
Therefore, the ratio E2/E1 is 16/9.
A bulb pile is driven to the ground with a 2.5 ton hammer. The drop height is 22 ft and the volume in last batch driven is 4 cu ft. Establish the safe load if the number of blows to drive the last batch is 36, volume of base and plug is 27 cu ft, and the soil K value is 28.
Answer:
159.1 ton
Explanation:
The solution is shown in the attached file
Answer:
The safe load is 159 ton
Explanation:
The safe load is equal to:
[tex]L=\frac{WHBV^{2/3} }{K}[/tex]
Where:
W = weight of hammer = 2.5 ton
H = drop height = 22 ft
B = number of blows used to drive the last batch = 36/4 = 9 ft³
K = dimensionless constant = 28
V = uncompacted volume = 27 ft³
Replacing values:
[tex]L=\frac{2.5*22*9*(27^{2/3}) }{28} =159ton[/tex]
The Acorn Insurance Company charges $3.50 for each unit of coverage under a block of 1-year term life insurance policies. The annual premium amount for a $50,000 policy in this block would be equal to ____.
Answer:
$175
Explanation:
Insurance premium is expressed as a rate $1000
($3.50 per $1000)
Therefore;
Annual premium= $50000x$3.50/$1000
= $175
Final answer:
The annual premium for a $50,000 policy at the rate of $3.50 per unit of coverage from the Acorn Insurance Company would be $175.00.
Explanation:
The Acorn Insurance Company charges $3.50 for each unit of coverage under a 1-year term life insurance policy. To calculate the annual premium for a $50,000 policy, we'll need to divide the total coverage amount by the unit price and then multiply by the cost per unit. Here's the math:
Total Coverage Needed: $50,000
Cost Per Unit of Coverage: $3.50
Number of Units: $50,000 / $1,000 = 50
Annual Premium: 50 units * $3.50 = $175.00
So, the annual premium amount for a $50,000 policy in this block would be $175.00.
A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the cylinder is held in place only by the pressure of the air. One day when the atmospheric pressure is 100 kPa, it takes a 173 N force to pull the cap off.
Explanation:
The given data is as follows.
Mass of oxygen present = 100 mg = [tex]100 \times 10^{-3}[/tex] g
So, moles of oxygen present are calculated as follows.
n = [tex]\frac{100 \times 10^{-3}}{32}[/tex]
= [tex]3.125 \times 10^{-3}[/tex] moles
Diameter of cylinder = 6 cm = [tex]6 \times 10^{-2}[/tex] m
= 0.06 m
Now, we will calculate the cross sectional area (A) as follows.
A = [tex]\pi \times \frac{(0.06)^{2}}{4}[/tex]
= [tex]2.82 \times 10^{-3} m^{2}[/tex]
Length of tube = 11 cm = 0.11 m
Hence, volume (V) = [tex]2.82 \times 10^{-3} \times 0.11[/tex]
= [tex]3.11 \times 10^{-4} m^{3}[/tex]
Now, we assume that the inside pressure is P .
And, [tex]P_{atm}[/tex] = 100 kPa = 100000 Pa,
Pressure difference = 100000 - P
Hence, force required to open is as follows.
Force = Pressure difference × A
= [tex](100000 - P) \times 2.82 \times 10^{-3}[/tex]
We are given that force is 173 N.
Thus,
[tex](100000 - P) \times 2.82 \times 10^{-3}[/tex] = 173
Solving we get,
P = [tex]3.8650 \times 10^{4} Pa[/tex]
= 38.65 kPa
According to the ideal gas equation, PV = nRT
So, we will put the values into the above formula as follows.
PV = nRT
[tex]38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T[/tex]
T = 462.66 K
Thus, we can conclude that temperature of the gas is 462.66 K.
A comet orbits a star in a strongly elliptical orbit. The comet and star are far from other massive objects. As the comet travels away from the star, how does the kinetic energy and potential energy of the system change?
Answer:
the kinetic energy decreases and the potential energy Increases.
Explanation:
as the comet travels away from the star it gains an energy which it posses because of its position or state.once the comet moves away form the star the kinetic decreases until its lost all together to where the potential energy starts increasing.
Answer:
The potential energy increase and the kinetic energy decrease.
What is the energy in the spark produced by discharging the second capacitor? 1. The same as the discharge spark of the first capacitor 2. More energetic than the discharge spark of the first capacitor 3. Less energetic than the discharge spark of the first capacitor
Complete Question
The two isolated parallel plate capacitors be- low, one with plate separation d and the other with D > d, have the same plate area A and
are given the same charge Q.
What is the energy in the spark produced by discharging the second capacitor?
1. The same as the discharge spark of the first capacitor
2. More energetic than the discharge spark of the first capacitor
3. Less energetic than the discharge spark of the first capacitor
Answer:
The correct option is 2
Explanation:
The formula for the energy stored in the capacitor is
[tex]U = \frac{Q^2}{2C}[/tex]
And generally the formula for finding the capacitance of a capacitor is
[tex]C = \frac{\epsilon_oA}{d}[/tex]
We can denote the capacitance of the first capacitor as [tex]C_1 = \frac{\epsilon_oA}{d}[/tex]
and denote the capacitance of the second capacitor as [tex]C_2 = \frac{\epsilon_oA}{D}[/tex]
Looking at this formula we can see that C varies inversely with d
as D > d it means that [tex]C_1 > C_2[/tex]
Since the charge is constant
[tex]U\ \alpha\ \frac{1}{C}[/tex] i.e U varies inversely with C
So [tex]C_1 > C_2[/tex] => [tex]U_2 >U_1[/tex]
This means that the energy of spark would be more for capacitor two compared to capacitor one
The correct option is 2
The energy in the spark produced by discharging a capacitor can vary between different capacitors based on their capacitance and potential difference.
When discharging a capacitor, the energy stored in the capacitor is released as a spark, which can be compared between different capacitors.
In this case, if the second capacitor discharges, the spark produced may have varying energy levels compared to the spark produced by the first capacitor, depending on the capacitance and potential difference.
The energy in the spark produced by discharging the second capacitor can be calculated using relevant formulas relating to capacitance, potential difference, and energy stored.
At a carnival, you can try to ring a bell by striking a target with a 7.76-kg hammer. In response, a 0.372-kg metal piece is sent upward toward the bell, which is 4.87 m above. Suppose that 29.7 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:
3.93 m/s
Explanation:
Let the kinetic energy of hammer be 'K' and speed of hitting the target be 'v'.
Given:
Mass of the hammer (M) = 7.76 kg
Mass of the metal piece (m) = 0.372 kg
Kinetic energy of the hammer used by the metal piece = 29.7% of 'K' = 0.297K
Vertical height traveled by the metal piece (h) = 4.87 m
From conservation of energy, the kinetic energy used by the metal piece is transformed to the gravitational potential energy when it reaches the height of the bell.
Gravitational potential energy of the piece is given as:
[tex]U=mgh\\\\U=0.372\times 9.8\times 4.87=17.754\ J[/tex]
Now, as per question:
[tex]0.297K=17.754\ J\\\\K=\frac{17.754}{0.297}\\\\K=59.78\ J[/tex]
Therefore, the kinetic energy of the hammer is 59.78 J.
We know that,
Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
So, [tex]K=\frac{1}{2}mv^2[/tex]
Expressing in terms of 'v', we get:
[tex]mv^2=2K\\\\v^2=\frac{2K}{m}\\\\v=\sqrt{\frac{2K}{m}}[/tex]
Plug in the given values and solve for 'v'. This gives,
[tex]v=\sqrt{\frac{2\times 59.78}{7.76}}\\\\v=3.93\ m/s[/tex]
Therefore, the hammer must move with a speed of 3.93 m/s when it strikes the target so that the bell just barely rings.