Answer:
temperature is usually uniform throughout a cell or temperature is usually uniform to do work.
Explanation:
Heat (thermal energy) is a kinetic energy. It s connected with the random movement of the atoms or molecules. The temperature is usually uniform throughout a cell, so most of the cells cannot harness heat to perform work.
Cells are unable to harness heat to perform work due to the second law of thermodynamics, which results in energy being lost in a form that is unusable, often as heat, during transfers and transformations. This heat energy is essentially lost to the cell for performing work. And as entropy increases, less energy becomes available for work.
Explanation:Most cells cannot harness heat to perform work because of the second law of thermodynamics. This law states that all energy transfers and transformations are never completely efficient, with some amount of energy being lost in a form that is unusable, commonly as heat energy. Strictly speaking, heat energy is defined as the energy transferred from one system to another that is not doing work.
For example, during cellular metabolic reactions, a portion of the energy is lost as heat energy. Despite it contributing to maintaining the body temperature of warm-blooded creatures, it is essentially lost to the cell for performing work. This demonstrates how the second law of thermodynamics makes the tasks of cells obtaining, transforming, and using energy to do work more difficult than they appear.
Taking into account the entropy, less and less energy in the universe is available to do work as entropy increases. Eventually, as all fuels are exhausted and temperatures equalize, it will become impossible for heat engines to function, or for work to be done. Hence, cells can't harness heat to perform work due to the inefficiencies described by the second law of thermodynamics.
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Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.
Yes, the voltage across a capacitor in an RLC series circuit with AC can be greater than the source voltage, especially at resonance where the capacitive and inductive reactances cancel each other out and energy oscillates between the capacitor and inductor.
Explanation:In an RLC series circuit with alternating current (AC), it is indeed possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source. This phenomenon occurs due to resonance in the circuit, where the capacitive and inductive reactances can cancel each other out at a certain frequency, known as the resonant frequency. When the circuit is at resonance, the voltages across the capacitor and inductor can be much greater than the source voltage because of the energy oscillating between the electric field of the capacitor and the magnetic field of the inductor.
According to the equation Vc = Q/C, where Q is the charge and C is the capacitance, we see that the voltage across the capacitor (Vc) is directly proportional to the amount of charge stored and inversely proportional to the capacitance. In an AC RLC circuit at resonance, the charge can oscillate at amplitudes resulting in voltages across the capacitor that exceed the source voltage.
The same can be true for the voltage across the inductor. If the circuit is driven at or near its resonant frequency, the inductive reactance and capacitive reactance can become equal in magnitude but opposite in phase, leading to a situation where the voltage across the inductor also exceeds the source voltage due to the energy stored in its magnetic field.
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Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degrees above the boiling point of water. This can happen because ________.
Answer:
High pressure inside the giant planet
Explanation:
As we move in the interior of the giant planet, the pressure and temperature in the interior of the planet increases. Since, the giant planets have hardly any solid surface and thus they are mostly constituted of atmosphere.
Also, the gravitational forces keep even the lightest of the matter bound in it contributing to the large mass of the planet.
If we look at the order of the magnitude of the temperature of these giant planets than nothing should be able to stay in liquid form but as the depth of the planet increases with the increase in temperature, pressure also increases which keeps the particle of the matter in compressed form.
Thus even at such high order of magnitude water is still found in liquid state in the interior of the planet.
A projectile is launched from the earth’s surface at initial speed v0 at angle θ0 with the horizontal. When the projectile is at its maximum height h, it has half the speed it had when it was at half its maximum height h 2 . At what angle was the projectile launched.
Answer:
θ₀ = 67.79°
Explanation:
Given info
we know that
Ymax = v₀y² / (2g)
v = v₀x (when Y = Ymax)
when the projectile was at half its maximum height (Y')
v' = 2v = 2*v₀x
we can use the equations
(v')²= v'x² + v'y² ⇒ (2*v₀x)² = (v₀x)² + v'y² ⇒ v'y² = 3*(v₀x)² (I)
if we know that
v'y² = v₀y² - 2*g*(y')
y' = Ymax /2 = (v₀y² / (2g)) / 2 = v₀y² / (4g)
then
v'y² = v₀y² - 2*g*(v₀y² / (4g)) = v₀y² - (v₀y² / 2) = v₀y² / 2
⇒ v'y² = v₀y² / 2 (II)
we can say that (I) = (II)
3*(v₀x)² = v₀y² / 2 ⇒ v₀y = √6*v₀x
Finally we apply
tan θ₀ = v₀y / v₀x
⇒ tan θ₀ = √6*v₀x / v₀x = √6
⇒ θ₀ = tan⁻¹ (√6) = 67.79°
A long straight rod experiences several forces, each acting at a different location on the rod. All forces are perpendicular to the rod. The rod might be in translational equilibrium, rotational equilibrium, both, or neither.
(1) If a calculation reveals that the net torque about the left end is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net force on the rod is also zero.
(C) might not be in rotational equilibrium even if the net force on the rod is also zero.
(D) might be in rotational equilibrium even if the net force is not zero.
(2) If a calculation reveals that the net force on the rod is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net torque about every axis through anyone point is found to be zero.
(C) might be in rotational equilibrium if the net torque about every axis through anyone point is found to be zero.
(D) might be in rotational equilibrium even if the net
torque about any axis through anyone point is not zero.
Answer:
1. C
2.C
Explanation:
1. The rod is perpendicular to every axis and forces are acting on every location. If the torque on the left side is zero, this indicates that forces with respect to their distance on the left side is zero and doesn't account for the net force at a point.
2. If the net torque about every point on every axis is zero, the rod will be rotational because each axis will yield a magnitude of zero which obeys the principle of rotation at a point.
When the net torque on a rod is zero about the left end, the rod is in rotational equilibrium. If the net force is zero, the rod is in translational equilibrium, and rotational equilibrium depends on the net torque being zero about every axis.
Equilibrium Conditions in Physics
To determine whether a rod is in equilibrium under the influence of several forces, we need to examine the conditions for both translational and rotational equilibrium. The conditions necessary for equilibrium are quite straightforward:
The net external force on the system must be zero (net F = 0), ensuring there is no linear acceleration.
The net external torque must also be zero (net T = 0), preventing any angular acceleration.
Addressing the questions:
If the net force on the rod is zero, it indicates that the rod is in translational equilibrium. However, for the rod to be in rotational equilibrium, we must also confirm that the net torque about every axis through anyone point is zero (Option B).
Each static equilibrium condition is critical in itself. While the zero net force ensures no linear acceleration, the zero net torque ensures no rotational acceleration. Both conditions are required for complete static equilibrium of an object.
A rocket ship starts from rest in turns on it's forward booster rocket causing it to have a constant acceleration of 4m/s*rightward. After 3s, what will be the velocity of the rocket ship
The velocity after 3 s is 12 m/s rightward
Explanation:
The motion of the rocket is a motion at constant acceleration, therefore we can apply suvat equations:
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time interval
For the rocket in this problem:
u = 0 (the rocket starts from rest)
t = 3 s
[tex]a=4 m/s^2[/tex] rightward is the acceleration
Solving for v, we find the final velocity:
[tex]v=0+(4)(3)=12 m/s[/tex] (rightward)
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A car and driver weighing 7130 N passes a sign stating...?
"Bridge Out 32 m Ahead." She slams on the brakes, and the car decelerates at a constant rate of 13.8 m/s^2. The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the work done stopping the car if the car just stops in time to avoid diving into the water?
Answer:
321 280 J
Explanation:
Work done = force * distance
The distance is 32 m
The force can be calculated using the second law of motion
F = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 N
Work done = force * distance
= 10 040 N * 32 m
= 321 280 J
Answer:
321 280
Explanation:
Work done = force * distanceThe distance is 32 mThe force can be calculated using the second law of motionF = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 NWork done = force * distance = 10 040 N * 32 m = 321 280 J
rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 seconds. Determine the distance Rickey slides across the ground before touching the base.
Answer:
15.825 m
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 6.75 m/s
v = Final velocity = 5.91 m/s
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m[/tex]
The distance Rickey slides across the ground before touching the base is 15.825 m
A chunk of aluminum at 91.4°C was added to 200.0 g of water at 15.5°C. The specific heat of aluminum is 0.897 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.9°C. Assuming no heat was lost to the surroundings, what was the mass of aluminum added?
The mass of the aluminum added is calculated through the principle of conservation of energy, specifically thermal energy. By considering the heat lost by the aluminum and gained by the water, we can rearrange the equation for heat transfer and find that the mass of the aluminum is approximately 37.9 grams.
Explanation:In this physics question, we're looking at a thermodynamic process involving a chunk of aluminum and water. Given the known values of their respective specific heats, the mass of water, and their final equilibrium temperature, we're aiming to find the mass of the aluminum.
We begin by understanding that in a closed system, the heat gained by one body is equal to the heat lost by another. In this case, the aluminum is losing heat, and the water is gaining it. The equation for heat transfer (Q = mcΔT), where m is mass, c is specific heat, and ΔT is change in temperature.
The heat gained by the water = mass of water * specific heat of water * change of temperature in water = 200g * 4.18J/g°C * (18.9°C - 15.5°C) = 2836.4J.
This is equal to the heat lost by the aluminum. Solving the analogous heat equation for the mass of the aluminum gives us the answer:
m = Q / (c * ΔT) = 2836.4J / (0.897J/g°C * (91.4°C - 18.9°C)) = 37.9g
So the mass of the aluminum is approximately 37.9 grams.
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Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation?
The subsequent velocities of the satellite and the remains of the launcher can be calculated using the principle of conservation of momentum. The resultant velocities are approximately 8.70×10−² m/s and 81.5 m/s.
Explanation:The subsequent velocities of the satellite and the remains of the launcher can be calculated using the principle of conservation of momentum. Since there are no external forces acting on the system, the initial momentum of the system is equal to the final momentum. Therefore, the final velocities of the satellite and the launcher can be calculated based on their masses and the kinetic energy supplied to them.
The mass of the satellite, m1 = 4800 kg, and the mass of the launcher, m2 = 1500 kg. The total initial kinetic energy supplied to the system is given as 5000 J. To calculate the velocities, we need to find the ratio of the kinetic energies of the satellite and the launcher.
Let v1 be the velocity of the satellite and v2 be the velocity of the launcher. According to the conservation of momentum, m1 * v1 + m2 * v2 = 0. Also, the total initial kinetic energy supplied to the system is given as 5000 J, where the kinetic energy of the satellite is (1/2) * m1 * v1^2 and the kinetic energy of the launcher is (1/2) * m2 * v2^2.
Using these equations, we can solve for v1 and v2. The subsequent velocities are approximately 8.70×10−² m/s in the direction of motion of the less massive satellite and 81.5 m/s, respectively.
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The satellite moves at approximately 0.704 m/s and the remains of the launcher move at approximately -2.253 m/s in the opposite direction.
To solve for the velocities of the satellite and the launcher remains after separation, we will use the principles of conservation of momentum and the given kinetic energy.
1. Conservation of Momentum
[tex]\[m_s v_s + m_l v_l = 0\][/tex]
Since the total momentum is zero:
[tex]\[4800 \, v_s + 1500 \, v_l = 0\][/tex]
Solving for one velocity in terms of the other:
[tex]\[v_l = - \frac{4800}{1500} v_s = -3.2 v_s\][/tex]
2. Conservation of Energy
The total kinetic energy provided to the system is 5000 J. The kinetic energy of the system is the sum of the kinetic energies of both parts:
[tex]\[\frac{1}{2} m_s v_s^2 + \frac{1}{2} m_l v_l^2 = 5000\][/tex]
Substituting [tex]\( v_l = -3.2 v_s \)[/tex]:
[tex]\[\frac{1}{2} (4800) v_s^2 + \frac{1}{2} (1500) (-3.2 v_s)^2 = 5000\][/tex]
[tex]\[2400 v_s^2 + 750 (10.24) v_s^2 = 5000\][/tex]
[tex]\[2400 v_s^2 + 7680 v_s^2 = 5000\][/tex]
[tex]\[10080 v_s^2 = 5000\][/tex]
Solving for [tex]\( v_s^2 \)[/tex]:
[tex]\[v_s^2 = \frac{5000}{10080} = 0.496 \, \text{m}^2/\text{s}^2\][/tex]
[tex]\[v_s = \sqrt{0.496} \approx 0.704 \, \text{m/s}\][/tex]
3. Determine [tex]\( v_l \)[/tex]
Using [tex]\( v_l = -3.2 v_s \)[/tex]:
[tex]\[v_l = -3.2 \times 0.704 \approx -2.253 \, \text{m/s}\][/tex]
The subsequent velocities of the satellite and the launcher remains after separation are:
- Velocity of the satellite: [tex]\( v_s \approx 0.704 \, \text{m/s} \)[/tex]
- Velocity of the launcher remains: [tex]\( v_l \approx -2.253 \, \text{m/s} \)[/tex]
Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the ice. Determine the magnitude and direction of their post-collision velocity.
Answer:
V = 3.6385 m/s
θ = 47.46 degrees
Explanation:
the important data in the question is:
Skater 1:
[tex]M_1[/tex]= 39.6 kg
direction: south (axis y)
[tex]V_{1iy}[/tex] = 6.21 m/s
Skater 2:
[tex]M_2[/tex] = 52.1 kg
direction: east (axis x)
[tex]V_{2ix}[/tex] = 4.33 m/s
Now using the law of the conservation of linear momentum ( [tex]P_i = P_f[/tex] and knowing that the collision is inelastic we can do the next equations:
[tex]M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex] (eq. 1)
[tex]M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)[/tex] (eq. 2)
Where [tex]V_{sx}[/tex] and [tex]V_{sy}[/tex] is the velocity of the sistem in x and y after the collision.
Note: the conservation of the linear momentum have to be make once by each axis.
Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace [tex]V_{1ix}[/tex] by 0 in the equation and get:
[tex]M_2V_{2ix} = V_{sx}(M_1+M_2)[/tex] (eq. 1)
also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace [tex]V_{2iy}[/tex] by 0 in the equation and get:
[tex]M_{1}V_{1iy} = V_{sy}(M_1+M_2)[/tex] (eq. 2)
Now, we just replace the data in both equations:
[tex](52.1)(4.33) = V_{sx}(39.6+52.1)[/tex] (eq. 1)
[tex](39.6)(6.21) = V_{sy}(39.6+52.1)[/tex] (eq. 2)
solving for [tex]V_{sx][/tex] and [tex]V_{sy}[/tex] we have:
[tex]V_{sx][/tex] = 2.46 m/s
[tex]V_{sy][/tex] = 2.681 m/s
using the pythagoras theorem we can find the magnitude of the velocity as:
V = [tex]\sqrt{2.46^2+2.681^2}[/tex]
V = 3.6385 m/s
For find the direction we just need to do this;
θ = [tex]tan^{-1}(\frac{2.681}{1.46})[/tex]
θ = 47.46 degrees
While a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of ________.
Answer:
Gravity
Explanation:
While a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of GRAVITY.
Gravity is a property of mass of objects. Pressure gravity is the resultant force felt by a particular mass object.
As gravity contact the star surface, density gradient is set up.
Pressure of gravity is a gravitational compression which increases the objects density, compresses the mass of objects and reduces the object's size.
A star might leave the sequence if the star run out of fuel for nuclear fusion in its core.
A 45 kg boy runs at a rate of 2.5 m/s and jumps on top of a stationary skateboard that has a mass of 4 kg. After jumping onto the board, the boy and the board travel off together. Assuming that momentum is conserved, what is the final speed of the boy and the board?
Answer:
v = 2.29 m/s
Explanation:
As we know that the external force on the system of mass of boy + board is ZERO
So here we can use momentum conservation
now we have
[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]
now we have
[tex]45 (2.5) + 4(0) = (45 + 4) v[/tex]
now we have
[tex]v = \frac{45}{49} (2.5)[/tex]
[tex]v = 2.29 m/s[/tex]
Answer:
V = 2.29 m/s
Explanation:
Given that,
Mass of the boy, [tex]m_1=45\ kg[/tex]
Mass of the skateboard, [tex]m_2=4\ kg[/tex]
Initial speed of the boy, v = 2.5 m/s
Let V is the final velocity of the boy and the board. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :
[tex]45\times 2.5=(45+4)V[/tex]
[tex]V=\dfrac{45\times 2.5}{(45+4)}[/tex]
V = 2.29 m/s
So, the velocity of the boat after Batman lands in it 2.29 m/s. Hence, this is the required solution.
The transfer of energy by electromagnetic waves is called
Radiation refers to the transfer of energy by electromagnetic waves, a process that can take place with no medium, such as the heat from the sun reaching Earth.
Explanation:The transfer of energy by electromagnetic waves is known as radiation. This process does not require a medium; the energy is carried by photons in the electromagnetic waves. Examples of this process include the heat produced by the sun, which reaches the Earth via the transfer of radiant energy, and microwaves heating food through radiation.
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The concept that large masses warp the shape of spacetime and thus change the path of an object moving through that spacetime applies to all objects. As such, Einstein’s theory of general relativity predicts that light will also curve around massive objects as it travels through space. Does Newton’s law of gravity agree with that prediction? Note that the mass of a photon is zero.
Answer: Newton's law of gravity theory disagrees with Einstein's theory. The last one were named Theory of General Relativity and was discovered and proposed by the physicist Einstein in the year of 1915.
Explanation: Until the beginning of 20 century, the physics were ruled by Isaac Newton's ideas. He believed that the gravity was a force caused by the objects mass on the space, made them to be draw towards each other. Newton thought that the greater the mass of the object, the more intense was its attraction, which would justify the planet's moovements around the sun and how the gravity between them maintain the planets on solar orbit. Concluding, he believed gravity was a immediate force of action, regardless of the distance of the bodies.
Contrary to Newton's law of gravity, in 1915, the physicist Einstein created the Theory of General Relativity, wich discovered that gravity was, in fact, the deformation caused by the attraction of massive celestial bodies. This deformation, related to the Sun, for example, creates a curvature on the space-time and this curvature are followed by the other planets.
So, we can conclude that Newton's law of gravity disagree's with the Theory of General Relativity, once the first believes that light force of attraction are transmitted instantly and, as Einstein already prooved, the gravity influency propagates in the speed of light.
holiday lights are often connected in series and use special lamps that short out when the potential difference across a lamp increases to the line voltage. generate an explanation why and explain why these light sets might blow their fuses after many bulb
Explanation:
If there isn't the shorting mechanism, the whole set will be blown if anyhow one lamp burns out. Since having blown out several lamps and then shorted, the overall resistance of the remaining operating lamps will be decreased resulting in an increased working current that is adequate to blast the fuse.
Transform boundaries are classified under which type of fault?
Answer:
Strike-slip fault
Explanation:
Transform boundaries play the role of connecting the other plate boundary segments.
When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.
Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement releases energy and takes place in a horizontal direction which is parallel to the force of compression.
Transform boundaries are classified under Transform Faults, where plates slide past each other causing stress and earthquakes, such as at the San Andreas Fault.
Explanation:Transform boundaries are classified under the type of fault known as Transform Faults. These happen at plate boundaries where the plates are not moving away or toward each other, but instead sliding past one another. This movement causes a huge amount of stress, leading to earthquakes. The most recognized transform fault is the San Andreas Fault in California, where the Pacific Plate and the North American Plate are sliding past each other.
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Oil explorers set off explosives to make loud sounds, then listen for the echoes from underground oil deposits. Geologists suspect that there is oil under 485-m-deep Lake Physics. It's known that Lake Physics is carved out of a granite basin. Explorers detect a weak echo 0.920 s after exploding dynamite at the lake surface.
The geologists use echo sounding for oil exploration, where they analyze the time taken by sound waves (created by explosives) to travel from the surface, reflect off oil deposits, and return. Given the speed of sound in granite (5000 m/s) and considering the travel time of the sound wave is two-way, we find the oil deposits to be approximately 2300m under the surface of the Lake Physics.
Explanation:The question revolves around the concept of echo sounding used for oil exploration. Exploders use explosives to create sounds waves that travel down into the earth and reflect back when they hit a boundary between different materials like rock or oil. In this case, the echo was detected 0.920 seconds after the explosion.
To find out the depth of the oil deposit, we need to know the speed of sound in the medium. Sound travels at different speeds in different materials, and in granite, it's typically about 5000 m/s.
We can use the formula for distance in relation to speed and time: d = vt. The time we use in this equation should be the time for the sound to travel to the deposits and back, so we halve the given time to 0.460 s. Substituting the values into the equation, we get: d = 5000 m/s * 0.460 s = 2300 m.
This means that the oil deposits are approximately 2300 m below the surface of the lake.
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Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Part A Determine the ratio of the masses of the two ropes (m1m2).
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
The ratio of the masses of the two ropes, given that the speed of a pulse on rope 1 is 1.4 times the speed on rope 2, is approximately 1:2. This conclusion is attained by applying the concept of pulse velocity in a string or rope, which is governed by the tension and the linear mass density of the material.
Explanation:The principle guiding the speed of a pulse on a rope or string is tied to the tension and linear density of the rope. The velocity of a pulse in a string or rope is given as v = √(T/μ), where T represents tension, and μ represents linear mass density (mass/length).
In the context of this question, if the speeds of pulses on two ropes are different but the tensions are the same (since the ropes are stretched the same way), the difference in speed must be due to a difference in linear mass density. Specifically, the speed of a pulse is inversely proportional to the square root of the rope's mass per unit length.
Given that the speed of the pulse on rope 1 is 1.4 times the speed on rope 2, the ratio of the masses of the two ropes (m1/m2) will be the square of the inverse ratio of the speeds. Hence, m1/m2 = (1/1.4)² = 0.51 or approximately 1:2.
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A 75.5-kg person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of 3.38 x 10-2 m3 and is completely submerged under the water. The volume of the person's body that is underwater is 6.42 x 10-2 m3. What is the density of the life jacket?
Answer:
Density of jacket will be [tex]680.4733kg/m^3[/tex]
Explanation:
We know that weight of water displaced= buoyant force=weight of man
Now volume of water displaced [tex]v=3.38\times 10^{-2}+6.42\times 10^{-2}=9.8\times 10^{-2}m^3[/tex]
Density of water [tex]d=100kg/m^3[/tex]
So weight of water displaced [tex]=9.8\times 10^{-2}\times 1000=98kg[/tex]
So weight of jacket = 98-75 = 23 kg
We have given volume of the jacket = [tex]3.38\times 10^{-2}m^3[/tex]
So density of jacket [tex]=\frac{mass}{volume}=\frac{23}{3.38\times 10^{-2}}=680.4733kg/m^3[/tex]
Old-fashioned pendulum clocks are powered by masses that need to be wound back to the top of the clock about once a week to counteract energy lost due to friction and to the chimes. One particular clock has three masses: 4.0 kg, 4.0 kg, and 6.0 kg. They can drop 1.3 meters. How much energy does the clock use in a week?
Answer:
Energy, E = 178.36 J
Explanation:
It is given that,
Mass 1, [tex]m_1=4\ kg[/tex]
Mass 2, [tex]m_2=4\ kg[/tex]
Mass 3, [tex]m_3=6\ kg[/tex]
Height from which they are dropped, h = 1.3 m
Let m is the energy used by the clock in a week. The energy is equal to the gravitational potential energy. It is given by :
[tex]E=(m_1+m_2+m_3)gh[/tex]
[tex]E=(4+4+6)\times 9.8\times 1.3[/tex]
E = 178.36 J
So, the energy used by the clock in a week is 178.36 Joules. Hence, this is the required solution.
This is the body of law that codifies what a state defines as legal and/or illegal as well as the punishments for the violations of the laws.
Answer:
Criminal Law
Explanation:
Criminal Law is the body of law that codifies what a state defines as legal and/or illegal as well as the punishments for the violations of the laws. Examples of Criminal laws are murder, assault, theft, or drunken driving. Under criminal comes that procedure of prosecution of individual who commit crime. It is different from the civil law.
Answer:
Criminal Law is the body of law that codifies what a state defines as legal and/or illegal as well as the punishments for the violations of the laws. Examples of Criminal laws are murder, assault, theft, or drunken driving. Under criminal comes that procedure of prosecution of individual who commit crime. It is different from the civil law.
Keller is driving at a speed of 60 MPH. At noon he sees the following sign: Belmont 3 miles Arville 17 miles Blison 35 miles Longwood 52 miles Swan's Lake 120 miles If he continues to drive the same speed, how far away will Longwood be at 12:20pm?
Answer:
32 miles
Explanation:
Assuming that the speed of 60 mph is constant
Distance = Speed × Time
Time is 20 minutes
Converting to hours
[tex]\frac{20}{60}=\frac{1}{3}\ h[/tex]
[tex]Distance=60\times \frac{1}{3}\\\Rightarrow Distance=20\ mi[/tex]
The car covered 20 mi in 20 minutes.
Distance to Longwood would be [tex]52-20=32\ mi[/tex]
Distance to Longwood is 32 miles
A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part of the pipe is 3.5 m/s, what is its speed when it moves through the narrow part of the pipe?
Answer:
v₂ = 7/ (0.5)= 14 m/s
Explanation:
Flow rate of the fluid
Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.
The formula for calculated the flow rate is:
Q= v*A Formula (1)
Where :
Q is the Flow rate (m³/s)
A is the cross sectional area of a section of the pipe (m²)
v is the speed of the fluid in that section (m/s)
Equation of continuity
The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:
Q₁= Q₂
Data
A₁ = 2m² : cross sectional area 1
v₁ = 3.5 m/s : fluid speed through A₁
A₂ = 0.5 m² : cross sectional area 2
Calculation of the fluid speed through A₂
We aply the equation of continuity:
Q₁= Q₂
We aply the equation of Formula (1):
v₁*A₁= v₂*A₂
We replace data
(3.5)*(2)= v₂*(0.5)
7 = v₂*(0.5)
v₂ = 7/ (0.5)
v₂ = 14 m/s
Using the Continuity Equation that states the mass flow rate must be constant in fluid flow, the speed of the fluid when it moves through the narrow part of the pipe is 14 m/s.
Explanation:The question pertains to the physics principle of fluid dynamics, specifically the concept of continuity in fluid flow. According to the Continuity Equation, the mass flow rate must remain constant across different sections of a pipe. In this particular scenario, the fluid is changing speed due to a reduction in pipe cross-sectional area. This principle is represented by the equation A₁V₁ = A₂V₂, where A₁ and A₂ represent the cross-sectional areas of the wide and narrow sections of the pipe respectively, and V₁ and V₂ represent the fluid's velocity in these sections.
Applying the equation to this scenario, we have A₁ (2.00 m²) multiplied by V₁ (3.5 m/s) equal to A₂ (0.50 m²) multiplied by V₂, the value we want to find. Solving for V₂, we get V₂ = A₁V₁ / A₂ = (2.00 m²)(3.5 m/s) / 0.50 m² = 14 m/s.
Hence, the speed of the fluid when it moves through the narrow part of the pipe is 14 m/s.
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Two platoons line up for a tug of war. If both platoons have the same number of participants on a team, what other factor is critical to either platoon's win?
A. age of men
b. height of men
c. total mass of the team
d. condition of the ground
Answer:
option (c)
Explanation:
In the game of tug of war, the Newton's third law is obeyed.
One team pulls the rope in one direction and the other team pulls the rope in another direction.
As the mass of one team is more, so it is harder to pull the rope by the another team.
So, it depends on the total mass of the team.
Option (c) is correct.
Answer:
The best answer would be C. Total mass of the team.
Explanation:
C would be the best answer because when looking in the formula of force then the 2 factors that are inputted are mass and acceleration but in this case acceleration would most likely not have much have an effect so that leaves the factor of mass.
Hope this helped!
Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s, piano A produces a wavelength of 0.766 m, while piano B produces a wavelength of 0.776 m. How much time separates successive beats?
Answer:
Time period between the successive beats will be 0.1703 sec
Explanation:
We have given speed of the sound v = 349 m/sec
Wavelength of piano [tex]A\lambda _A=0.766m[/tex]
Wavelength of piano [tex]B\lambda _B=0.776m[/tex]
So frequency of piano A [tex]f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz[/tex]
Frequency of piano B [tex]f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz[/tex]
So beat frequency f = 455.61 - 449.74 = 5.87 Hz
So time period [tex]T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec[/tex]
So time period between the successive beats will be 0.1703 sec
As sources of electrical power, windmills now account for only about 2,500 megawatts nationwide, but production is almost expected to double by the end of the year, which would provide enough electricity for 1.3 million households.
a. almost expected to double by the end of the year, which would provide
b. almost expected that it will double by the end of the year, thus providing
c. expected that it will almost double by the end of the year to provide
d. expected almost to double by the end of the year and thus to provide
e. expected almost to double by the end of the year, which would thus be providing
Answer: C
Explanation:
Production is expected that it will almost double by the end of the year is the answer because production in this context relates to a numerical quantity which is a function of time.
Therefore the probability there is on whether production will almost double to provide enough electricity and not if production will occur.
Windmills account for about 2,500 megawatts of electrical power nationwide, and the production is expected to almost double by the end of the year, providing enough electricity for 1.3 million households.
Windmills account for about 2,500 megawatts of electrical power nationwide, and the production is expected to almost double by the end of the year. This doubling of production would provide enough electricity for 1.3 million households.
The correct answer choice is c. expected that it will almost double by the end of the year to provide because it accurately represents the expectation of the doubling of production and the purpose of providing enough electricity for households.
A student sees her physical science professor approaching on the sidewalk that runs by her dorm. She gets a water balloon and waits. When the professor is 2 s from being directly under her window 11 m above the idewalk, she drops the balloon. You finish the story. (Assume g # 10 m/s*)
Answer:Misses the Target
Explanation:
Given
Distance between window and Professor is h=11 m
she dropped the balloon 2 sec earlier
time taken by balloon to cover the 11 m
using
[tex]h=ut+\frac{at^2}{2}[/tex]
where h=height
u=initial velocity
t=time
a=acceleration
[tex]11=0+\frac{9.8t^2}{2}[/tex]
[tex]t=\sqrt{\frac{2\times 11}{10}}[/tex]
[tex]t=\sqrt{2.2}[/tex]
[tex]t=1.483 s[/tex]
Therefore she misses the Professor by 0.51 s
Final answer:
The water balloon will hit the ground before the professor is directly underneath the window since it takes approximately 1.48 seconds to fall 11 meters, which means the student is unlikely to hit the professor.
Explanation:
The student's question involves calculating the time it takes for a water balloon dropped from a height to hit a target. This is a problem in projectile motion and free fall within the context of physics, specifically under the topic of mechanics. As the professor is 2 seconds away from the point directly below the student's window, which is 11 meters above the sidewalk, we can determine if the balloon will hit the target by calculating how long it takes for the balloon to reach the ground.
Using the equation of motion for free fall s = 1/2gt^2, where g is the acceleration due to gravity (approximated to 10 m/s^2), and s is the distance (11 meters), we can solve for t (time): s = 1/2gt^2 implies t = sqrt(2s/g). Substituting the given values, we get t = sqrt(2*11/10), which gives t = sqrt(2.2), or approximately t = 1.48 seconds. Because the professor is 2 seconds away from being underneath the window, the water balloon will hit the ground before the professor reaches the point below the student's window, meaning the student's attempt to hit the professor with the balloon will likely fail unless the professor's speed changes.
During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0 m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.
The projectile reaches a maximum height of 26.4 m.
The question is demanding us to find the;
i) Time of flight:
T = 2usinθ/g
u = initial velocity
θ = angle of projection
g = acceleration due to gravity
T = 2 × 30.0 × sin 50.0°/10
T = 4.6 s
ii) The range;
R = u^2sin2θ/g
R = (30.0)^2 × sin 2( sin 50.0°)/10
R = 88.6 m
ii) The maximum height;
H = u^2sin^2θ/2g
H = (30.0)^2 × sin^2(50.0°)/2 × 10
H = 26.4 m
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At what displacement of a sho is the energy half kinetic and half potential?
Answer:
Displacement = 0.707A
Explanation:
To solve for the displacement we know that
Potential energy PE = 1/2Total energy (Etotal)
Therefore 1/2kx^2 = 1/2(1/2KA^2)
Solving for x we have
x^2 = √A^2/2
x = A/√2
x= 0.707A
Blaise Pascal duplicated Torricelli's barometer using a red Bordeaux wine of density 965 kg/m3 as the working liquid (see figure below).
1. What was the height h of the wine column for normal atmospheric pressure?
2. Would you expect the vacuum above the column to be as good as that for mercury?
A. Yes
B. No
Answer:
a) h=10.701m
b) No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.
Explanation:
The Torricelli's experiment "was invented by the Italian scientist Evangelista Torricelli and the most important purpose of this experiment was to prove that the source of vacuum comes from atmospheric pressure"
Pressure is defined as "the force that is applied on any object in the direction perpendicular to the surface of the object in the unit area is known as the pressure. There are various types of pressure".
Part a
We have the density for the red Bordeaux wine given [tex]\rho=965\frac{kg}{m^3}[/tex], the atmospheric pressure on the Toriccelli's barometer is given by:
[tex]P_{atm}=\rho g h[/tex]
Solving for the height of wine in the column we have this:
[tex]h=\frac{P_{atm}}{\rho g}[/tex]
And replacing we have:
[tex]h=\frac{101300Pa}{965\frac{kg}{m^3} 9.81\frac{m}{s^2}}=10.701 m[/tex]
So the height of the red Bordeaux wine would be h=10.701m. A very high value on this case if we compare with the usual values for this variable.
Part b
No. On this case a liquid like wine is not good as the mercury, because the wine is composed of water, alcohol and other elements, but specially the alcohol evaporates much easier than the mercury, and that will cause malfunction in the vacuum of the baroemter used for the experiment.
The height of the wine column for normal atmospheric pressure can be calculated using the equation P = ρgh. The vacuum above the column in the wine barometer would not be as good as that for mercury. The correct answer is No (B).
Explanation:The height h of the wine column for normal atmospheric pressure can be calculated using the equation P = ρgh, where P is the atmospheric pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height of the column. The atmospheric pressure can be taken as P = 101325 Pa and the acceleration due to gravity is approximately g = 9.8 m/s^2. Plugging in these values and the density of the wine, ρ = 965 kg/m^3, we can solve for h. Hence, h = P / (ρg).
To compare the vacuum above the column in the wine barometer with that of a mercury barometer, it is important to note that the height of the liquid column in a barometer is determined by the atmospheric pressure pushing down on the liquid. Since both the wine and mercury barometers are subjected to the same atmospheric pressure, the height of the columns will be the same if the densities of the liquids are the same. However, since the density of wine is less than that of mercury, the height of the wine column will be greater than that of the mercury column. Therefore, No (B) is the correct answer.
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