Answer:
.864 M
Explanation:
For first order decomposition,
rate constant k = 1/t x ln a / (a - x )
given , a = 1.33 M , t = 644 s , k = 6.7 x 10⁻⁴ , a - x = ? = b( let )
6.7 x 10 ⁻⁴ = 1/644 x ln 1.33/b
ln 1.33/b = 6.7 x 10⁻⁴ x 644 = .4315
1.33 / b = e⁰ ⁴³¹⁵ = 1.5395
b = 1.33 / 1.5395 = .864 M.
Question Part Points Submissions Used If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v = mg c (1 − e−ct/m) where g is the acceleration due to gravity and c is a positive constant describing air resistance. (a) Calculate lim t→∞ v.
Answer:
[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]
Explanation:
the velocity as a function of time is
[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]
[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]
[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]
The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.
Explanation:
To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.
As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:
∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.
The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.
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Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s). Give your answer in thousands of km.
Answer: 768000km
Explanation:
Velocity is given by the relation between the distance [tex]d[/tex] and the time it takes to travel that distance [tex]t[/tex]:
[tex]V=\frac{d}{t}[/tex] (1)
In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":
[tex]t=2.56s[/tex] (2)
In addition, we know radio waves are electromagnetic waves (light), and its velocity is:
[tex]V=3(10)^{8}m/s[/tex] (3)
Substituting (2) and (3) in (1):
[tex]3(10)^{8}m/s=\frac{d}{2.56s}[/tex] (4)
And finding [tex]d[/tex]:
[tex]d=(3(10)^{8}m/s)(2.56s)[/tex] (5)
Finally we can obtain the distance:
[tex]d=768000000m=768000km[/tex]
Answer:
384,000 km
Explanation:
Given
Velocity of radio wave [tex]v = 3.00 \times 10^{8}m/s[/tex]
Duration of echo T = 2.56 s
Solution
Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s
Since any time delays in the electronic equipment can be ignored
time taken for the radio wave to reach moon
[tex]t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s[/tex]
[tex]v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km[/tex]
Consider a mass-spring system. The spring has a spring constant 2.17e 3 N/m. On the end is a mass of 4.71 kg. You start at equilibrium with an initial velocity of 1.78 m/s. What is the maximum displacement
Answer:
7.4 cm
Explanation:
K = 2.17 x 10^3 N/m
m = 4.71 kg
v = 1.78 m/s (It is maximum velocity)
The angular velocity
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]\omega =\sqrt{\frac{2.71\times 10^{3}}{4.71}}[/tex]
ω = 24 rad/s
Maximum velocity, v = ω x A
Where, A be the maximum displacement
1.78 = 24 x A
A = 0.074 m = 7.4 cm
Final answer:
The maximum displacement of the mass-spring system is 0.035 m.
Explanation:
If we have a mass-spring system with a spring constant of 2.17e3 N/m and a mass of 4.71 kg attached to it, we can calculate the maximum displacement using the formula for potential energy in a spring: U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium. At maximum displacement, the potential energy is converted to kinetic energy, which is given by K = (1/2)mv^2, where m is the mass and v is the velocity. Equating the two equations and solving for x, we can find the maximum displacement.
First, let's find the equilibrium position by taking the weight of the mass as force due to gravity: F = mg, where m is the mass and g is the acceleration due to gravity. Then, we can calculate the maximum displacement using the formula x = (U/mg)^0.5.
Plugging in the given values, we have: x = ((1/2)(2.17e3 N/m)(x)^2) / (4.71 kg)(9.8 m/s^2), which gives us x = 0.035 m.
An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 24 mm horizontally.
Explanation:
It is given that,
Speed of the electron in horizontal region, [tex]v=1.9\times 10^7\ m/s[/tex]
Vertical force, [tex]F_y=4.9\times 10^{-16}\ N[/tex]
Vertical acceleration, [tex]a_y=\dfrac{F_y}{m}[/tex]
[tex]a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}[/tex]
[tex]a_y=5.37\times 10^{14}\ m/s^2[/tex]..........(1)
Let t is the time taken by the electron, such that,
[tex]t=\dfrac{x}{v_x}[/tex]
[tex]t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}[/tex]
[tex]t=1.26\times 10^{-9}\ s[/tex]...........(2)
Let [tex]d_y[/tex] is the vertical distance deflected during this time. It can be calculated using second equation of motion:
[tex]d_y=ut+\dfrac{1}{2}a_yt^2[/tex]
u = 0
[tex]d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2[/tex]
[tex]d_y=0.000426\ m[/tex]
[tex]d_y=0.426\ mm[/tex]
So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.
Answer:
[tex]y = 4.24 *10^{-4} m[/tex]
Explanation:
the vertical accerlaration acting on the electron
[tex]a = \frac{f_y}{m}[/tex]
[tex]a = \frac{4.9*10^{-16}}{9.11*10^{-31}} = 5.37 *10^{14} m/s^{2}[/tex]
the time taken by the electron to cover the horizontal distance is
[tex]t =\frac{24*10^{-3}}{1.9*10^{7}}[/tex]
[tex]t = 1.26 *10^{-9} s[/tex]
[tex]v_i = o[/tex]
the vertical distance trvalled in time t is
[tex]y = v_i*t +\frac{1}{2} a_y*t^{2}[/tex]
[tex]y = 0 +\frac{1}{2}*(5.37 *10^{14})(1.26 *10^{-9})^{2}[/tex]
[tex]y = 4.24 *10^{-4} m[/tex]
Which type of energy is thermal energy a form of?
Answer:
Kinetic Energy
Explanation:
Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.
The pressure inside a latex balloon is nearly the same as the pressure outside. If you let a helium balloon go, and if, as it rises, it stays at a constant temperature, the volume of the balloon will:A. stay the sameB. IncreaseC. Decrease
Answer:
Option (B)
Explanation:
As the balloon is filled with helium, it rises up because the density of helium is less than the density of air. As it rises up the outside pressure means atmospheric pressure goes on decreasing and thus the inside pressure increases. It results teh volme of the balloon increases.
A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water flow is 6.31kg/s, and the nozzle speed is 12.5 m/s. The hose passes vertically between the fireman’s feet, which are 1.30 m below the nozzle. Choose the origin to be inside the hose between the fireman’s feet. What torque must the fireman exert on the hose? (This could also be stated, what is the rate of change of the angular momentum of the water?)
Answer:
A torque of 102.5375 Nm must be exerted by the fireman
Explanation:
Given:
The rate of water flow = 6.31 kg/s
The speed of nozzle = 12.5 m/s
Now, from the Newton's second law we have
The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction
thus we have,
F = 6.31 kg/s x 12.5m/s
or
F = 78.875 N
Now,
The torque (T) exerted by water force about the fireman's will be
T = (F x d)
or
T = 78.875 N x 1.30 m
T = 102.5375 Nm
hence,
A torque of 102.5375 Nm must be exerted by the fireman
The torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.
What is Newtons second law of motion?Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
For the flow of water, the second law of motion can be given as,
[tex]F=Q\times v[/tex]
As he rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. Thus the force of this can be given as,
[tex]F=6.31\times12.5\\F=78.875\rm N[/tex]
The torque for the fireman exert on the hose is equal to the product of force applied and the distance traveled. Therefore the value of torque is,
[tex]\tau=78.875\times1.30\\\tau=102.5375\rm Nm[/tex]
Thus, the torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.
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if we want to describe work, we must have
Energy and time since,
[tex]W=\dfrac{E}{t}[/tex]
Hope this helps.
r3t40
A 55.6-kg skateboarder starts out with a speed of 2.44 m/s. He does 80.4 J of work on himself by pushing with his feet against the ground. In addition, friction does -244 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 7.24 m/s. (a) Calculate the change (PEf - PE0) in the gravitational potential energy. (b) How much has the vertical height of the skater changed? Give the absolute value.
Applying the work-energy theorem, which states that the work done by nonconservative forces equates to the change in kinetic plus potential energy, the skateboarder's change in potential energy results in -18.4J, and the absolute change in height becomes 0.032 m.
Explanation:The student is asking about the concept of conservation of energy in physics, particularly in the context of the work-energy theorem. This theorem states that work done by nonconservative forces (like friction) is equal to the change in the kinetic and potential energy of the system. According to the given problem, the initial kinetic energy of a 55.6-kg skateboarder is KE0 = 0.5 * m * v², and after performing 80.4J of work on himself, while friction does -244J of work on him, the final kinetic energy becomes KEf = m * v² / 2 where 'v' is the final speed of 7.24 m/s.
(a) Considering that there are no conservative forces, the work-energy theorem states that Wnc = ΔKE + ΔPE. From here, we can calculate that ΔPE or PEf - PE0 results in -18.4J.
(b)The change in vertical height can be obtained by Δh = ΔPE / (m*g), where 'g' is the gravitational constant. Thus, Δh =-18.4J divided by (55.6 * 9.8), which results in -0.032 m, but since we are asked for the absolute value, the height change is 0.032 m.
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Suppose that on a hot (33.0°C) and sticky (80% humidity) afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab (volume of 180.0 m³) was 1.00 atm before the storm and 0.800 atm after the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
1. 211 m3
2. 134 m3
3. 1800 m3
4. 7,150 m3
Answer:
V₂ = 225 m^{3}
so no exact option is match with the calculated answer.
Explanation:
According to Boyle's Law, we have
P_1 V_1 = P_2 V_2 ----------- (1)
Data Given;
P_1 = 1.0 atm
V_1 = 180 m³
P_2 = 0.80 atm
V_2 = ?
Solving equation 1 for V₂,
[tex]V_2= \frac{P_1 V_1}{P_1}[/tex]
Putting values,
[tex]V_2 = \frac{1.0*180}{0.80}[/tex]
[tex]V_2 = 225 m^{3}[/tex]
so no exact option is match with the calculated answer.
Answer:
E
Explanation:
A sensor on a traffic light is most likely to produce electromagnetic waves at which of these frequencies?
Answer:
Option-(D): 10¹¹ waves per second.
Explanation:
Electromagnetic waves:The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).
Answer:
answer is A
Explanation:
A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank (Fig. P11.43). The width of the box is 75.0 cm, and sand is to be distributed uniformly throughout it. The center of gravity of the nonuniform plank is 50.0 cm from the right end. What mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed just below its midpoint?
Answer:
Required mass of sand is 20 kg
Explanation:
Given:
Mass of the plank = 25 kg
Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]
Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]
Balancing the torque due to the plank and the sand box with respect to the fulcrum
Torque = Force × perpendicular distance
thus, we get
(25 × g) × 0.5 = weight of sand × 0.625
where, g is the acceleration due to gravity
or
(25 × g) × 0.5 = (mass of sand × g) × 0.625
or
mass of sand = 20 kg
Hence, the required mass of the sand is 20 kg
Answer:
20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.
Explanation:
Use the second condition of equilibrium.
[tex]$\sum} \tau=0$[/tex]
[tex]MgL-$M g x_{c m}=0$[/tex]
[tex]$M=\frac{m x_{c m}}{L}[/tex]
[tex]=\frac{25(0.50)}{0.625}[/tex]
[tex]=20 \mathrm{~kg}$[/tex]
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According to the ___________________ hypothesis, each emotion comes with its own specific profile of autonomic activity (heart rate, skin conductance, and finger temperature).
Answer:
autonomic specificity
Explanation:
The hypothesis which states that the different emotions in our body involve different and unique physiological profiles is called autonomic specificity hypothesis.
Viewed from this hypothesis perspective, emotions in the human body can be seen as time-tested solutions to timeless problems and challenges. With emotions, evolution in the human body has provided the human with at least one generalized response to tackle these problems.
When a balloon is deflating, why does air leave the balloon?
Answer:
When a balloon deflates air moves out of the balloon because the pressure inside the balloon is higher than the pressure outside the balloon.
Explanation:
An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.
The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.
This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.
Final answer:
Air leaves a deflating balloon as the cooler air inside has lower pressure than the outside, causing the higher pressure external air to push its way in, equalizing the pressure until it's the same inside and outside the balloon.
Explanation:
When a balloon is deflating, air leaves the balloon due to the difference in pressure between the air inside the balloon and the air outside. The air inside the balloon is initially at a higher pressure than the surrounding air. According to the principles of thermal expansion, the air in the balloon cools and its pressure decreases. As the pressure inside becomes lower than the pressure outside, air naturally flows out to equalize the pressure.
The buoyancy that lifts a hot-air balloon is based on the air's density inside being lower than outside. When a hot air balloon is inflating, the air inside becomes less dense due to heating, and it floats. However, as the air cools or if there is a leak in the balloon's material, the air inside loses its buoyancy and starts to deflate, releasing air until pressure equilibrium is reached or until the material cannot hold the air any longer.
easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.
Answer:
1.034 m above the floor
Explanation:
The location of center of body for a compound body, when the weights are given is calculated as:
[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]
where,
[tex]\bar x[/tex] is the center of gravity of the entire body
W = weight of the individual body
x = center of gravity of the individual body
Thus on substituting the values we get,
[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]
or
[tex]\bar x = \frac{691.83}{669}[/tex]
or
[tex]\bar x =1.034m[/tex]
Hence, the center of gravity of the entire body lies 1.034 m above the floor
Which ion channels mediate the falling phase of an action potontial?
Answer:
Voltage-gated K+ channels
The correct answer is that potassium ion channels mediate the falling phase of an action potential.
During the falling phase of an action potential, the membrane potential of the neuron must return to its resting state after being depolarized. This is primarily achieved through the efflux of potassium ions (K^+) out of the cell. The opening of voltage-gated potassium channels allows for this efflux, which repolarizes the membrane potential back towards the resting membrane potential.
Here is the sequence of events during an action potential:
1. Resting potential: The neuron is at its resting membrane potential ,typically around -70 mV, due to the concentration gradients of ions across the membrane and the selective permeability of the membrane to potassium ions via leak channels.
2. Rising phase (Depolarization): When a stimulus reaches the threshold level, voltage-gated sodium channels open, allowing sodium ions (Na^+) to rush into the cell. This influx of positive charges depolarizes the membrane potential towards +30 mV.
3. Peak of the action potential: The membrane potential reaches its peak when the sodium channels become inactivated, stopping the influx of sodium ions.
4. Falling phase (Repolarization): Voltage-gated potassium channels open, and potassium ions move out of the cell, restoring the membrane potential towards the resting potential. The movement of potassium ions out of the cell is slower than the initial sodium influx, which is why the falling phase is slower than the rising phase.
5. Overshoot: Sometimes, the membrane potential temporarily overshoots the resting potential, becoming more negative than at rest, due to the continued efflux of potassium ion.
6. Return to resting potential: The potassium channels close, and the sodium-potassium pump restores the original ion distribution across the membrane, bringing the membrane potential back to the resting potential.
In summary, the falling phase of an action potential is mediated by the opening of voltage-gated potassium channels, which allows for the efflux of potassium ions, thereby repolarizing the neuron's membrane potential."
A 26.0 kg wheel, essentially a thin hoop with radius 1.30 m, is rotating at 297 rev/min. It must be brought to a stop in 23.0 s. (a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.
Answer:
Work done on the loop to stop it = 21,269.1496 J
Average power P = 924.7456 watt
Explanation:
Mass of the wheel M = 26.0 Kg
Radius r = 1.30 m
The wheel is rotating at a speed of 297 rev/min
1 minute = 60 seconds,
So,
The wheel is rotating at a speed of 297/60 rev/sec
Initial angular speed (ω₁) = 2π(297/60) = 31.1143 rad/s
Final angular speed (ω₂) = 0 rad/s
Time taken to stop , (t) = 23 s econds
Moment of inertia I of a circular hoop around its central axis = Mr²
Where m is the mass of the wheel and r is the radius of the wheel
Thus, I = 26.0×(1.30)² Kgm² = 43.94 Kgm²
(a)
Work done to stop it is the difference in the kinetic energy of the initial and the final system. So,
Work done W = (1/2)I(ω₂² - ω₁²) = 0.5*43.94(0 - (31.1143)²) = -21,269.1496 J
Thus, work done on the loop to stop it = - 21,269.1496 J
The answer has to be answered in absolute value so, Work = |-21,269.1496 J| = 21,269.1496 J
(b)
Average power P = |W|/t = 21,269.1496 J/23.0 s = 924.7456 watt
Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow ________ about the center. a.counterclockwise and inward b.clockwise and outward
In a Northern Hemisphere extratropical cyclone, the surface winds blow counterclockwise and inward due to the Coriolis effect, a result of the Earth's rotation.
Explanation:Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow counterclockwise and inward about the center. This is due to the Coriolis effect, a phenomenon caused by the Earth's rotation which influences the direction that wind travels across the globe. It results in winds in the Northern Hemisphere curving to the right, thus causing cyclone winds to move counterclockwise.
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In a Northern Hemisphere extratropical cyclone, the surface winds blow in a counterclockwise and inward direction. This pattern is influenced by the Coriolis force, which causes winds to be deflected to the right, leading to this counterclockwise rotation. The phenomenon also aligns with the fact that air is attracted towards low-pressure centers like cyclones.
Explanation:In the Northern Hemisphere, when viewed from above, surface winds around an extratropical cyclone flow in a counterclockwise and inward direction. This phenomenon is influenced by the Coriolis force, which causes winds to be deflected to the right in the Northern Hemisphere. This deflection results in a counterclockwise rotation. The Coriolis force similarly influences tropical cyclones, causing them to display the same pattern of rotation.
A key point to note is that this rotation is associated with areas of low pressure, such as the center of these cyclone systems. This low-pressure center attracts air, causing winds to flow inward. As the air rises within these low-pressure zones, it cools and forms clouds, making such cyclonic weather patterns visible from space.
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A gas is compressed by an adiabatic process that decreases its volume by a factor of 2.In this process, the pressurea) increases by a factor of more than 2.b) increases by a factor of 2.c) does not change.d) increases by a factor of less than 2.
Answer:
a) Increase by a factor of more than 2
Explanation:
The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.
Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:
[tex]P=\frac{RT}{v}[/tex]
Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.
If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.
Julie and Eric row their boat (at a constant speed) 63 miles downstream for 7 hours helped by the current. Rowing at the same rate, the trip back against the current takes 9 hours. Find the rate of the boat in still water.
Answer:
Boat speed = 8 miles/hr
Explanation:
Let the speed of the boat be U
Let the speed of the current be V.
Therefore, the downstream speed is U+V
and the upstream speed is U-V
Now we know that Speed = distance / time
Therefore, downstream speed, U+V = 63 / 7
U+V = 9 miles/hr ------(1)
Upstream speed, U-V = 63 / 9
U-V = 7 miles/hr --------(2)
Therefore subtracting (2) from (1), we get
( U+V) - ( U-V ) = 9-7
2V = 2
V = 1 miles/hr
Therefore the speed of the current is V = 1 mile/hr
Now from (1) we get
U+V = 9
U+1 = 9
U = 8
Therefore, the speed of the boat is U = 8 miles/hr
A box at rest on a ramp is in equilibrium, as shown.
What is the force of static friction acting on the box? Round your answer to the nearest whole number.
_______N
What is the normal force acting on the box? Round your answer to the nearest whole number.
_______ N
Answer:
Ffs = 251 N
Fn = 691 N
Explanation:
Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.
The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°. Therefore:
Fgx = Fg sin 20°
Fgy = Fg cos 20°
Sum of the forces in the x direction (parallel to the ramp):
∑F = ma
Ffs − Fgx = 0
Ffs = Fgx
Ffs = Fg sin 20°
Ffs = 735 sin 20°
Ffs ≈ 251
Sum of the forces in the y direction (normal to the ramp):
∑F = ma
Fn − Fgy = 0
Fn = Fgy
Fn = Fg cos 20°
Fn = 735 cos 20°
Fn ≈ 691
To answer that question we need to apply equations of movement ( from Newton´s laws )
In equilibrium:
∑F = 0 or ∑Fx = 0 ; ∑Fy = 0
Solution is:
a) F(sf) = 251 [N]
b) Fn = 691 [N]
From the attached drawings we can see: ( Body free diagram)
∑ Fₓ = F(sf) - Pₓ = 0 where P = m×g = 735 [N] ( the weigth)
and Pₓ = P× cos20°
Then F(sf) = Pₓ × sin 20°
F(sf) = m×g×cos20° = 735× 0.34202 [N]
F(sf) = 251.3847
F(sf) = 251 [N]
rounding to the nearest number
F(sf) = 691 [N]
∑ Fy = 0
∑ Fy = Fn - Py = 0 Py = P×cos20° Py = m×g×cos20°
Py = 735×0.939693 [N] Py = [N]
Fn = Py = 690.674 [N]
rounding to the nearest whole number
Fn = 691 [N]
Related question Brainly.com/question/ 11544804
Bob can row 14 mph in still water. The total time to travel downstream and return upstream to the starting point is 4 hours. If the total distance downstream and back is 42 miles, determine the speed of the river (current speed).
Answer:
7 mph
Explanation:
Let the speed of river is c mph.
The effective downstream speed = 14 + c
The effective upstream speed = 14 - c
Total time = 4 hrs
Total distance = 42 miles
Time for upstream + time for downstream = 4 hrs
21 / (14 + c) + 21 / (14 - c) = 4
21 (14 - c + 14 + c) = 4 (196 - c^2)
21 x 28 = 4 (196 - c^2)
c^2 = 196 -147 = 49
c = 7 mph
The speed of the river is 7 miles/h.
Speed of Bob in still water is 14 miles/h.
Speed of Bob downstream = (14 + x) miles/h
Speed of Bob upstream = (14 - x) miles/h
here x = speed of the river.
The total distance downstream and back is 42 miles. Hence,
Distance upstream = Distance downstream = 21 miles
Given that total time of journey upstream and downstream is 4 hours. Then, using the formula:
[tex]t = \frac{distance}{speed}[/tex], we get:
[tex]\frac{21 \hspace{0.5mm} miles} {(14 + x) \hspace{0.5mm} miles/h} + \frac{21 \hspace{0.5mm} miles}{(14 - x)\hspace{0.5mm}miles/h} =4 \hspace{0.5mm} hours[/tex]
or, [tex]\frac{2 \times 21 \times 14 \hspace{0.5 mm} h}{14^{2} -x^{2} } = 4 \hspace{0.5mm} hours[/tex]
or, 21 × 7 h = 14² - x²
or, x² = 49
or, x = 7 miles/h
A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 3 feet per minute. Find the rate at which the area is changing at the instant the radius is 4 feet. When the radius is 4 feet, the area is changing at approximately nothing square feet per minute. (Type an integer or a decimal. Round to the nearest thousandth as needed.)
Answer:
Rate of change of area is [tex]75.398ft^{2}/sec[/tex]
Explanation:
[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]
Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]
Two airplanes leave an airport at the same time and travel in opposite directions. One plane travels 87 km/h faster than the other. If the two planes are 11,865 km apart after 7 hours, what is the rate of each plane?
Answer:
Speed of A = 891 km/h
Speed of B = 804 km/h
Explanation:
Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.
So, the speed of aeroplane A is 87 + v.
Distance traveled by A after 7 hours, d1 = (87 + v) x 7
Distance traveled by B after 7 hours, d2 = v x 7
Total distance traveled = 11865 km
So, d1 + d2 = 11865
(87 + v) x 7 + v x 7 = 11865
609 + 14 v = 11865
14 v = 11256
v = 804 km/h
So, the speed of A = 87 + 804 = 891 km/h
Speed of B = 804 km/h
Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?
A. Weather data are misinterpreted.
B. Weather instruments are extremely precise.
C. Interference from the sun causes data to be collected inaccurately.
D. Local variations in weather are too small for weather instruments to replicate.
Answer:
C. Interference from the sun causes data to be collected inaccurately.
Explanation:
Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.
A neutron star and a white dwarf have been found orbiting each other with a period of 28 minutes. If their masses are typical, what is their average separation? Compare the separation with the radius of the sun, or about 0.005 AU. (Hints: Refer to Kepler's third law with regard to mass. Assume the mass of the neutron star is 2.5 solar masses and the mass of the white dwarf is 0.3 solar mass.)
Answer:
The average separation is 0.002041 AU
Explanation:
Given data:
Mass of the neutron star, M₁ = 2.5[tex]M_{solar}[/tex]
Mass of the White dwarf, M₂ = 0.3[tex]M_{solar}[/tex]
Orbiting period (P)= 28 minutes
1 year = 365 × 24 × 60 minutes = 525600 minutes
or
1 minute = 1/525600 years
thus, 28 minutes = 28/525600 years = 5.327 × 10⁻⁵ years
now from the Kepler's third law we have,
MP² = a³
where, P is the period
M is the mass = M₁ + M₂
a is the size of the orbit
thus, by substituting the values in the equation we get
(2.5[tex]M_{solar}[/tex]+0.3[tex]M_{solar}[/tex])(5.327 × 10⁻⁵ years)² = a³
Also,
[tex]M_{solar}=1[/tex] when planets orbiting sun
thus,
2.8 ×(5.327 × 10⁻⁵ years)² = a³
or
a³ = 7.94 × 10⁻⁹
or
a = 1.99 × 10⁻³ AU
thus, the average separation is 0.001995 AU
Now
1 AU = 1.5 × 10⁸ km
thus,
0.001995 AU = 299281.61 km = 2.99 × 10⁵ km
in terms of sun's radius = (2.99 × 10⁵ km)/(7 ×10⁵) = 0.427
Thus, the this orbit system will fit inside the sun
A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic field = 0.75 T . The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?
Answer:
i = 7.777 × 10⁻⁴ A = 0.77 mA
Explanation:
Given:
loop radius, r = 3.0 cm = 0.03 m
Area, A = π x r² = π x 0.03² = 0.0028 m²
Magnetic Field, B = 0.75 T
Loop resistance, R = 18 Ω
time, t = 0.15 seconds
Now,
the induced emf is given as:
EMF = [tex]-\frac{BA}{t}[/tex]
also
EMF = i x R
Where, i is the current flowing
equating both the formulas for EMF, we get
[tex]{i}{R}=-\frac{BA}{t}[/tex]
or
[tex]{i}=-\frac{BA}{tR}[/tex]
substituting the values in the above equation we get
[tex]{i}=-\frac{0.75\times 0.0028}{0.15\times 18}[/tex]
or
the magnitude of the current, i = 7.777 × 10⁻⁴ A = 0.77 mA
The electric current produced in a loop due to a changing magnetic field can be calculated using Faraday's law of electromagnetic induction. Based on the given data, the magnitude of the induced current in the loop is approximately 1.18 mA.
Explanation:The scenario described in the question depicts a situation where a magnetic field decreases uniformly in magnitude, thus inducing an electric current within a circular loop according to Faraday's law of electromagnetic induction.
The electric current produced in the loop as a result of the change in the magnetic field can be calculated using the expression, I = ΔΦ/ Δt * R, where ΔΦ is the change in magnetic flux, Δt is the time interval, and R is the resistance of the loop.
In this case, the initial magnetic flux, Φ1 = B1 * A, where B1 is the initial magnetic field and A is the area of the loop. Since the magnetic field decreases to zero, the final magnetic flux, Φ2, is zero. So, ΔΦ = Φ2 - Φ1 = - B1 * A. Here, A = πr², where r is the radius of the loop.
So, I = - [ (B1* πr²) / ( Δt * R ) ]. Substituting the given values into the equation, we get I = (0.75 T * π * (0.03 m)²) / ( 0.15 s * 18 Ω ) = 0.00118 A or 1.18 mA.
So, the magnitude of the induced current is approximately 1.18 mA.
Learn more about Magnetic Flux Change and Induced Current here:https://brainly.com/question/31955117
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A 73 kg man in a 7.4 kg chair tilts back so that all his weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area of radius 1.5 cm. Find the pressure exerted on the floor by each leg. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.A 73 kg man in a 7.4 kg chair tilts back so that all his weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area of radius 1.5 cm. Find the pressure exerted on the floor by each leg. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.
Answer:
55738.539 Pa
Explanation:
Gvien:
The mass of the man = 73 kg
Mass of the chair = 7.4 kg
radius of the leg of the chair = 1.5 cm = 0.015 m
Now,
the force of gravity due to both the masses, F = (73+7.4) kg x 9.8 = 787.92 N
Also,
Pressure = force / area
Area of a circle of leg = πr² = π x 0.015² = 7.068 x10⁻⁴ m2
Now,
there are 2 legs so the force will be divided evenly in each leg
thus,
F' = [tex]\frac{787.92}{2}=393.96N[/tex]
Hence, the pressure on each leg will be
Pressure = [tex]\frac{393.96N}{7.068\times 10^{-4}}=55738.539 Pa[/tex]
When the temperature goes up 3^\circ on the Cantor scale, it goes up 8^\circ on the Frobenius scale. On both scales, 18^\circ is the same temperature. How many Frobenius degrees are equal to 30^\circ Cantor?
Answer:
50°
Explanation:
It is given that the both scales are the same at 18˚
Now,
at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above
the 18˚ mark.
also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,
mathematically, we can write it as (by unitary method)
3˚ increase in the temperature on the Cantor = 8˚ increase on the Frobenius scale
or
1˚ increase in the temperature on the Cantor = (8/3)˚ increase on the Frobenius scale
thus, for x˚ increase in the temperature on the Cantor = ((8/3)˚ × x) increase on the Frobenius scale
hence, for 12° increase we have
12˚ increase in the temperature on the Cantor = ((8/3)˚ × 12) increase on the Frobenius scale
or
12˚ increase in the temperature on the Cantor = 32° increase on the Frobenius scale
hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.
Car A
Mass: 1,500 kg
Velocity: 10 m/s
Car B
Mass: 1,500 kg
Velocity: 25 m/s
Car C
Mass: 1,000 kg
Velocity: 10 m/s
which order shows decreasing momentum?
~A, B, C
~B, A, C
~C, B, A
Answer:
B,A,C
Explanation:
Hope this helps!!!
Answer:
B, A, C
Explanation:
Momentum:
[tex]P = m*v[/tex]
[tex]P = momentum[/tex]
[tex]m = mass[/tex]
[tex]v = velocity[/tex]
Momentum of car A
[tex]P_{A} = 1500kg*10m/s = 15,000kg*m/s[/tex]
Momentum of car B
[tex]P_{B} = 1500kg*25m/s = 37,500kg*m/s[/tex]
Momentum of car C
[tex]P_{C} = 1000kg*10m/s = 10,000kg*m/s[/tex]
In decreasing order:
B, A, C