Answer: plate tectonics
Explanation: plate tectonics move constantly movement in narrow zones along plate boundaries causes earthquakes and plate tectonics come together and create pressure upon the magma deposit underground forcing it up into a volcanic eruption
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 22 s interval? Assume constant angular acceleration while it is starting and stopping.
Answer:
22 revolutions
Explanation:
2 rev/s = 2*(2π rad/rev) = 12.57 rad/s
The angular acceleration when it starting
[tex]\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2[/tex]
The angular acceleration when it stopping:
[tex]\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2[/tex]
The angular distance it covers when starting from rest:
[tex]\omega^2 - 0^2 = 2\alpha_a\theta_a[/tex]
[tex]\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad[/tex]
The angular distance it covers when coming to complete stop:
[tex]0 - \omega^2 = 2\alpha_o\theta_o[/tex]
[tex]\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad[/tex]
So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions
You stop for a cappuccino at a coffee shop and notice that the tiny white bubbles of steamed milk remain on the surface of the coffee. These air-filled bubbles stay where they are, rather than descending into the coffee or rising into the air, because they are:
A. more dense than the coffee but less dense than the air above thecoffee.
B. thicker than the coffee but less thick than the air above thecoffee.
C. less dense than the coffee but more dense than the air above thecoffee.
D. lighter than the cup of coffee but heavier than the column ofair above the coffee.
Answer:
C. less dense than the coffee but more dense than the air above thecoffee.
Explanation:
How many steps in glycolysis have atp as a substrate or product?
Answer:
4
Explanation:
1. Phosphorylation of Glucose
2. Production of Fructose-6 Phosphate
3. Production of Fructose 1, 6-Diphosphate
4. Splitting of Fructose 1, 6-Diphosphate
5. Interconversion of the Two Sugars
6. Formation of NADH and 1,3-Diphoshoglyceric acid
7. Production of ATP and 3-Phosphoglyceric Acid
8. Relocation of Phosphorus Atom
9. Removal of Water
10. Creation of Pyruvic Acid and ATP
The process of breaking down glucose into pyruvate while oxygen is present is known as glycolysis.
Thus, It is a multi-step process that the cytoplasm's many enzymes catalyze. It is recognized as the initial stage of the cellular respiration process that all living things go through.
Glucose is converted to glyceraldehyde-3-phosphate during the preliminary phase, and glyceraldehyde-3-phosphate is converted to pyruvate during the pay-off phase.
The glycolysis process produces the chemicals NADH and ATP. The ten steps of glycolysis include the partial oxidation of glucose to pyruvic acid.
Thus, The process of breaking down glucose into pyruvate while oxygen is present is known as glycolysis.
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Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
potential energy
kinetic energy
mechanical energy
Answer:
The answer is potential energy
Explanation:
The potential energy is the energy possessed by a body by virtue of it position
For example the water at the top of the dam is being held at a height h above the bottom of the dam
Then the potential energy
PE= weight of the water* the height
PE= m*g*h
"Describe how increasing the stimulus frequency affected the force developed by the isolated whole skeletal muscle in this activity. How well did the results compare with your prediction
Answer is seen below
Explanation: Stimulus frequency refers to the rate that stimulating voltage pulses are applied to an isolated whole skeletal muscle.
When a stimulus frequency is at the lowest ( let's say 50stimuli/second) the force will be at its lowest level out of all of the experiments. As the stimulus frequency was increased to 130 stimuli/second the force increased slightly but fused tetanus( tetanus refers to a sustained muscle tension due to very frequent stimuli) developed at the higher frequency. When the stimulus frequency was increased to the amounts of 146-150 stimuli/second, a maximum tetanic tension occurred, where no further increases in force occur from additional stimulus frequency.
By increasing the stimulus frequency if it resulted in increasing the muscle tension generated by each successive force and it had limit that was eventually reached. Then the results equaled to your prediction.
Un cuerpo de 10N de peso esta apoyado sobre una superficie horizontal. Se le ata una cuerda y se tira de él con una fuerza de 15 N que forma un ángulo de 60° con la horizontal. ¿Cuál es la fuerza normal sobre el cuerpo?
La fuerza normal sobre el cuerpo es de [tex]\(5 \, N\)[/tex].
Explanation:El problema implica un cuerpo de [tex]\(10 \, N\)[/tex] de peso que está siendo sometido a una fuerza de tracción de [tex]\(15 \, N\)[/tex] a un ángulo de [tex]\(60^\circ\)[/tex] con la horizontal. La fuerza normal es la componente perpendicular de la fuerza peso, ya que no hay movimiento vertical. Utilizamos la relación trigonométrica [tex]\(F_{\text{normal}} = F_{\text{peso}} \cdot \cos(\theta)\)[/tex], donde [tex]\(F_{\text{peso}}\)[/tex] es el peso del cuerpo y [tex]\(\theta\)[/tex] es el ángulo entre la fuerza peso y la horizontal.
En este caso, la fuerza peso [tex]\(F_{\text{peso}}\) es \(10 \, N\)[/tex] hacia abajo, y el ángulo [tex]\(\theta\)[/tex] es [tex]\(60^\circ\)[/tex]. Aplicando la fórmula, obtenemos [tex]\(F_{\text{normal}} = 10 \, N \cdot \cos(60^\circ)\)[/tex]. Calculando esto, encontramos [tex]\(F_{\text{normal}} = 10 \, N \cdot 0.5 = 5 \, N\)[/tex].
La fuerza normal es, por lo tanto, [tex]\(5 \, N\)[/tex], lo que significa que la superficie horizontal ejerce una fuerza hacia arriba de [tex]\(5 \, N\)[/tex] para equilibrar la componente vertical de la fuerza aplicada.
Este resultado es consistente con el principio de equilibrio en el plano horizontal. La fuerza normal contrarresta la componente vertical de la fuerza aplicada, manteniendo el cuerpo en equilibrio sin movimiento vertical. Este enfoque, basado en las leyes de la trigonometría y el equilibrio, proporciona una solución clara y precisa para el problema.
In this section we considered a circular parallel-plate capacitor with a changing electric field. Describe the induced magnetic field lines?
Answer:
The study of Maxwell's equations and Ampere's law are used for analysis
Explanation:
Though There's a magnetic field associated with a changing electric field in TEM propagation of an EM wave through space (which is how it propogates, the changing E field begets the M field, the changeing M field begets the E field, leapfrogging each other).
But between capacitor plates, according to Maxwell and Ampere law, the displacement current in Ampere law was exactly to solve cases like that of a capacitor. A magnetic field cannot have discontinuities, unlike the electric field because there are electric charges, but there are no magnetic monopoles, at least as far as we know in the Universe in its current state. We can therefore conclude that there cannot be a magnetic field outside the capacitor and nothing inside
Consider a coin which is tossed straight up into the air. After it is released it moves upward, reaches its highest point and falls back down again. Air resistance can be neglected. What force acts on the coin while it moves upward after it is released
Answer:
GRAVITATIONAL FORCE
Explanation:
We may have noticed that a body thrown upward in air falls back down again after attaining a particular height. The object was able to fall down back due to the effect of gravity acting on it. If there are no force of gravity acting on the body, the body will not fall back but rather disappears into the thin air.
A coin tossed upward in the air which falls back down when released is therefore under the influence of gravity i.e GRAVITATIONAL FORCE while it moves upward after it is released
When an object is approached electrostatically charged to a neutral body it is possible to charge it by induction, however, if it is touched for a moment the object will be charged by contact, explain this phenomenon, why this happens
Answer:
When the charge body touches the other body, charge pass from it to the other body until get the charge is distributed over the largest possible surface, when separating the two bodies they are charged, but with half the initial charge, this phenomenon is called contact charge
Explanation:
When a charged body approaches another body, it attracts or repels the electrons of the other body, until the net charge is zero, that effect disappears when the body moves away.
When the charge body touches the other body, charge pass from it to the other body until get the charge is distributed over the largest possible surface, when separating the two bodies they are charged, but with half the initial charge, this phenomenon is called contact charge
A cylindrical insulated wire of diameter 5.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adjacent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center
Final answer:
To find the magnetic field inside a solenoid, use the formula B = μ₀ * n * I, where B is the magnetic field, n is the number of turns per unit length, and I is the current through the solenoid.
Explanation:
To find the magnetic field inside the solenoid, we can use the formula B = μ0* n * I, where B is the magnetic field, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current through the solenoid.
Given that the solenoid has a diameter of 5.0 mm, we can calculate the radius by dividing the diameter by 2, which is 2.5 mm or 0.0025 m. Using the radius, we can find the number of turns per unit length by dividing the total number of turns (200) by the length of the solenoid (unknown in the given information).
Once we know the number of turns per unit length and the current (0.10 A), we can calculate the magnitude of the magnetic field using the formula mentioned earlier.
Final answer:
To calculate the magnitude of the magnetic field on the axis of the solenoid near its center, we can use the formula B = μ0 * n * I.
Explanation:
To calculate the magnitude of the magnetic field on the axis of the solenoid near its center, we can use the formula:
B = μ0 * n * I
Where B is the magnetic field, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.
In this case, the solenoid has a diameter of 5.0 mm, so the radius is 2.5 mm. The circumference of the solenoid is 2 π * r, and the length is 200 * this circumference. The number of turns per unit length (n) can be calculated by dividing the number of turns by the length of the solenoid. Finally, plug in the values into the formula to calculate the magnetic field.
Any measurement that includes both magnitude and direction is called
A measurement that includes both magnitude and direction is called a vector, which is essential for analyzing motion and forces in physics.
Any measurement that includes both magnitude and direction is called a vector. Vectors are physical quantities that have both an amount, known as magnitude, and a specified direction in space. For example, velocity is a vector because it describes not only how fast an object is moving, but also the direction of movement.
This contrasts with a scalar, which is a quantity that has only magnitude and no direction, such as mass or time. In physics, understanding vectors is crucial for analyzing motion, forces, and other concepts that depend on both the amount and the direction of a quantity.
The function x = (5.2 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 5.3 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?
Answer:
(a) Displacement = - 3.0576 m
(b) Velocity [tex]=-66.48[/tex] m/s
(c)Acceleration = -753.39 m²/s
(d)The phase motion is 26.7 [tex]\pi[/tex].
(e)Frequency =2.5 Hz.
(f)Time period =0.4 s
Explanation:
Given function is
[tex]x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
(a)
The displacement includes the parameter t, so,at time t=5.3 s
[tex]x|_{t=5.3}= (5.2 m)cos[ (5\pi \ rad/s)5.3+ \frac\pi5][/tex]
[tex]= (5.2 m)cos[ 26.5\pi+ \frac\pi5][/tex]
=(5.2)(-0.588)m
= - 3.0576 m
(b)
[tex]x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
To find the velocity of simple harmonic motion, we need to find out the first order derivative of the function.
[tex]v=\frac{dx}{dt}[/tex]
[tex]=\frac{d}{dt} (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= (5.2 m)(-5\pi)sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
Now we can plug our value t=5.3 into the above equation
[tex]v= -26\pi sin[ (5\pi \ rad/s)5.3\ s+ \frac\pi5][/tex]
[tex]=-66.48[/tex] m/s
(c)
To find the acceleration of simple harmonic motion, we need to find out the second order derivative of the function.
[tex]v= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]a=\frac{d^2x}{dt^2}[/tex]
[tex]=\frac{dv}{dt}[/tex]
[tex]=\frac{d}{dt}( -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5])[/tex]
[tex]= -26\pi (5\pi)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= -130\pi^2cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
Now we can plug our value t=5.3 into the above equation
[tex]a= -130\pi^2cos[ (5\pi \ rad/s)5.3 \ s+ \frac\pi5][/tex]
= -753.39 m²/s
(d)
The general equation of SHM is
[tex]x=x_mcos(\omega t+\phi)[/tex]
[tex]x_m[/tex] is amplitude of the displacement, [tex](\omega t+\phi)[/tex] is phase of motion, [tex]\phi[/tex] is phase constant.
So,
[tex](\omega t+\phi)=5\pi t+\frac\pi5[/tex]
Now plugging t=5.3s
[tex](\omega t+\phi)=5\pi \times 5.3+\frac\pi5[/tex]
=26.7 [tex]\pi[/tex]
The phase motion is 26.7 [tex]\pi[/tex].
The angular frequency [tex]\omega = 5\pi[/tex]
(e)
The relation between angular frequency and frequency is
[tex]\omega =2\pi f[/tex]
[tex]\therefore f=\frac{\omega}{2\pi}[/tex]
[tex]=\frac{5\pi}{2\pi}[/tex]
[tex]=\frac52[/tex]
= 2.5 Hz
Frequency =2.5 Hz.
(f)
The relation between frequency and time period is
[tex]T=\frac1 f[/tex]
[tex]=\frac1{2.5}[/tex]
=0.4 s
Time period =0.4 s
A mass of 1.0 kg is initially held in place on a ramp of angle 45o at height of 20. meters above the ground. The mass is released and slides to the bottom of the ramp (h=0). There is friction between the mass and the ramp. At the bottom of the ramp, the object has a speed of 10. m/s. The thermal energy was generated by the friction as the mass slides down the ramp is closest to which value?
Answer:
The thermal energy generated by the friction as the mass slides down the ramp is [tex]\bf{146~J}[/tex].
Explanation:
Given:
The mass of the object is, [tex]m = 1.0~kg[/tex]
The angle of the ramp is, [tex]\theta = 45^{0}[/tex]
The initial height of the object on the ramp is, [tex]h = 20~m[/tex]
The final velocity of the object is, [tex]v = 10~m/s[/tex]
When the object is at rest on the ramp, its total energy is potential energy. When it moves down the ramp its kinetic energy is increased and potential energy is decreased and a part of its energy is lost to overcome the force of friction. Finally, when it is at the bottom of the ramp, its total energy becomes only kinetic energy.
The total energy of the object at a height [tex]20~m[/tex] on the ramp is given by
[tex]E_{1} &=& mgh\\~~~~&=& (1.0~kg)(9.8~m/s^{2})(20~m)\\~~~~&=& 196~J[/tex]
When the object is at the bottom of the ramp, its total energy is given by
[tex]E_{2} &=& \dfrac{1}{2}mv^{2}\\~~~~&=& \dfrac{1}{2}(1.0~kg)(10~m/s)^{2}\\~~~~&=& 50~J[/tex]
So, the energy that is lost as thermal energy is given by
[tex]E &=& E_{1} - E_{2}\\~~~~&=& 196~J - 50~J\\~~~~&=& 146~J[/tex]
what does a electric field of two positive charges look like
Hope this helped you
:)
Have a great day!
Dolphin echolocation is similar to ultrasound. Reflected sound waves
allow a dolphin to form an image of the object that reflected the waves.
Dolphins can produce sound waves with frequencies ranging from
0.25 kHz to 220 kHz, but only those at the upper end of this spectrum
are used in echolocation. Explain why high-frequency waves work better
than low-frequency waves.
Answer:
Waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
Explanation:
To understand why high-frequency waves work better than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.
The relation between frequency and wavelength is given by
λ = c/f
Where λ is wavelength, c is the speed of light and f is the frequency.
Since the speed of light is constant, the wavelength and frequency are inversely related.
So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.
Thus, waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
A speed skater is travelling at 2 m/s and accelerates uniformly to 4 m/s in 5 seconds. What is her acceleration?
Answer:
The answer to your question is a = 0.4 m/s²
Explanation:
Data
speed 1 = 2 m/s
speed 2 = 4 m/s
time = 5 s
Acceleration measures the change of speed over a unit of time.
Formula
Acceleration = (speed 2 - speed 1) / time
Substitution
Acceleration = (4 - 2) / 5
Simplification
Acceleration = 2/5
Result
Acceleration = 0.4 m/s²
Answer: her acceleration is 0.4 m/s^2
I need some help with my science homework about energy, work, and power. It would be greatly appreciated :)
Answer:
1. B
Explanation:
Work = Force × Distance
Work = 4N × 1.5M
Calculate the electrical energy expended in a device across which the circuit voltage drops by 20.0 volts in moving a charge of 4 coulombs.
A)5 J
B)80 J
C)0.2 J
Answer:
The answer is 80 Joules
Explanation:
Electrical energy = Q x V
Energy = 2 x 40
= 80
I just took the test and it was right :D
The correct answer is B.
I hope this helped! :D
why is it painful to lift a heavy load with a thin piece of string?
Explanation:
A thin piece of string has lower force than of a heavy load.
When the load is being lifted, the force exerted by the load is much more greater than the string, which eventually hits our hands.
For example, a heavy bag. You tie a thin string to it and try to lift it. The force exerted by the bag will hit the string harder, reaching for your plams. As the thin string has less force, its reaction force is not enough to hit back the greater force. Also, the less the surface area, the more difficult the grip gets. But, if you attach a thick rope or belt, the force exerted my the bag is automatically minimized as the reaction force of the thick rope is near about the action force. Hence, greater the surface area, better the grip.
*action force: force exerted by the bag
*reaction force: the force hit back by the rope
It is painful to lift a heavy load with a thin piece of string because the string has a small cross-sectional area and cannot withstand a large amount of tension without breaking.
What is tensile strength?Tensile strength is the maximum amount of tensile stress that a material can withstand before it breaks or undergoes permanent deformation. It is a measure of the material's ability to resist external forces that try to pull it apart or elongate it.
Tensile strength is an important mechanical property of materials, and it is commonly used in engineering and design to select materials for specific applications. The tensile strength of a material is typically determined through a tensile test, where a sample of the material is subjected to a gradually increasing tensile force until it breaks.
The tensile strength of a material depends on its composition, microstructure, and processing conditions. For example, materials that have a high degree of crystallinity and are free from defects or impurities generally have a higher tensile strength than materials with a less ordered microstructure or defects. In addition, the processing conditions, such as temperature and strain rate, can also affect the tensile strength of a material.
The tensile strength is typically reported in units of force per unit area, such as megapascals (MPa) or pounds per square inch (psi). The higher the tensile strength of a material, the greater its ability to withstand tensile stress without breaking or undergoing permanent deformation.
Here in the Question,
It is painful to lift a heavy load with a thin piece of string because the string has a small cross-sectional area and cannot withstand a large amount of tension without breaking. When a heavy load is attached to the string and lifted, the weight of the load creates a tension force in the string that is equal to the weight of the load. If the tension force in the string exceeds its maximum tensile strength, it will break.
The maximum tensile strength of a string depends on its material properties and cross-sectional area. When a thin piece of string is used to lift a heavy load, the tension force in the string can easily exceed its maximum tensile strength, causing it to break. This sudden release of tension can also cause the load to drop suddenly, which can lead to injury.
This can be explained using the concept of stress and strain. Stress is the force applied to an object per unit area, while strain is the deformation of the object due to the applied stress. When a heavy load is attached to a thin piece of string, the weight of the load creates a large stress in the string, which can cause it to undergo plastic deformation and break. This is because the string has a small cross-sectional area, which means that the stress is concentrated over a small area, leading to a high level of strain.
Therefore, a thicker piece of string or rope has a larger cross-sectional area, which means that the stress is spread out over a larger area. This allows it to withstand a greater amount of tension before breaking. Therefore, it is important to use a string or rope with a sufficient cross-sectional area when lifting heavy loads to prevent injury and damage.
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The rate of change of angular momentum of a particle equals the torque of the net force acting on it is called
Answer:
Explanation:
The rate of change of angular momentum of a particle equals the torque of the net force acting on it is called CONSERVATION OF ANGULAR MOMENTUM
ΣIα = τ
τ(net) = I•α
Where,
α is angular acceleration
τ is Torque
Answer:
Conservation of angular momentum
Explanation:
The rate of change of angular momentum of a particle equals the torque of the net force acting on it is called conservation of angular momentum.
L = ΣF x r = τ
where;
L is the angular momentum of the particle
τ is torque on the particle
r is the distance through which the force act on the particle
ΣF is the net force on the particle
Thus, the rate of change of angular momentum of a particle equals the torque of net force acting on it, (L = τ) and it is called conservation of angular momentum.
A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are 1200 kg and 1800kg; the magnitude of the impulse on each par from the bolts is 300 N s. With what relative speed do the twe twe parts separate because of the detonation?
Answer:
Explanation:
Given that,
A space ship is separated into two part
Mass of first part
M1 = 1200kg
Mass of second part
M2 = 1800kg
Magnitude of impulse on each part is
I = 300Ns
We want to find the relative velocity at which the two parts separate
Now,
Impulse is give as
I = ft = mv - mu
Since the body is considered to be at rest before detonation
Then, I = mv
So for first part
I = M1•V1
V1 = I / M1
V1 = 300 / 1200
V1 = 0.25m/s
Also, for second part
V2 = —I / M2
V2 = —300 / 1800
V2 = —0.167 m/s
NOTE: the negative sign is due to the fact that the two bodies are moving away from each other.
The relative speed of the two masses because of detonation is
V = V1 — V2
V = 0.25 — (—0.167)
V = 0.25 + 0.167
V = 0.417 m/s
The relative speed of the two masses because of detonation is 0.417 m/s
Answer:
(1) [tex]\frac{5}{12} m/s[/tex]
(2) [tex]-\frac{5}{12} m/s[/tex]
Explanation:
Lets say two parts went left and right, one with 1200Kg, call it A ,went right and one with mass 1800Kg went left. Let's establish right as positive and left as negative for our speed convention.
Both have initial acceleration of 300N, assuming for one second, by newton's second law F=ma, 'A' will have acceleration of 1/4m/s^ which will induce velocity of 0.25m/s as well.
So A has velocity of [tex]\frac{1}{4} m/s[/tex] towards right direction.
and B has velocity of - [tex]\frac{1}{6} m/s[/tex] towards left direction.
Relative Velocities.
and 'B' will have acceleration of [tex]\frac{1}{6} m/s^2[/tex] which will also produce velocity of [tex]\frac{1}{6} m/s[/tex].
[tex]V_{AB} = V_{A} -V_{B}[/tex] = [tex]$\frac{1}{4} --\frac{1}{6}= \frac{5}{12}m/s[/tex] (Read as Velocity A with respect to B) (1)
[tex]V_{BA} =V_{B} -V_{A} = -\frac{1}{6}-\frac{1}{4} =-\frac{5}{12} m/s[/tex] (Read as Velocity B with respect to B). (2).
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark. From this information, the mass of the meterstick is _________.
A) 1/4 kg.B) 1/2 kg.C) 3/4 kg.D) 1 kg.E) none of the above
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
[tex]9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}[/tex]
W = 3.266 N
The mass of the meters stick is :
[tex]m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg[/tex]
So, the mass of the meter stick is 0.333 kg.
The mass of the meterstick is calculated using the principle of torque balance. The meterstick must have a mass that creates an equal opposite torque to the one created by the 1-kg rock. The calculation shows that the meterstick has a mass of 1/3 kg, which is not listed in the options, so the answer is (E) none of the above.
Explanation:The mass of the meterstick can be found using the principle of moments, also known as the principle of leverage, which states that for an object to be in rotational equilibrium, the sum of the clockwise moments about the pivot (fulcrum) must equal the sum of the counterclockwise moments. In this case, the 1-kg rock is hanging from the 0-cm mark and the meterstick balances when the fulcrum is at the 12.5-cm mark. The torque created by the 1-kg mass is given by its mass multiplied by its distance from the fulcrum, or 1 kg × 12.5 cm.
For the meterstick to balance, its center of mass must be located directly above the fulcrum. Since the meterstick is uniform, its center of mass is in the middle, at the 50-cm mark. This means that the mass of the meterstick acts 50 cm - 12.5 cm = 37.5 cm away from the fulcrum. Let's denote the mass of the meterstick as 'M'. The counterbalance torque provided by the meterstick is given by M kg × 37.5 cm.
Setting the torques equal for a balanced system:
M kg × 37.5 cm = 1 kg × 12.5 cm, which simplifies to M = 1/3 kg. Since 1/3 kg is equivalent to approximately 0.33 kg, the correct answer is (E) none of the above, as none of the provided options matches this result.
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A record player runs at 78 RPMs (revolutions per minute). That means that it spins 78 times (cycles) in a minute. work each one out.
a. Find the frequency of the record player.
b. Find the period of the record player.
Answer:
a. the frequency of the record player
f = 1.3 rev per second = 1.3 Hz
b. the period of the record player.
T = 0.77 seconds
Explanation:
Given;
Speed of record player v = 78 rpm
a) frequency of the record player is the number of revolutions the record player makes per second.
f = 78 rev/minute = 78/60sec = 1.3 rev per second
f = 1.3 Hz
b) period of the record player is the amount of time needed by the record player to complete one revolution.
T = 1/f (or 60s/78rev = 0.77 s)
T = 1/1.3
T = 0.77 seconds
Answer:
(a) The frequency of the record player is 1.3 Hz
(b) The period of the record player is 0.77 s
Explanation:
Given number of revolution per minute = 78 RPM
Part (a) the frequency of the record player
One revolution per minute, 1 RPM = ¹/₆₀ Hz
78 RPM = ?
Thus, 78 RPM = 78 ( ¹/₆₀ HZ) = 1.3 Hz
The frequency of the record player is 1.3 Hz
Part (b) the period of the record player
Period is inverse of frequency
T = 1 / f
T = 1 / 1.3 Hz
T = 0.77 s
The period of the record player is 0.77 s
Dan is gliding on his skateboard at 3.00 m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.00 m/s. Dan's mass is 70.0 kg and the skateboard's mass is 6.00 kg. How fast is Dan going as his feet hit the ground?
Answer:
The velocity of Dan is 2.57m/s as his feet hit the ground
Explanation:
The impact experience by Dan and the skate board is an elastic collision
Collision is elastic when the kinetic energy is not conserved and if there is rebound after collision
Given that
U= initial velocity of Dan and the skate board 3m/s
M1 =mass of Dan 70kg
M2= mass of Skate board 6kg
V1= final velocity of Dan?
V2= Final velocity of skate board 8m/s
The expression for Dan and the skate board collision can be expressed as
Momentum before impact momentum after impact
(M1+M2)U=M1V1+M2V2
Substituting our data we have
(70+6)3=(70*V1)+(6*8)
228=70V1+48
Solving for V1
228-48=70V1
V1=180/70
V1=2.57m/s
How do the three opsin molecules respond to different wavelengths of light when their retinal molecular structures are identical?
Answer:
The three opsin molecules respond to different wavelengths of light because of the different structure of the protein bound to the opsin molecules.
Explanation:
Opsin is a protein that is released by the action of light and forms part of the visual pigment rhodopsin. There are two groups of protein termed opsins, type I opsins, which are employed by prokaryotes and by some algae and fungi, and type II opsins which are used by animals.
Though the retinal molecular structures of the opsin molecules are identical, they respond to different wavelengths of light because of the different structure of the protein bound to the opsin molecules.
Conventional current is the rate at which positive charge flows in a circuit. In atoms, only the electrons are free to move. What can you conclude from these statements?
A) Electrons must be positively charged.
B) The definition of current must be changed.
C) Electrons do not flow in electric circuits.
D) Charges actually flow opposite the conventional current.
Answer:
In a metal circuit, the charges which are free electrons flow opposite to the flow of conventional current(which is assumed as the flow of positive charges)
Explanation:
Conventional current is defined as the direction that positive charge would flow, which is opposite to the actual flow of electrons in a circuit.
From the statements given, we can conclude that charges actually flow opposite the conventional current. This is because conventional current assumes that positive charge is moving in the direction of the electric field, whereas in actuality, in metal wires, it is the electrons— which have a negative charge—that are moving.
Electrons flow in a direction opposite to the defined conventional current. The designation of the direction of conventional current dates back to Benjamin Franklin's time, and despite later discovery that electrons are the primary charge carriers in circuits, the convention has remained the same.
The speed of propagation equals the frequency times the wavelength.
O
True
O
False
Answer:
true
Explanation:
because I say it's true
Answer:help me with my physics please
Explanation:
5. Suppose a cup of cocoa cooled from 90° C to 60° C after 10 minutes in a room whose temperature was 20° C . Use Newton’s law of cooling [T − Ts = (T0 − Ts )⋅ e−k⋅t ] to answer the following questions: a) How much longer would it take the cocoa to cool to 35° C ? b) Instead of being left to stand in a room, the cup with initial temperature 90° C is placed in a freezer whose temperature is −15° C . How long will it take the cocoa to cool from 90° C to 35° C ?
Answer:
a) t = 1051.6 sec = 17.5 min
b) t = 795.5 sec = 13.25 min
Explanation:
First of all we use the initial data to find out constant 'K'.
T - Ts = (T₀ - Ts) e^(-kt)
Here, we have:
T = Final Temperature = 60° C
Ts = Surrounding Temperature = 20° C
T₀ = Initial Temperature = 90° C
t = time = 10 min = 600 sec
k = constant = ?
Therefore,
60° C - 20° C = (90° C - 20° C).e^(-k600)
40° C/70° C = e^(-k600)
ln (0.57142) = -600k
k = 9.327 x 10⁻⁴ sec⁻¹
a)
Now, for this case we have:
T = Final Temperature = 35° C
Ts = Surrounding Temperature = 20° C
T₀ = Initial Temperature = 60° C
t = time = ?
k = constant = 9.327 x 10⁻⁴ sec⁻¹
Therefore,
35° C - 20° C = (60° C - 20° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
15° C/40° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
ln (15/40) = - 9.327 x 10⁻⁴ sec⁻¹ x t
t = 1051.6 sec = 17.5 min
b)
Now, for this case we have:
T = Final Temperature = 35° C
Ts = Surrounding Temperature = -15° C
T₀ = Initial Temperature = 90° C
t = time = ?
k = constant = 9.327 x 10⁻⁴ sec⁻¹
Therefore,
35° C + 15° C = (90° C + 15° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
50° C/105° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
ln (50/105) = - 9.327 x 10⁻⁴ sec⁻¹ x t
t = 795.5 sec = 13.25 min
A 2.9 kg ball strikes a wall with a velocity of 8.6 m/s to the left. The ball bounces off with a velocity of 7.3 m/s to the right. If the ball is in contact with the wall for 0.33 s, what is the constant force exerted on the ball by the wall? Answer in units of N.
Answer:
The constant force exerted by the wall is F=11.4N
Explanation:
The problem bothers on the impulse of a force
Which is given as
Ft=mv-mu
Ft=m(v-u)
Given data
mass of ball m =2.9kg
Final speed v=8.6m/s
Initial speed u=7.3m/s
Time t= 0.33s
Substituting to find F
F*0.33=2.9(8.6-7.3)
F*0.33=2.9*1.3
F=3.77/0.33
F=11.4N
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the assembly expands from an initial volume of 0.90 L to a final volume of 5.92 L against an external pressure of 1.0 atm. Use the wattage of the lightbulb and the time it is on to calculate ΔE in joules (assume that the cylinder and lightbulb assembly is the system and assume two significant figures). Calculate w and q.
Answer:
[tex]w = - 508.53[/tex] joules
[tex]q = - 3091.47[/tex] joules
Explanation:
Let us convert the time in hours into seconds
[tex]0.010* 3600\\= 36[/tex]
Change in internal energy
[tex]\delta E = p * \delta t[/tex]
where E is the internal energy in Joules
p is the power in watts
and t is the time in seconds
[tex]\delta E = - 100 * 36\\[/tex]
[tex]\delta E = - 3600[/tex] Joules
Amount of work done by the system
[tex]w = - P * \delta V[/tex]
where P is the pressure and V is the volume
Substituting the given values in above equation, we get -
[tex]w = - 1 * ( 5.92 -0.90)\\[/tex]
[tex]w = -5.02[/tex] liter-atmospheres
Work done in Joules
[tex]- 5.02 * 101.3\\[/tex][tex]= 508.53[/tex]Joules
[tex]q = \delta E - w\\[/tex]
Substituting the given values we get -
[tex]q = - 3600 - (-508.53)\\q = - 3091.47[/tex]
Thus
[tex]w = - 508.53[/tex] joules
[tex]q = - 3091.47[/tex] joules
The change in energy (ΔE) is 1 joule (J), the work (w) is -41 J, and the heat (q) is 42 J.
Explanation:To calculate the change in energy (ΔE) in joules, we can use the formula ΔE = power (P) x time (t). In this case, the power of the lightbulb is 100 W and the time it is on is 0.010 hour. Therefore, ΔE = 100 W x 0.010 hour = 1 joule (J).
To calculate work (w), we can use the equation w = -PΔV, where ΔV is the change in volume of the assembly. Here, ΔV = final volume - initial volume = 5.92 L - 0.90 L = 5.02 L. Given that the external pressure is 1.0 atm, we can convert the volume to liters-atmospheres (L·atm) by multiplying by 0.0821. Therefore, w = -100 W x (5.02 L x 0.0821) = -41 J.
Finally, to calculate heat (q), we can use the first law of thermodynamics, which states that q = ΔE - w. Substituting the values, q = 1 J - (-41 J) = 42 J.
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