Answer:
The internal energy of the gas remains constant,
Explanation:
The isothermal process can be defined as a thermodynamic process in which the temperature of the system remains constant.In isothemal process the process of transfer of heat energy from the surrounding to the system or to the system from the surrounding takes place at constant temperature.
in isothermal process internal energy of the system remains unchanged or constant.
For ideal gas internal energy of the system depends on the temperature.
For ideal gas when it is compressed the work is done by the surrounding on the gas is positive and heat added to the system is negative .When the heat energy is absent,both temperature and internal energy increases. As it is a isothermal process and temperature remains constant the gas must transfer the positive amount of heat to the system.
Hence internal energy of the system of gas remains constant.
Final answer:
During an isothermal compression of a gas, energy is transferred into the gas by heat to maintain its constant temperature, while work is done on the gas and its internal energy remains constant.
Explanation:
When a gas is compressed isothermally, specific events related to energy take place. Isothermal processes occur at a constant temperature, implying that the thermal energy of the gas does not change. If we refer to the first law of thermodynamics, stating that the change in internal energy (\(\Delta E_{int}\)) is equal to heat added to the system (Q) minus the work done by the system (W), we can understand these events. Since the temperature remains constant during an isothermal compression, the internal energy of an ideal gas, which depends only on temperature, also remains constant (\(\Delta E_{int} = 0\)). Work is done on the gas during compression, meaning that there is a transfer of energy into the gas by way of work (W > 0).
To maintain the constant temperature, heat must be removed from the gas equal to the work done on it (Q = W), so there is effective energy transfer into the gas by heat to maintain isothermality. Therefore, the correct statement among the given options is: Energy is transferred into the gas by heat.
Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.
The resistance means resist, means to resist the current, so we are here given that, three resistors are connected in series, so as we know that, if n resistors are connected in series combination, then the total resistance is the individual sum of all n Resistors, so if we apply the same formula here, we will be having :
[tex]{:\implies \quad \sf R_{Total}=3+4+5}[/tex]
[tex]{:\implies \quad \boxed{\bf{R_{Total}=12\Omega}}}[/tex]
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https://brainly.com/question/27101567Which statement describes Newton's law of universal gravitation?
Mass has hittle effect on gravity between objects.
Gravity pushes objects away from Earth's center,
Gravity does not act between Earth and the moon.
Every object in the universe attracts every other object
Final answer:
Newton's law of universal gravitation states that all objects in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
Explanation:
Newton's Law of Universal Gravitation:
Sir Isaac Newton's law of universal gravitation states that every object in the universe attracts every other object with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational force is universally attractive and only depends on mass and distance, following the equation F = G × (m1 × m2) / r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the centers of the two masses.
A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rotate. It is supported by a cable that makes an Angle of 53 degrees with the horizontal. If the person of weight 350 N stands 1.50 m from the wall. Find the magnitude of the tension T in the cable. (Show your workout instead of giving the final answer).
The magnitude of the tension T in the cable is approximately 468.164 N.
To find the tension T in the cable supporting the beam, we can analyze the forces acting on the beam in equilibrium.
First, let's consider the forces acting on the beam:
1. The weight of the beam, acting downward at its center (2.0 m from the wall).
2. The tension T in the cable, acting upward and at an angle of 53 degrees with the horizontal.
3. The reaction force at the pin connection, acting horizontally to the left.
4. The vertical force exerted by the person standing on the beam, which contributes to the vertical component of the tension in the cable.
The beam is in equilibrium, so the sum of the torques about any point must be zero. Let's take moments about the pin connection at the wall.
Clockwise torques:
- Weight of the beam: [tex]\(200 \, \text{N} \times 2.0 \, \text{m}\)[/tex]
- Vertical component of the tension: [tex]\(T \times \cos(53^\circ) \times 1.5 \, \text{m}\)[/tex]
Counterclockwise torques:
- Tension in the cable: [tex]\(T \times \sin(53^\circ) \times 1.5 \, \text{m}\)[/tex]
- Vertical force exerted by the person: [tex]\(350 \, \text{N} \times 1.5 \, \text{m}\)[/tex]
Since the beam is in equilibrium, these torques must balance:
[tex]\[ 200 \, \text{N} \times 2.0 \, \text{m} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 350 \, \text{N} \times 1.5 \, \text{m} \][/tex]
Now, we can solve for T:
[tex]\[ 400 \, \text{N} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ 400 \, \text{N} = T \times \sin(53^\circ) \times 1.5 \, \text{m} - T \times \cos(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ 400 \, \text{N} = T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} = 125 \, \text{N} \][/tex]
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
Now, we can calculate T:
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{(0.7986 - 0.6206) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{0.178 \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{0.267 \, \text{m}} \][/tex]
[tex]\[ T \approx 468.164 \, \text{N} \][/tex]
So, the magnitude of the tension T in the cable is approximately 468.164 N.
During which phase of the moon may a lunar eclipse occur?
Answer: New moon
Explanation: A solar eclipse is only possible during a new moon phase
Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude E0 , frequency w, and phase angle zero that is (a) traveling in the negative x direction and polarized in the z direction; (b) traveling in the direction from the origin to the point (1, 1, 1), with polarization parallel to the xz plane. In each case, sketch the wave, and give the explicit Cartesian components ofk and ii.
Answer:
a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
b) the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
Explanation:
a) the polarization the determined wave oscillates the electric field, which is the z axis
As the wave travels on the negative x-axis and the magnetic field is perpendicular, this field goes on the positive y-axis
the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
be) in the case of a polarization in the xi plane the magnetic field must go in the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
During a redox reaction the molecule that gains an electron is
In a redox (reduction-oxidation) reaction, the molecule that gains an electron is reduced. Reduction and oxidation always occur together in such reactions. Reduced molecules often act as energy carriers in metabolic pathways.
Explanation:During a redox (reduction-oxidation) reaction, the molecule that gains an electron is referred to as being reduced. This process is accompanied by an energy transfer. When a molecule loses an electron (a process termed as oxidation), energy is released; this energy and the electron are then transferred to another molecule, resulting in reduction of that molecule. These two processes, oxidation and reduction, always occur together in a redox reaction.
For example, considering a metabolic pathway involving Nicotinamide adenine dinucleotide (NAD+), when it accepts a hydride ion (H-) from a hydrogen atom, it gets reduced to form NADH. Here, NAD+ is the molecule that gets reduced in this redox reaction.
In summary, in redox reactions, the molecule that gains an electron and is reduced can act as an energy carrier, an important function in metabolism, facilitating the controlled transfer of energy within the cell. The concepts of oxidation and reduction are integral to understanding energy extraction and utilization in cells.
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A technician is checking refrigerant system pressures. Both high- and low-side service ports are located on the A/C compressor. When the compressor is engaged, the high-side pressure instantly goes over 375 psig. What is the most likely cause of this condition
Answer:
Two major causes are outline bellow
1. The presence of air in the system
2. Clogged condenser
Explanation:
1. The presence of air in the system
One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.
2. Clogged condenser
Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve
The most likely cause of a refrigerant system's high-side pressure exceeding 375 psig is a blockage, overcharging, or a malfunctioning expansion valve. These issues prevent proper refrigerant flow and pressure regulation within the vapor compression system.
If a technician observes that the high-side pressure of a refrigerant system instantly exceeds 375 psig once the compressor is engaged, the most likely cause for this condition could be a blockage in the condenser or the liquid line, overcharging of the refrigerant, or a malfunctioning expansion valve that is not allowing the liquefied refrigerant to flow properly into the evaporator. The blockage could result in an excessively high pressure build-up because the refrigerant cannot circulate properly. Overcharging of the system adds too much refrigerant, which also leads to high pressure. Additionally, if the expansion valve is stuck closed or not functioning correctly, it would prevent the refrigerant from moving into the evaporator, causing the pressure to build up on the high side.
In a typical vapor compression system, as the electrically driven compressor introduces work (W), it raises the temperature and pressure of the refrigerant, forcing it into the condenser coils. Here, heat transfer occurs as the gas inside the coils is higher in temperature than the room, leading to condensation of the gas back into a liquid. The cooled pressurized liquid passes through the expansion, or pressure-reducing valve, and goes to the evaporator coils located outdoors where it is further cooled by expansion.
Air pollution from a power plant is being monitored for levels of nitrogen dioxide and ground-level ozone. The levels are measured daily at the same time. What is the impact of a dark and cloudy day on the readings for the levels of air pollutants measured?
An increase in the level of nitrogen dioxide and a decrease in the level of ground-level ozone occurs.
Explanation:
Ozone gas is normally found in stratosphere, it protects us from solar radiation. Under certain circumstances, it can be formed on the ground level.
Conditions that have to be met in order for this to happen are existing of nitrogen oxides and volatile organic compounds and their reaction catalyzed by heat and light.
During dark, cold and cloudy days, due to the lack of heat and light from the Sun, formation of ozone will be decreased.
Also, this will enable building up of nitrogen dioxide, due to the same reason, leading to increased concentration.
How many significant figures are present in the number 50,670?
A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 55 ms. (a) How far below the release point is the center of mass of the two stones at t = 470 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?
Answer:
Part(a): At [tex]\bf{t = 470~ms}[/tex] the center of mass will travel [tex]\bf{0.903~m}[/tex].
Part(b): The velocity of the first stone is [tex]\bf{4.606~m.s^{-1}}[/tex] and the velocity of the second stone is [tex]\bf{4.067~m.s^{-1}}[/tex].
Explanation:
Given:
The first stone is dropped at [tex]t_{1}=0~s[/tex].
The second stone is dropped at, [tex]t_{2}=55~ms=0.055~s[/tex]
The mass of the second stone is 3 times the mass of the first.
Both the stones are dropped from the same point.
Consider the mass of the first stone be [tex]m[/tex]. So the mass of the second stone is [tex]3m[/tex].
(a)
The formula to calculate the distance traveled by each stone is given by
[tex]y = \dfrac{1}{2}gt^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by each stone.
Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{1}[/tex] for [tex]y[/tex] and [tex]470~ms[/tex] or [tex]0.47~s[/tex] for first stone in equation (1), we have
[tex]y_{1}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.47~s)^{2}\\~~~~&=& 1.08~m[/tex]
where [tex]y_{1}[/tex] is the distance traveled by the first stone.
Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{2}[/tex] for [tex]y[/tex] and [tex](470-55)~ms = 415~ms[/tex] or [tex]0.415~s[/tex] for second stone in equation (1), we have
[tex]y_{2}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.415~s)^{2}\\~~~~&=& 0.844~m[/tex]
The formula to calculate the distance traveled by the center of mass is given by
[tex]y_{c} &=& \dfrac{my_{1}+3my_{2}}{m+3m} \\~~~~&=& \dfrac{1.08m + 0.844(3m)}{4m}\\~~~~&=& 0.903~m[/tex]
(b)
The formula to calculate the velocity of each stone is given by
[tex]v=gt~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Substituting [tex]v_{1}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.47~s[/tex] for [tex]t[/tex] in equation (2), we have
[tex]v_{1} &=& (9.8~m.s^{-2})(0.47~s)\\~~~~&=& 4.606~m.s^{-1}[/tex]
where [tex]v_{1}[/tex] is the velocity of the first stone after [tex]470~ms[/tex].
Substituting [tex]v_{2}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.415~s[/tex] for [tex]t[/tex] in equation (2), we have
[tex]v_{2} &=& (9.8~m.s^{-2})(0.415~s)\\~~~~&=& 4.067~m.s^{-1}[/tex]
where [tex]v_{2}[/tex] is the velocity of the second stone after [tex]470~ms[/tex].
El vector a tiene componented ax= 2.70 cm, ay=2.25 cm; y el vector b tiene componentes bx= 0.30 cm, by= 1.75 cm. Calcular las componentes de la resultante a+b
Answer:
Explanation:
Given that,
A vector A has x component to be 2.7cm and y component to be 2.25cm
Then,
A = 2.7•i + 2.25•j
A vector B has x component of 0.30cm and y component of 1.75cm
B = 0.3•i + 1.75•j
So, we want to find A+B
Addition of vectors
Generally
(a•i + b•j) + (c•i + d•j) = (a+c)•i +(b+d)•j
Vectors are added component wise.
So,
A + B = (2.7•i + 2.25•j) + (0.3•i + 1.75•j)
A + B = (2.7 + 0.3)•i + (2.25 + 1.75)•j
A + B = 3•i + 4•j
We can also find it magnitude and direction
Generally,
A = a•i + b•j
|A| = √(a²+b²)
<A = arctan(b/a)
So,
|A+B| = √(3²+4²) = √9+16 = √25
|A+B| = 5
And it's direction
< = arctan(y/x)
< = arctan(4/3)
< = 53.13°
A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds?
Answer:
43 m/s
Explanation:
Mass, m = 5 kg
Force, F(t) = 6t² - 4t + 3
To find the speed, we first need to get the acceleration, a.
Force is the product of mass and acceleration. It is given as:
F = ma
Therefore, acceleration is:
a(t) = F(t)/m
a(t) = (6t² - 4t + 3) / 5
a(t) = 1.2t² - 0.8t + 0.6
Acceleration is the differentiation of velocity with respect to time, t. Therefore, to get velocity, v, we integrate a(t):
a(t) = dv(t) / dt
=> v(t) = (1.2/3)t³ - (0.8/2)t² + 0.6t
v(t) = 0.4t³ - 0.4t² + 0.6t
Therefore, at time t = 5secs, velocity is:
v(5) = 0.4 * (5³) - 0.4 * (5²) + 0.6 * 5
v(5) = 50 - 10 + 3
v(5) = 43 m/s
The velocity at time, t = 5 secs is 43 m/s
Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you the tube from a toilet-paper roll (12 cm long, 4.0 cm diameter), and asks how many turns she should wind on the full length of the tube.
Answer:
N = 195 turns
Explanation:
The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H
The length of the tube, l = 12 cm = 0.12 m
The diameter of the tube, d = 4 cm = 0.04 m
Radius, r = 0.04/2 = 0.02 m
Area of the tube, A = πr² = 0.02²π = 0.0004π m²
[tex]\mu_{0} = 4\pi * 10^{-7}[/tex]
The inductance of a solenoid is given by:
[tex]L = \frac{\mu_{0}N^{2} A }{l}[/tex]
[tex]500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns[/tex]
When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain constant force acts upon an object with mass , the acceleration of the object is 26m/s^2 . If the same force acts upon another object whose mass is 13 , what is this object's acceleration
If the same force acts upon another object whose mass is 13 kg the acceleration is [tex]4m/s^2[/tex].
Force is a fundamental concept in physics that describes the interaction between objects that can cause a change in their motion or shape. It's a vector quantity, meaning it has both magnitude and direction. In simpler terms, force is what can make things move, stop, change direction, or deform.
Given:
Masses, [tex]m_1 = 2\ kg[/tex] and [tex]m_2 = 13\ kg[/tex]
Acceleration, [tex]a = 26\ m/s^2[/tex]
The force is computed as:
[tex]F = m_1a\\F = 2\times26\\F = 52\ N[/tex]
The acceleration for mass 52 kilograms is:
[tex]F = m_2a\\a = F/m_2\\a = 52/ 13\\a = 4\ m/s^2[/tex]
Hence, if the same force acts upon another object whose mass is 13 kg the acceleration is [tex]4m/s^2[/tex].
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Applying Newton's second law of motion, the acceleration of an object is directly proportional to the force acting on it, but inversely proportional to its mass. The acceleration of a 13kg object, when acted upon by the same force that gives a 2kg object an acceleration of 26m/s^2, would be 4 m/s^2.
Explanation:The question basically asks us to apply Newton's second law of motion, which states that the acceleration of a system is directly proportional to the net external force acting on the system, but inversely proportional to its mass. In equation form, this can be written as F=ma, where F is the force, m is the mass, and a is the acceleration. This is also referred to as the principle of constant acceleration, in which the force applied to an object will either accelerate or decelerate it at a constant rate, provided the mass of the object remains unchanged.
In this case, we know that a 2kg object accelerates at 26m/s^2 under the influence of a constant force. Therefore, the force (F) acting on it can be computed as F = m * a = 2kg * 26m/s^2 = 52N. Now, if the same force acts on another object with a mass of 13kg, the acceleration (a) of this object can be computed by rearranging the equation as a = F/m = 52N / 13kg = 4m/s^2. So, the acceleration of the second object, under the same force, would be 4 m/s^2.
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__________ is a type of molding in which thermoplastics are heated above melting temperature and, using an extruder, are forced into a closed die to produce a molding.
a. Injection molding
b. Blow molding
c. Thermoforming
d. Extrusion
Answer: The correct and is
D. Extrusion
Explanation:
What is extrusion?
Extrusion is a manufacturing process used to make basically pipes and hoses.
The pvc granules are melt into a liquid which is forced through a die, forming a long 'tube like' shape. The shape of the die determines the shape of the tube.
The extrusion is then cooled and forms a solid shape
Extrusion moulding is used to create products with a consistent cross-section.
A form of charge designated as negative positive or neutral is called
Answer:
Electrical charge
Explanation:
Electrical Charge is a form of charge, designated negative, positive, or neutral (without charge) that is found on the subatomic particles that make up all atoms
A block of unknown mass is attached to a spring of spring constant 6.1 N/m and undergoes simple harmonic motion with an amplitude of 10.3 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 31.3 cm/s. Calculate the mass of the block. Answer in units of kg
Answer:
Explanation:
Given that, .
Spring constant.
K = 6.1 N/m
Amplitude of oscillation
A = 10.3cm = 0.103m
Half wave, it speed is
V = 31.3cm/s = 0.313m/s
Mass of block
M =?
The is half of the distance between the equilibrium position and endpoint
Then,
x = A/2
Where,
A is the amplitude
x is the position of the block
x = A / 2 = 0.103/2
x = 0.0515m
The velocity if the block at any given position is given as
v = ω√(A²—x²)
Then,
ω = v / √(A²—x²)
Where
ω is angular frequency
v Is the velocity of the block
The force constant is given as
k = mω²
Where,
K is spring constant
ω is angular frequency
Substitute ω into k
Then,
k = m (v / √(A²—x²))²
k = mv² / (A²—x²)
Make m subject of formula since we want to find m
m = k(A²—x²) / v²
m = 6.1 (0.103²—0.0515²) / 0.313²
m = 6.1 × 7.96 × 10^-3 / 0.313²
m = 0.495 kg
The mass of the block is approximately 0.5kg
Answer:
The mass of the block is 496 g
Explanation:
Here we have by the principle of conservation of energy;
Energy in spring, E = 0.5·k·A²
Where:
k = Spring constant = 6.1 N/m
A = Amplitude of motion = 10.3 cm = 0.103 m
E = 0.5×6.1×0.103² = 0.03235745 J= 3.24 × 10⁻² J
At half way, we have
[tex]E = \frac{1}{2} k(\frac{A}{2} )^{2} + \frac{1}{2} mv^{2} = \frac{E}{4} + \frac{1}{2} mv^{2}[/tex]
Where:
m = Mass of the block
v = Velocity of block at the instant (Halfway between its equilibrium position and the endpoint)
Therefore,
[tex]\frac{3}{4}E = \frac{1}{2} mv^{2}[/tex] or
m = [tex]\frac{3}{2v^2}E[/tex] = [tex]\frac{3}{2\times 0.313^2} \times 3.24 \times 10^{-2}[/tex]= 0.496 kg = 496 g.
Will water flow more easily through a wide pipe or a narrow pipe? Will current flow more easily through a thick wire or a thin wire?
Answer:
Explained below.
Explanation:
The water will flow more easily through a wide pipe, because as we know that if the diameter of the pipe is wider than the flow of water will be more.
And same case will be applied with the electric current, if the is wire thicker then the flow of the current will be more easy and the charge will also flow easily.
A bat locates insects by emitting 25.3 kHz ultrasonic chirps and then listening for echoes from insects flying around it. Humans can hear sounds with frequencies up to 20 kHz. How fast would a bat have to fly and in which direction (positive away from and negative towards a stationary human listener) in order for a person to hear these chirps
Answer:
Speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.
Explanation:
Given:
Frequency of the bat that is source here, [tex]f_S[/tex] = 25.3 kHz = 25.3 * 10^3 Hz
Frequency of the listener (human), [tex]f_L[/tex] = 20 kHz = 20*10^3 Hz
We have to identify how fast the bat have to fly in order for a person to hear these chirps .
Let the velocity of bat that is source is "Vs" and "Vs" = "Vbat".
Doppler effects formulae :
When the source is receding (moving away) [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex] When the source is approaching [tex]f_L=(\frac{V+V_L}{V-V_S}) f_S[/tex] Speed of sound in [tex]V_S =433.895-343[/tex]the air (medium), [tex]V = 343\ ms^-^1[/tex]Using the above formula and considering that the bat is moving away so that the human can listen the chirps also [tex]V_L=0[/tex] as listener is stationary.
⇒ [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex] ⇒ [tex]f_L=(\frac{V+0}{V+V_S}) f_S[/tex]
Re-arranging in terms of Vs.
⇒ [tex]V+V_S =\frac{V\times f_S}{f_L}[/tex]
⇒ [tex]V_S =\frac{V\times f_S}{f_L}-V[/tex]
⇒ [tex]V_S =\frac{343\times 25.3\times 10^3}{20\times 10^3}-343[/tex]
⇒ [tex]V_S=90.895[/tex] m/s
The speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.
Identify which type of natural selection is described in each case below.
Results in the evolution of the species:
Results in new species:
Favors traits at the two extremes:
No increase in diversity:
One extreme trait is favored:
Diversity decreases:
Diversity increases:
Answer:
1.) A
2.) B
3.) B
4.) A
5.) A
6.) C
7.) B
Explanation:
Answer:
A directional selection
B disruptive selection
B disruptive selection
A directional selection
A directional selection
C stabilizing selection
B disruptive selection
Explanation:
A 0.100-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone
Answer:
Change in momentum of the stone is 3.673 kg.m/s.
Explanation:
Given:
Mass of the ball on the horizontal the surface, m = 0.10 kg
Velocity of the ball with which it hits the stone, v = 20 m/s
According to the question it rebounds with 70% of the initial kinetic energy.
We have to find the change in momentum i.e Δp
Before that:
We have to calculate the rebound velocity with which the object rebounds.
Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".
⇒ [tex]KE_1=0.7\times \frac{mv^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times (20)^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times 400}{2}[/tex]
⇒ [tex]KE_1=14[/tex] Joules (J).
Rebound velocity "v1".
⇒ [tex]KE_1=\frac{m(v_1)^2}{2}[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2KE_1}{m} }[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2\times 14}{0.10} }[/tex]
⇒ [tex]v_1=16.73[/tex]
⇒ [tex]v_1=-16.73[/tex] m/s ...as it rebounds.
Change in momentum Δp.
⇒ [tex]\triangle p= m\triangle v[/tex]
⇒ [tex]\triangle p= 0.10\times (20-(-16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (20+16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (36.73)[/tex]
⇒ [tex]\triangle p = 3.673[/tex] Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
The starting conditions of an oscillator are characterized by the
Answer:
The phase constant
Explanation:
The phase constant tells how much a signal is shifted along the x-axis. A phase constant of ϕ means that each value of the signal happens ϕ amount of time earlier. If the signal has a beginning, then a phase constant of ϕ means the signal occurs that much sooner.
The starting conditions of an oscillator are characterized by the phase constant. Option B is correct.
Phase Constant:
This represents the number of oscillation per cycle of wave. It is denoted by [tex]\bold {\phi}[/tex]. It is also known as Propagation constant. The phase constant can be calculated using the formula.
[tex]\bold {\phi = \dfrac {2\pi }{\lambda}}[/tex]
Where, lambda is wavelength.
Frequency is the number of waves passing from a point per unit time.Amplitude is the height of the wave from its mid point.Therefore, option B is correct. The starting conditions of an oscillator are characterized by the phase constant.
To know more about phase constant,
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A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?
(A) 20 m/s
(B) 1.0 m/s
(C) 10 m/s
(D) 5.0 m/s
(E) 115 m/s
Answer:
(c) 10m/s
Explanation:
to find the speed of the waves you can use the following formula:
[tex]v=\frac{\lambda}{T}[/tex]
λ: wavelength of the wave
T: period
the wavelength is the distance between crests = 48m
the period is the time of a complete oscillation of the wave. In this case you have that the float takes 2.4 s to go from its highest to the lowest point. The period will be twice that time:
T = 2(2.4s)=4.8s
by replacing you obtain:
[tex]v=\frac{48m}{4.8s}=10\frac{m}{s}[/tex]
the answer is (c) 10m/s
An ideal transformer has 50 turns in its primary and 250 turns in its secondary. 12 Vac is connected to the primary. Find: (a) volts ac available at the second- ary; (b) current in a 1O-ohm device connected to the secondary; and (c) power supplied to the primary.
Answer:
a. 60 V b. 6 A c. 360 W
Explanation:
a. Voltage in secondary
For an ideal transformer,
N₁/N₂ = I₂/I₁ = V₁/V₂ where N₁ = turns in primary = 50 turns, N₂ = turns in secondary = 250 turns, V₁ = voltage in primary = 12 V, V₂ = voltage in secondary = ? V
N₁/N₂ = V₁/V₂
V₂ = V₁N₂/N₁ = 250 × 12/50 = 60 V
b. Since V₂ = I₂R,
I₂ = V₂/R R = 10 Ω
I₂ = 60 V/10 Ω
I₂ = 6 A
c. We first calculate the current in the primary from
N₁/N₂ = I₂/I₁ where I₁ = primary current
I₁ = N₂I₂/N₁ = 250 × 6 A/50 = 30 A
The power supplied to the primary is thus
P = I₁V₁ = 30 A × 12 V = 360 W
(a) The voltage supply at the secondary coil is of 60 V.
(b) The current flow through the 10- ohm device is 6 A.
(c) The power supply to the primary coil is 360 W.
Given Data:
Number of turns of primary coil is, n1 = 50.
Number of turns of secondary coil is, n2 = 250.
The voltage supply to primary coil is, V1 = 12 V.
The resistance of device is, R = 10 ohm.
(a)
The first part of the given problem is based on the Transformer equation relating the number of turns of coils at primary and secondary and voltage supply at each coils.
Therefore,
n1 / n2 = V1 / V2
Here,
V2 is the voltage supply to the secondary coil.
Solving as,
50 / 250 = 12 / V2
V2 = 250 / 50 × 12
V2 = 60 V.
Thus, we can conclude that the voltage supply at the secondary coil is of 60 V.
(b)
Now in order to find the current flow through the 10 - ohm device, we can use the Ohm's law as,
V2 = I' × R
Solving as,
60 = I' × 10
I ' = 60 / 10
I' = 6 A
Thus, we can conclude that the current flow through the 10- ohm device is 6 A.
(c)
Now, we first calculate the current in the primary from
n1 / n2 = I' / I
here,
I is the primary current
Solving as,
I = n2 × I' / n1
I = 250 × 6 /50
I = 30 A
The power supplied to the primary is thus
P1 = I × V1
P1 = 30 × 12
P1 = 360 W
Thus, we can conclude that the power supply to the primary coil is 360 W.
Learn more about the transformer equation here:
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Luis is trying to push a box of new soccer balls across the floor. In the illustration, the arrow on the box is a vector representing the force that Luis exerts . If the box is not moving, which of the filling must be true
Answer:
Complete question
Luis is trying to push a box of new soccer balls across the floor.
If the box is not moving, which of the following must be true?
A. The box is exerting a larger force on Luis than he is exerting on the box
B. There is another force acting on the box that balances Luis's force.
C. Luis is applying a force that acts at a distance.
D. There is no force of friction acting on the box.
Explanation:
Using newton second law of motion
ΣF = ma
Now, for a body not to move when a force is acting on it means that the body is in equilibrium and acceleration a = 0
Therefore,
Fnet = 0
So, if he is applying a force F to push the box and the box is not moving then, there is an external force that is pushing the force back opposite the direction he his pushing and this force counterbalance is own force.
F—F' = 0
F' = F
So, F' is the counter balance force and it is equal to the force applied by Luis
Or it might be frictional force, because if the static friction is not overcome, then, the body will not leave it's state of rest. So if the fictional force is very high, then the box will not leave it rest position and we also know that frictional force opposes motion,
F—Fr=0
F = Fr
So using this explanation,.
The answer is B
B. There is another force acting on the box that balances Luis's force.
why hot air balloon rises up in air n large ships Floyd in water
Answer:
Hot air is less dense than cool air; the heated air causes the balloon to rise simply because it is lighter than an equal volume of cold air.So hot air balloon rises up in air.
Large ships float in water as their density is lesser than the water so.
When convection currents flow in the mantle they also move the
Answer:
Crust
Explanation:
When the convection currents flow in the mantle they also move the crust. This is best explained as that the crust gets a free ride with the convection currents in the mantle; just as a conveyor belt in a factory moves boxes.
the difference between an electric motor and an electric generator is that a motor converts _____ energy into ______ energy, while a generator converts _______ energy into ________ energy
Answer:electrical,kinetic,kinetic,electrical
Explanation:
Answer:
Electrical, kinetic, kinetic, electrical
Explanation:
Just did it
A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
mb is
placed, as indicated by the dot in the left figure above. The compartment moves with a constant angular speed
about the center of the ride along a circular path of radius R. The bench remains horizontal throughout the
compartment’s motion. The right figure above shows a magnified view of the compartment.
The graph below shows the horizontal ( ) x component of the book’s position as a function of time, where the
+x-direction is to the right.
(a)
i. Determine the period of revolution of the book.
ii. Calculate the tangential speed vb (not the angular speed) of the book.
a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:
[tex]L=2\pi R[/tex]
where R is the radius of the wheel.
The period of revolution is:
[tex]T=120 s[/tex]
Therefore, the tangential speed of the book is:
[tex]v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R[/tex]
In physics how much energy is required to raise the temp of 9.78kg of water from 40.82c to 52.07c? The specific heat of water in this state is 4186j/kg-c.
Answer:
469.6KJ
Explanation:
Heat energy required can be calculated using the formula
H = mc∆t where
m is the mass of the water
c is the specific heat capacity of the water
∆t is the change in temperature of the water
Given m = 9.78kg
c = 4186j/kg-c.
∆t = 52.07°C - 40.82°C
∆t = 11.25°C
H = 9.78 × 4186 × 11.25
H = 460,564.65Joules
= 460.6KJ
Answer:
the amount of energy required to raise the temperature of the water is 460564.65 J
Explanation:
The energy required to raise the temperature of water can be calculated as follows;
Q = mcΔθ
where;
Q is the quantity of heat or energy required to raise the temperature of water
m is mass of water
c is specific heat capacity of water
Δθ is change in temperature = T₂ - T₁
Given;
m = 9.78kg
c = 4186j/kg-c
Δθ = T₂ - T₁ = 52.07°C - 40.82°C = 11.25°C
Q = mcΔθ
Q = (9.78)(4186)(11.25)
Q = 460564.65 J
Therefore, the amount of energy required to raise the temperature of the water is 460564.65 J