12-inch wide conveyor unloads 5000 bushels of wheat per hour. At this same speed, the number of bushels that will be unloaded per hour by an 18-inch wide conveyor can be calculated below
[tex]\begin{gathered} 12\text{ = 5000} \\ 18=\text{?} \\ u\text{nload amount=}\frac{5000\times18}{12}=\frac{90000}{12}=7500\text{ bushels per hour } \end{gathered}[/tex]the position of an object is given by: x(t)=2t^3-35t^2+10 what is the acceleration of the object at t=250 seconds?
We are given that a particle's position is determined by the function:
[tex]x\left(t\right)=2t^3-35t^2+10[/tex]We are asked to determine the acceleration of the object. To do that we will first determine the first derivative of the position with respect to time, i. e. the velocity of the particle:
[tex]\frac{dx}{dt}=\frac{d}{dt}(2t^3-35t^2+10)[/tex]Now, we distribute the derivative operator:
[tex]\frac{dx}{dt}=\frac{d}{dt}(2t^3)-\frac{d}{dt}(35t^2)+\frac{d}{dt}(10)[/tex]Now, we use the following rule to determine each derivative:
[tex][/tex]An acorn is dropped and reaches the ground in 2.0s. What is the acorn’s velocity just as it hits the ground? How high above the ground was it released?
Take into account that the vertical distance traveled by the acorn can be calculated by using:
[tex]h=\frac{1}{2}gt^2[/tex]where,
g: gravitational acceleration constant = 9.8m/s^2
t: time acorn takes to hit the ground = 2.0s
h: height at which acorn is initially = ?
Replace the given values of the parameters into the previous formula and simplfiy:
[tex]h=\frac{1}{2}(9.8\frac{m}{s^2})(2.0s)^2=19.6m[/tex]Hence, the acorn was released from a high of 19.6m above the ground.
The velocity just after acorn hits the ground is given by:
[tex]\begin{gathered} v=gt \\ v=(9.8\frac{m}{s^2})(2.0s)=19.6\frac{m}{s} \end{gathered}[/tex]Hence, the speed is 19.6 m/s
Unbeknownst to most students, every time the school floors are waxed, the physics teachers get together to have a barrel of phun doing friction experiments in their socks (uhm - they do have clothes on; its just that they don't have any shoes on their feet). On one occasion, Mr. London applies a horizontal force to accelerate Mr. Schneider (mass of 84 kg) rightward at a rate of 1.2 m/s/s. If the coefficient of friction between Mr. Schneider 's socks and the freshly waxed floors is 0.35, then with what force (in Newtons) must Mr. London be pulling?
The frictional force is taken away from the net force. Mr. London will pull with a force of 388.9 N
What is Force ?Force can simply be defined as a pull or push. It is a product of mass and acceleration. It is a vector quantity.
Given that Mr. London applies a horizontal force to accelerate Mr. Schneider (mass of 84 kg) rightward at a rate of 1.2 m/s/s. If the coefficient of friction between Mr. Schneider 's socks and the freshly waxed floors is 0.35, then the force F in which Mr. London is pulling will be;
F - Fr = ma
Where
F = ?m = 84 kga = 1.2 m/s²N = ?μ = 0.35N = mg
N = 84 × 9.8
N = 823.2 N
Fr = μN
Fr = 0.35 × 823.2
Fr = 288.12 N
F - 288.12 = 84 × 1.2
F - 288.12 = 100.8
F = 100.8 + 288.12
F = 388.92 N
Therefore, Mr. London must be pulling with a force of 388.9 N
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i need help asap i just need the vocab i need help with this please help
ANSWERS
0. First law of Thermodynamics:, ,B,. Thermal energy can change form and location, but it cannot be created or destroyed
,1. Thermal Energy: ,H,. Kinetic energy in transit from one object to another due to temperature difference
,2. Temperature: ,D,. the average kinetic energy of particles in an object - not the total amount of kinetic energy particles
,3. Absolute zero: ,J, occurs when all kinetic energy is removed from an object
,4. Thermal Equilibrium: ,E, obtained when touching objects within a system reach the same temperature
,5. Zeroth Law of Thermodynamics: ,C, if two systems are separately found to be in thermal equilibrium with a third system, the first two systems are in thermal equilibrium with each other.
,6. Conduction: ,F, ,the transfer of thermal energy within an object or between objects from molecule to molecule
,7. Coefficient of Heat Conductivity: ,G, is the measure of a material's ability to conduct heat.
,8. Thermal resistance of a material: ,A, the measure of a material's ability to resist heat
,9. Radiation: ,I, the process by which energy is transmitted through a medium, including empty space, as electromagnetic waves.
EXPLANATION
The zeroth and first laws of thermodynamics are known statements, as well as the definition of thermal equilibrium, so there is nothing to explain about these three.
Thermal energy is a form of kinetic energy. It is "generated" by the vibrations of the molecules or atoms that form the object, causing the object to increase its temperature. As a result, the temperature is the average kinetic energy of these particles.
Absolute zero is a temperature of 0 degrees Kelvin - approximately -273 degrees Celsius. When an object reaches this particular temperature it is considered as if the object has "no temperature" - that is why it is called absolute zero. As explained above, the temperature is the average kinetic energy of the object's particles, so if the particles have no kinetic energy, then the object has zero temperature.
Heat conductivity and thermal resistance are basically opposites. The first is the object's ability to conduct heat while the second is the ability to resist the conduction of heat.
Radiation and Conduction are two ways energy can be transferred. The first does not need a medium - that is why energy can be transmitted through empty space, and the second is by contact between two objects or an object and a medium.
Question 28 of 30The graph below shows the conservation of energy for a skydiver jumping outof a plane and landing safely on the ground. Which energy is represented byline A?сAEnergy (kJ)BDistance (m)A. Potential energyB. Kinetic energy
Answer:
Explanation:
Potential energy depends on the height of the object above the ground. As the height decreases, the potential energy decreases. Looking at line A, the energy is decreasing as the distance is increasing. Thus, the energy represented by line A is
A. Potential energy
Light of wavelength 589 nm in vacuum passes through a piece of fused quartz of index of refraction n = 1.458. (a) Find the speed of light in fused quartz. (b) What is the wavelength of this light in fused quartz? (c) What is the frequency of the light in fused quartz?
The speed of light in fused quartz 2.058 * [tex]10^{8}[/tex] m/s
The wavelength of the light in fused quartz 404.322 * [tex]10^{-9}[/tex] m
The frequency of the light in fused quartz 0.00509 * [tex]10^{17}[/tex] hz
refractive index of a material = speed of light in vacuum / speed of light in that medium
1.458 = 3 * [tex]10^{8}[/tex] / speed of light in that medium
speed of light in fused quartz = 3 * [tex]10^{8}[/tex] / 1.458 = 2.058 * [tex]10^{8}[/tex] m/s
frequency = c / lambda = 3 * [tex]10^{8}[/tex] / 589 * [tex]10^{-9}[/tex] = 0.00509 * [tex]10^{17}[/tex] hz
wavelength of light in fused quartz = speed of light in fused quartz / frequency
= 2.058 * [tex]10^{8}[/tex] / 0.00509 * [tex]10^{17}[/tex] = 404.322 * [tex]10^{-9}[/tex] m = 404.322 nm
wavelength = wave velocity in fused quartz / frequency
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URGENT!! ILL GIVE
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A feather and a bowling ball are each dropped from an equal height in a vacuum and land at the same time. Which graph shows the total mechanical energy of the bowling ball as it falls?
The total mechanical energy of the bowling ball and the feather is shown by the graph in option D
What is the total mechanical energy?We know that the mechanical energy is the energy that is possessed by a body by virtue of its motion or by virtue of its staying at a place. Thus mechanical energy is possessed by an object that is moving or by an object that is at rest.
In this case, we have a feather and a bowling ball are each dropped from an equal height in a vacuum and land at the same time. We know that the mechanical energy of the two objects must be constant. This is because, the potential energy of the feathers and the ball at a height is converted to kinetic energy as the two objects begin to move.
The graph that would show the total energy must be one in which the energy axis of the graph is constant as we see in option D.
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Answer:
the answer is D
Explanation:
of the following and draw a neat, properly labelled free body diagram. Indicate the size of forces with the lengths of the arrows. i.e., if two forces are the same size draw them the same length and if one is much bigger draw it much longer. a) The Space Shuttle accelerating vertically just after leaving the launch pad.
We will have that the diagram will be the following:
Here "F" is the force the shuttle is using to take off, friction is the friction with the air and "W" is the weight of the shuttle.
A 0.24 kg blob of clay strikes a wall with an Initial velocity of 25 m/s the clay come to a stop against the wall in 91 ms what is the average impact force on the clay
ANSWER
65.9N
EXPLANATION
The impact force of an object is
[tex]F=\frac{\Delta p}{\Delta t}[/tex]And the change in momentum is
[tex]\Delta p=m(v_f-v_i)[/tex]We know that the mass of the blob of clay is 0.24kg, its initial velocity is 25m/s and its final velocity is 0m/s. Also, the time of the collision is 91 ms.
The force is:
[tex]F=\frac{m(v_f-v_i)}{\Delta t}[/tex]To be consistent with the units, we have to write the time in seconds, not milliseconds:
[tex]F=\frac{0.24\operatorname{kg}(0-25)m/s}{0.091s}[/tex][tex]F\approx-65.9N[/tex]This result is negative because this is the force the clay exerts on the wall. By Newton's third law, the impact force on the clay is the force with the same magnitude and opposite direction. Therefore, the average impact force on the clay is 65.9N
Classify the types of forces below into four groups in the table provided. Write only the letters to represent the forces. Give a heading for each group in the table.
A The force which prevents a glass bottle from slipping out of your hand
B The force which makes it difficult to climb a steep mountain
C The force which gives you weight
D The force which makes it difficult for you to push down a float in water
E The force which causes a parachutist to fall when he jumps out of a helicopter
F The force which slows down an airplane moving in air
G The force which makes a compass turns
Gravity force, friction force and buoyant force are the types of forces given below.
A) The force which prevents a glass bottle from slipping out of your hand- friction force
B) The force which makes it difficult to climb a steep mountain- gravity
C) The force which gives you weight- gravity
D) The force which makes it difficult for you to push down a float in water- buoyant force
E) The force which causes a parachutist to fall when he jumps out of a helicopter- gravity
F) The force which slows down an airplane moving in air- friction force
G) The force which makes a compass turns- gravity
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Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved 12 m?24 m16 m9.0 m28 m21 m
We will have the following:
First, we recall that momentum is given by:
[tex]m_1v_0+m_2v_0=m_1v_1+m_2v_2[/tex]So:
[tex]\begin{gathered} (60kg)(0m/s)+(80kg)(0m/s)=(60kg)v_1+(80kg)v_2 \\ \\ \Rightarrow v_1=\frac{-(80kg)v_2}{(60kg)}\Rightarrow v_1=-\frac{4}{3}v_2 \end{gathered}[/tex]Then:
[Let's assume that when astronaut 2 moved 12 meters = t]
[tex]\begin{gathered} x_1=v_1t\Rightarrow x_1=-\frac{4}{3}v_2t \\ \\ x_2=v_2t\Rightarrow t=\frac{12}{v_2} \end{gathered}[/tex]Then:
[tex]x_1=-\frac{4}{3}v_2\ast\frac{12}{v_2}\Rightarrow x_1=-16m[/tex]Finally, the total distance will be:
[tex]\begin{gathered} d=|x_1|+|x_2|\Rightarrow d=16m+12m \\ \\ \Rightarrow d=28m \end{gathered}[/tex]in a particle accelerator a proton moves at constant speed 0.750c in a circle of radius 663 m. what is the net force on the proton?
In a particle accelerator a proton moves at constant speed 0.750c in a circle of radius 663 m. The net force on the proton will be .0127 * [tex]10^{-11}[/tex] N
Any object moving in a circle (or along a circular path) experiences a centripetal force. That is, there is some physical force pushing or pulling the object towards the center of the circle.
Force = m ( [tex]v^{2} / R[/tex])
m= mass
v = speed
R = radius
force = 1.67 * [tex]10^{-27}[/tex] * [tex](0.75 * 3 * 10^{8}) ^{2}[/tex] / 663 = 0.0127 * [tex]10^{-11}[/tex] N
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A solenoid is wound with 259 turns per cm. An outer layer of insulated wire with 51 turns per cm is wound over the first layer of wire. The inner coil carries a current of 7.577 A, and the outer coil carries a current of 21.68 A in the opposite direction. What is the magnitude of the magnetic field, in microTeslas, at the central axis ?
Answer tab
For this question, we'll first define a number, called the "linear density of the coil" which is exactly the one the exercise gives us, in a unit of turns per length. The magnetic field can be calculated as:
[tex]B=\mu_0in[/tex]Where n is the linear density.
In our case, as the coils carry current in opposite directions, the generated magnetic fields will be opposed, and we'll have:
[tex]B=B_{outer}-B_{inner}=\mu_0*(51*10^2*21.68-259*10^2*7.577)[/tex]Please note that we had to multiply by 10^2, in order to convert turns/cm to turns/m
Then, our final magnetic field will be:
[tex]|B|=0.107664T=107664\mu T[/tex]Our final answer is B=107664uT
Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength (in T) magnetic field is needed to hold antiprotons, moving at 6.10 ✕ 107 m/s in a circular path 1.70 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (Enter the magnitude.)
Data:
B=?, v=6.10*10^7 m/s, R=1.70m, m=1.67*10^(-27)kg, q=-1.6*10^(-19)C
Answer:
Firstly, we need to remember the formula for the Magnetic force, which is:
[tex]F_m=|q|vB[/tex]However, in this scenario the magnetic force will act as a centripetal force, thus:
[tex]F_m=\frac{mv^2}{R}[/tex]If we equal both of them:
[tex]|q|vB=\frac{mv^2}{R}\Rightarrow B=\frac{mv}{|q|R}[/tex]Replacing our values:
[tex]B=\frac{1.67*10^{-27}*6.1*10^7}{1.6*10^{-19}*1.7}=0.3745T[/tex]Then, our magnetic field will be B=0.3745T
A bicycle wheel, of radius 0.300 m and mass 2.07 kg (concentrated on the rim), is rotating at 4.00 rev/s. After 59.0 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?
ANSWER:
0.079 J
STEP-BY-STEP EXPLANATION:
The average torque exerted due to friction force is:
[tex]\tau=I\alpha[/tex]Here, I is the moment of inertia and α is the angular acceleration.
The mass of wheel is concentrated at the rim. Therefore, the moment of inertia of wheel is,
[tex]I=m\cdot r^2[/tex]Here, m is the mass of wheel and r is the radius of wheel.
replacing:
[tex]\begin{gathered} I=2.07\cdot0.3^2 \\ I=0.1863\text{ kg}\cdot s^2 \end{gathered}[/tex]The angular speed is decreased from 4 rev/s to 0 rev/s in 59 sec. Therefore, the angular acceleration is calculated as:
[tex]\begin{gathered} \alpha=\frac{0-4}{59} \\ \alpha=0.068\cdot\frac{rev}{\sec}\cdot\frac{2\pi\text{ rad/rev}}{1rev/sec^2} \\ \alpha=-0.426rad/sec^2 \end{gathered}[/tex]The negative sign represents that the wheel is de-accelerating.
Replacing:
[tex]\begin{gathered} \tau=I\alpha \\ \tau=0.1863\cdot-0.426 \\ \tau=-0.079\text{J} \end{gathered}[/tex]The negative sign represents the direction of torque i.e. clockwise direction. Thus, the magnitude of average torque is 0.079 J.
A massless scaffold is held up by a wire at each end. The scaffold is 12 m long. A300-N box sits 4.0 m from the left end. What is the tension in each wire?1) left wire = 100 N; right wire = 200 N2) left wire = 200 N; right wire = 100 N3) left wire = 900 N; right wire = 2700 N4) left wire = 2700 N; right wire = 900 N
Free body diagram:
Given data:
Length of massless scaffold (end to end) L=12 m.
Weight of box m=300 N.
Length of massless scaffold (center to end) l=6 m.
As, the box sits 4.0 m from the left end, the distance of the box from the center of massless scaffold is given as,
[tex]\begin{gathered} r=6.0\text{ m}-4.0\text{ m} \\ =2.0\text{ m} \end{gathered}[/tex]Balancing force in y direction,
[tex]T_1+T_2=300\text{ N}\ldots(1)[/tex]The torque is given as,
[tex]\tau=perpendicular\text{ distance}\times force[/tex]Therefore, torque along the center of massless scaffold is given as,
[tex]\begin{gathered} \Sigma\tau=0 \\ l\times T_1+r\times(300\text{ N})-l\times\tau_2=0 \\ 6\times T_1+2\times(300\text{ N})-6\times T_2=0 \\ 6T_1+600\text{ N}-6T_2=0 \\ 6(T_1+100\text{ N}-T_2)=0 \\ T_1+100\text{ N}-T_2=0 \\ T_1-T_2=-100\text{ N}\ldots(2) \end{gathered}[/tex]Adding equation (1) and (2),
[tex]\begin{gathered} (T_1+T_2)+(T_1-T_2)=300\text{ N}-100\text{ N} \\ T_1+T_1+T_2-T_2=200\text{ N} \\ 2T_1=200\text{ N} \\ T_1=\frac{200\text{ N}}{2} \\ T_1=100\text{ N} \end{gathered}[/tex]Substituting T1 in equation (1) we get,
[tex]\begin{gathered} 100\text{ N}+T_2=300\text{ N} \\ T_2=300\text{ N}-100\text{ N} \\ T_2=200\text{ N} \end{gathered}[/tex]Therefore, tension in left wire is 100 N and tension on right wire is 200 N. Hence, option (1) is the correct choice.
A concave mirror of focal length 10 cm forms an upright and diminished image of a real object placed at a distance of 5 cm from the mirror. Is this true or false?
Given,
The focal length of the concave mirror, f=10 cm
The object distance, o=5 cm
From the mirror formula, we have,
[tex]\frac{1}{f}=\frac{1}{o}+\frac{1}{i}[/tex]Where i is the image distance.
On substituting the known formula,
[tex]\begin{gathered} \frac{1}{10}=\frac{1}{5}+\frac{1}{i} \\ \Rightarrow\frac{1}{i}=\frac{1}{10}-\frac{1}{5} \\ =-\frac{1}{10} \\ \Rightarrow i=-10\text{ cm} \end{gathered}[/tex]And the magnification is given by,
[tex]m=\frac{-i}{o}[/tex]On substituting the known values,
[tex]\begin{gathered} m=\frac{-(-10)}{5} \\ =2 \end{gathered}[/tex]Thus the object is not diminished, it is magnified.
Thus the given statement is false.
The wire feed controls the amperage during the welding process. A control knob labeled in metric units shows a maximum wire-feed speed of 160 mm/sec. What is this maximum speed in inches per minute?The maximum speed is ___ inches per minute.
Answer: 378 inches per minute
Explanation:
We want to convert 160 mm/sec to inches per minute.
Recall,
1 mm = 0.03937 inches
1 second = 0.01667 minutes
Thus,
160 mm/sec = 160 x 0.03937 inches/0.01667 minutes
= 378 inches per minute
A spaceship travels at an enormous speed to go from one star system to another. As it reaches its destination, its “hyperdrive” deactivates, bringing the ship to an immediate halt in orbit around a planet.
Why does this defy physics? Give Equations/ Laws to justify yourself
This situation defies the laws of physics because it has no physical basis to explain the immediate stoppage of the ship.
Why does this situation defy physics?This situation defies the laws of physics because it describes an event in which an ship stops abruptly without any apparent physical explanation.
From the above, it can be inferred that this does not have scientific support. Additionally, it is defying the laws of physics because, according to Newton's second law, every body in motion continues to move at a constant speed. Therefore, if the ship goes to a hypervelocity it must continue moving, by Newton's second law. Additionally, if there were a force such as friction or gravity that causes it to stop, this would not be immediately but progressively.
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A diagram of a student on a playground swing is shown below. At which point is the kinetic energy the greatest?A. Point 1B. Point 2C. Point 3D. Point 4
To find:
The point where the kinetic energy is the greatest.
Explanation:
The kinetic energy of an object is the energy possessed by the object due to its motion. Thus the kinetic energy of an object is proportional to the square of the velocity of the object.
The velocity of the wing will be zero at the extreme points of the path of the swing. But as the swing approaches the mean point, i.e., point 3, the velocity of the swing increases. The velocity will be maximum at the mean point.
Thus the kinetic energy will be the greatest at point 3.
Final answer:
Therefore the correct answer is option C.
Describe in simple terms each of Kepler's three laws. Why were Tycho Brahe's observations so important for Kepler to develop these laws?
Kepler's first law of planetary motion states that every planet’s orbit is elliptical and has the Sun at a focus
Kepler's second law of planetary motion states that the line joining the Sun and a planet sweeps out equal areas in equal times
Kepler's third law of planetary motion states that the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit
Tycho Brahe provided some accurate astronomical data which makes it possible for Kepler to develop the laws of planetary motion. He devised the best instruments available before the invention of the telescope. These data and instrument provided by Tycho Brahe was able to assist Kepler in formulating these laws
a given mass of a gas at - 73°C exerts pressure of 40 cm of mercury. what pressure will be exerted at 127 °C if the volume remains constant
The pressure that will be exerted at 127 °C if the volume remains constant is 80 cm of mercury
How do I determine the new pressure?First, we shall list out the given parameters from the question. This is shown below:
Initial temperature (T₁) = -73 °C = -73 + 273 = 200 K Initial pressure (P₁) = 40 cm of mercuryNew temperature (T₂) = 127 °C = 127 + 273 = 400 KVolume = ConstantNew pressure (P₂) = ?Thus, we can obtain the new pressure of the gas at 127 °C as illustrated below:
P₁ / T₁ = P₂ / T₂
40 / 200 = P₂ / 400
Cross multiply
200 × P₂ = 40 × 400
200 × P₂ = 16000
Divide both sides by 200
P₂ = 16000 / 200
P₂ = 80 cm of mercury
Therefore, the new pressure is 80 cm of mercury
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I need help with this practice problem. Observe what happens in each of these three spheres with rain?Give at least two sentences each HYDROSPHERE:BIOSPHERE:GEOSPHERE:
The hydrosphere is the amount of water that there is on the planet. Water follows a cycle called the cycle of water where the heat from the sun evaporates water and is collected in clouds that eventually condensate and precipitates back to the surface in form of rain.
The biosphere is the part of the earth where there is life. When it rains this life receives the water and is used for various cycles that sustain life.
The geosphere is the soil of the earth. This soil receives the water and usually, this sinks in the soil sometimes accumulating and eventually evaporating again. This water is essential to sustain life.
Objects with masses M1=12.0 kg and M2= 7.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, M2 falls 1.00m in 1.33s, determine the coefficient of Kinect friction between M1 and the table
We are asked to determine the coefficient of friction in the sysmtem. First we will do a free body diagram of the first mass, like this:
Where:
[tex]\begin{gathered} N=\text{ Normal Force} \\ m_1=\text{ mass} \\ g=\text{ acceleration of gravity} \\ T=\text{ tension} \\ F_f=\text{friction force} \end{gathered}[/tex]Now, we add the forces in the horizontal direction, we get:
[tex]T-F_f=m_1a[/tex]Now, to determine the friction force we need to use the following relationship:
[tex]F_f=\mu N[/tex]To determine the normal force we will add the forces in the vertical direction. Since there is no movement in the vertical direction this sum must add up to zero:
[tex]\begin{gathered} N-m_1g=0 \\ N=m_1g \end{gathered}[/tex]Now, we substitute the value of the normal force in the equation for the friction force:
[tex]F_f=\mu m_1g[/tex]Now, we substitute the friction force in the sum of horizontal forces:
[tex]T-\mu m_1g=m_1a[/tex]Now, we turn our attention to the second mass. We add the forces in the vertical direction:
[tex]m_2g-T=m_2a[/tex]Now, since the acceleration "a" and the tension "T" is the same for both masses we will solve fot "T" in the sum of forces for the second mass, like this:
[tex]m_2g-m_2a=T[/tex]Now, we substitute the value in the sum of forces of the first mass, we get:
[tex]m_2g-m_2a-\mu m_1g=m_1a[/tex]Now, we solve for the coefficient of friction. To do that we will subtract "m2g" to both sides:
[tex]-m_2a-\mu m_1g=m_1a-m_2g[/tex]Now, we add "m2a" to both sides:
[tex]-\mu m_1g=m_1a-m_2g+m_2a[/tex]Now, we divide both sides by "-m1g":
[tex]\mu=\frac{m_1a-m_2g+m_2a}{-m_1g}[/tex]We have gotten an expression for the coefficient of friction but we need to determine the acceleration of the system. To do that we will use the fact that the second mass moves 1 meter in 1.33 seconds. Assuming constant acceleration we can use the following equation of motion:
[tex]y=v_0t+\frac{1}{2}at^2[/tex]Where:
[tex]\begin{gathered} y=\text{ distance} \\ v_0=\text{ initial velocity} \\ t=\text{ time} \\ a=\text{ acceleration} \end{gathered}[/tex]Since the mass starts from rest we have that the initial velocity is zero, therefore:
[tex]y=\frac{1}{2}at^2[/tex]Now, we solve for the acceleration. First, we multiply both sides by 2:
[tex]2y=at^2[/tex]Now, we divide both sides by the time squared:
[tex]\frac{2y}{t^2}=a[/tex]Now, we plug in the values:
[tex]\frac{2(1m)}{(1.33s)^2}=a[/tex]Solving the operations:
[tex]1.13\frac{m}{s^2}=a[/tex]Now, we substitute the values in the equation for the coefficient of friction:
[tex]\mu=\frac{(12kg)(1.13\frac{m}{s^2})-(7kg)(9.8\frac{m}{s^2})+(7kg)(1.13\frac{m}{s^2})}{-(12kg)(9.8\frac{m}{s^2})}[/tex]Solving the operations we get:
[tex]0.4=\mu[/tex]Therefore, the coefficient of friction is 0.4
A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizontally enters uniform, vertical 0.0800 T magnetic field and follows a circular arc of radius of 38.5 cm. What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuterium?
_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:
[tex]r=\frac{mv}{qB}[/tex]Where:
[tex]\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}[/tex]We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:
[tex]\frac{r_d}{m_d}=\frac{v}{q_B_}[/tex]Now, we multiply both sides by the magnetic field "B":
[tex]\frac{Br_d}{m_d}=\frac{v}{q}[/tex]Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:
[tex]r_p=\frac{m_pv}{qB}[/tex]And substitute the quotient between velocity and charge:
[tex]r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})[/tex]Now, we cancel out the magnetic field:
[tex]r_p=\frac{m_pr_d}{m_d}[/tex]Now, we substitute the values:
[tex]r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}[/tex]Solving the operations:
[tex]r_p=0.193m=19.3cm[/tex]Therefore, the radius is 19.3 cm.
A sample of silver at 10°C is heated to 90°C when 1 kcal of heat is added. What is the mass of the silver?
We can use the following formula:
[tex]q=m\cdot c\cdot(T2-T1)[/tex]Where:
q = heat = 1 kcal = 4184000J
c = specific heat (of silver) = 0.235 J °C /g
m = mass
T1 = Initial temperature = 10°C
T2 = Final temperature = 90°C
Therefore, solve for m:
[tex]\begin{gathered} m=\frac{q}{c(T2-T1)} \\ m=\frac{4184000}{0.235(90-10)} \\ m=222553.1915g \end{gathered}[/tex]Answer:
222553.1915g
Could really some help answering this question I don’t really understand! :)
Given:
Object A is at a higher temperature than object B.
Here, there is a difference in temperature.
Temperature flow is similar to water flow.
The water stops flowing if the water level becomes the same in botht containers.
Similarly, thermal energy transfers between the objects until the objects have changed the temperature by the same amount.
Thus, option B is correct.
Two 4.952 cm by 4.952 cm plates that form a parallel-plate capacitor are charged to +/- 0.576 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 2.97 mm?
In order to solve this question, we will need to know the E-field of a capacitor
E = sigma/epsilon naught
To find sigma (charge density), we will need to divide the total charge by the area
sigma = (.576 x 10^-9)/(4.952x10^-2)^2 = 2.34 x 10^-7
E = 2.34 x 10^-7/ (8.85x10^-12) = 26541.04 V/m
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. If, instead of separating the plates, the empty space between the plates is filled with a slab of insulating material that has a dielectric constant κ= 2, which of the following correctly describes the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor? 1) Voltage halvesElectric field halves Energy halves 2. Voltage halves Electric field halves Energy does not change 3. Voltage does not change Electric field does not change Energy halves 4. Voltage does not changeElectric field halves Energy halves 5. Voltage does not change Electric field does not change Energy does not change
The capacitance of a capacitor of a parallel plate is given by:
[tex]C=\frac{kA\cdot\epsilon_o}{d}[/tex]Where:
[tex]\begin{gathered} \epsilon_o=_{\text{ }}Vacuum_{\text{ }}permittivity \\ k=_{\text{ }}Dielectric_{\text{ }}constant=2 \\ A=_{\text{ }}Area_{\text{ }}of_{\text{ }}the_{\text{ }}plates \\ d=_{\text{ }}distance_{\text{ }}betwen_{\text{ }}the_{\text{ }}plates \end{gathered}[/tex]We also know:
[tex]\begin{gathered} Q=\frac{q}{V} \\ so\colon \\ \frac{q}{V}=\frac{kA\cdot\epsilon_o}{d} \\ V=\frac{qd}{kA\cdot\epsilon_o} \\ if \\ k=2 \\ V=\frac{1}{2}(\frac{qd}{kA\cdot\epsilon_o}) \end{gathered}[/tex]As we can see the voltage halves
And since:
[tex]\begin{gathered} E=\frac{V}{d} \\ W=qV \end{gathered}[/tex]We can conclude that electric field and also the energy halve
How far will a bike travel if you pedal at 13 m/s for 7 seconds?
Given
[tex]\begin{gathered} v=13\text{ m/s} \\ t=7\text{ s} \end{gathered}[/tex]The distance covered is given as,
[tex]s=vt[/tex]Here, s is the distance travelled, v is the speed and t is the time .
Substituting all known values,
[tex]\begin{gathered} s=(13\text{ m/s)}\times(7\text{ s)} \\ =91\text{ m} \end{gathered}[/tex]Therefore, the bike will travel 91 m is 7 second.