Answer:
Gravitational potential energy, kinetic energy and mechanical energy
Explanation:
The body is at a displacement to the gravity and therefore has energy stored in it as a result of its displacement against gravity. When this body is released the gravitational potential energy is converted to translational kinetic energy as well as rotational kinetic energy as it rolls down the the ramp. The sum of the kinetic energies and potential energy of the body amount to the mechanical energy of the ball down the ramp.
When the following equation is balanced, the sum of all the coefficients is ____________. CO + NO → CO2 + N2.
Answer:
7
Explanation:
The balanced reaction is as follows:
[tex]2CO+2NO \rightarrow 2CO_2 + N_2[/tex]
Coefficient of CO = 2
Coefficient of NO = 2
Coefficient of CO2 = 2
Coefficient of N2 = 1
Sum of all coefficient = 2 + 2+ 2+ 1
= 7
A stubborn 130 kg pig sits down and refuses to move. To drag the pig to the barn, the exasperated farmer ties a rope around the pig and pulls with his maximum force of 800 NN. The coefficients of friction between the pig and the ground are μs=0.80μs=0.80 and μk=0.50μk=0.50.
Answer:
Explanation:
Given
mass [tex]m=130 kg[/tex]
Force [tex]F=800 N[/tex]
coefficient of static friction [tex]\mu _s=0.8[/tex]
coefficient of static friction [tex]\mu _k=0.5[/tex]
maximum Static Friction will be given by
[tex]F_s=\mu _smg[/tex]
[tex]F_s=0.8\times 130\times 10=1040 N[/tex]
initially we need to provide a force of 1040 N to move the Pig . As soon Pig starts moving kinetic friction will come into play.
But maximum force is less than Maximum static friction force so it is impossible to move pig.
Answer:
C
Explanation:
The distance that the pig moves up will be less than the distance that the strongman pulls down.
Explanation: The strongman is further from the fulcrum than the pig is, so the distance that he pulls down will be greater than the distance that the pig moves up.
You must dim your high beams for oncoming vehicles by the time they are within 500 feet of your vehicle.
Answer:
True
Explanation:
when oncoming vehicle driver is at a distance of 500 feet from you, then it is advisable to dim your high beam light so that it cannot blind oncoming vehicle driver. However, of the oncoming driver is at 200 to 300 ft then you must use low beam light so to pass the oncoming driver safely from your side. Hence the statement is true.
A mass resting on a shelf 10.0 meters above the floor has a gravitational potential energy of 980. joules with respect to the floor. The mass is moved to a shelf 8.00 meters above the floor. What is the new gravitational potential energy of the mass?
a. 960. J
b. 784 J
c. 490. J
d. 196 J
Answer:
New potential energy will be 784 J
So option (b) will be correct answer
Explanation:
We have given height of the shelf h = 10 m
Potential energy E = 980 J
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We know that potential energy [tex]E=mgh[/tex]
So [tex]980=m\times 9.8\times 10[/tex]
m = 10 kg
Now new height h = 8 m
So new potential energy [tex]E=mgh=10\times 9.8\times 8=784J[/tex]
So option (B) will be the correct answer
Answer:
option (b) is correct.
Explanation:
h1 = 10 m
U 1 = 980 J
Let mg be weight of mass.
U1 = m x g x h1
980 = mg x 10
mg = 98
Now h = 8 m
U2 = mg x h = 98 x 8
U2 = 784 J
thus, the potential energy is 784 J.
Bird bones have air pockets in them to reduce their weight–this also gives them an average density significantly less than that of the bones of other animals. suppose an ornithologist weighs a bird bone in air and in water and finds its mass is 43.0 g and its apparent mass when submerged is 3.60 g (the bone is watertight).a. what mass of water is displaced? b. what is the volume of the bone? c. what is its average density?
Answer:
39.4 g
39.4 cm³
1.09137 g/cm³
Explanation:
[tex]\rho[/tex] = Density of water = 1 g/cm³
Mass of water displaced will be the difference of the
[tex]m=43-3.6\\\Rightarrow m=39.4\ g[/tex]
Mass of water displaced is 39.4 g
Density is given by
[tex]\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{39.4}{1}\\\Rightarrow v=39.4\ cm^3[/tex]
So, volume of bone is 39.4 cm³
Average density of the bird is given by
[tex]\rho=\dfrac{43}{39.4}\\\Rightarrow \rho=1.09137\ g/cm^3[/tex]
The average density is 1.09137 g/cm³
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0°.
Answer:
Ф1=295.96Nm^2/C
Ф2=469.73Nm^2/C
Explanation:
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0
we have two surfaces where we know angle of surface 1 lets call it
s1= 34.
Therefore to find s2
s2 (the angle from surface 2) we have that
s2=180-(90+s1),
so s2=180-(90+34),
thus s2=56 degrees.
Flux equation reads as Φ=ΕΑ,
where Φ is the flux,
E is the electric field and
A is the surface area.
So with respect to the angles and the figure provided,
we have Φ=EAcos(s).
So we can solve further by writing
. For Surface 1 we have
Φ1=EAcos(s1)=210 x 1.7 x cos 34,
so Φ1=295.96,
approximately to an whole number
Φ1=296 Nm^2/c.
Similarly for Φ2, we have
Φ2=EAcos(s2)
=210 x 4 x cos34=469.7,
thus Φ2=469.7Nm^2/c.
which cruising altitude is appropriate for vfr flight on a magnetic course of 135°
Answer:odd thousands plus 500 feet.
Explanation:
On a magnetic course of zero through 179 degrees, select an odd thousand foot cruising altitude plus 500 feet, such as 3,500, 5,500, up to and including 17,500. Even and odd thousands are reserved for those aircraft on an active instrument flight plan. Even thousands plus 500 feet are for aircraft flying between 180 and 359 degrees.
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C at an average speed of 60 kilometers per hour. If the train did not stop at Station B, what was the average speed at which the train traveled from Station A to C?
(1) The distance that the train traveled from Station A to Station B was 4 times the distance that train traveled from Station B to Station C.
(2) The amount of time it took to the train to travel from Station A to Station B is 3 times the amount of time that it took the train to travel from Station B to Station C.
Answer:
1)
75 kmh⁻¹
2)
75 kmh⁻¹
Explanation:
1)
[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹
[tex]d_{ab}[/tex] = distance traveled from station A to station B
[tex]t_{ab}[/tex] = time of travel between station A to station B
we know that
[tex]Time = \frac{distance}{speed}[/tex]
[tex]t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}[/tex]
[tex]d_{bc}[/tex] = distance traveled from station B to station C
[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹
[tex]t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}[/tex]
Total distance traveled is given as
[tex]d = d_{ab} + d_{bc}[/tex]
Total time of travel is given as
[tex]t = t_{ab} + t_{bc}[/tex]
Average speed is given as
[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }[/tex]
Given that :
[tex]d_{ab} = 4 d_{bc}[/tex]
So
[tex]v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}[/tex]
2)
[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹
[tex]t_{ab}[/tex] = time of travel between station A to station B
[tex]d_{ab}[/tex] = distance traveled from station A to station B
we know that
[tex]distance = (speed) (time)[/tex]
[tex]d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}[/tex]
[tex]d_{bc}[/tex] = distance traveled from station B to station C
[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹
[tex]t_{bc}[/tex] = time of travel for train from station B to station C
we know that
[tex]distance = (speed) (time)[/tex]
[tex]d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}[/tex]
Total distance traveled is given as
[tex]d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}[/tex]
Total time of travel is given as
[tex]t = t_{ab} + t_{bc}[/tex]
Average speed is given as
[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}[/tex]
Given that :
[tex]t_{ab} = 3 t_{bc}[/tex]
So
[tex]v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}[/tex]
A stationary source S generates circular outgoing waves on a lake. The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. A person in a motorboat heads directly toward S at 3.0 m/s. To this person, the frequency of these waves is:
Answer:4 Hz
Explanation:
Speed of wave [tex]v=5 m/s[/tex]
crest to crest distance [tex]\lambda =2 m[/tex]
velocity of observer [tex]v_0=3 m/s[/tex]
actual frequency [tex]f=\frac{velocity}{\lambda }[/tex]
[tex]f=\frac{5}{2}=2.5 Hz[/tex]
Apparent frequency [tex]f'=f(\frac{v+v_0}{v})[/tex]
[tex]f'=2.5\times \frac{5+3}{5}[/tex]
[tex]f'=4 Hz[/tex]
The frequency of these waves is 4hz.
Given information:The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. Alo, A person in a motorboat head directly toward S at 3.0 m/s.
Calculation of wave frequency:The actual frequency is [tex]= 5 \div 2[/tex] = 2.5Hz
Now
Apparent frequency is
[tex]= 2.5 \times (5 + 3) \div 5[/tex]
= 4hz
Hence, the frequency of these waves is 4hz.
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Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds Number Re is less than about 2000: Re = 2v Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 rhoR Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 /η. Here Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 v, rho, and η are, respectively, the average speed, density, and viscosity of fluid, and R is the radius of the pipe. Calculate the highest average speed that blood (rho = 1060 kg/m3, η = 4.0 x 10-3 Pa.s) could have and still remain in laminar flow when it flows through the aorta (R = 8.0 x 10-3 m)
Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4
Answer: 0.471 m/s
Explanation:
We are given the followin equation:
[tex]Re=\frac{D v \rho}{\eta}[/tex] (1)
Where:
[tex]Re[/tex] is the Reynolds Number, which is adimensional and indicates if the flow is laminar or turbulent
When [tex]Re<2100[/tex] we have a laminar flow
When [tex]Re>4000[/tex] we have a turbulent flow
When [tex]2100<Re<4000[/tex] the flow is in the transition region
[tex]D=2R[/tex] is the diameter of the pipe. If the pipe ha a radius [tex]R=8(10)^{-3} m[/tex] its diameter is [tex]D=2(8(10)^{-3} m)=0.016 m[/tex]
[tex]v[/tex] is the average speed of the fluid
[tex]\rho=1060 kg/m^{3}[/tex] is the density of the fluid
[tex]\eta=4(10)^{-3} Pa.s[/tex] is the viscosity of the fluid
Isolating [tex]v[/tex]:
[tex]v=\frac{Re \eta}{D \rho}[/tex] (2)
Solving for [tex]Re=2000[/tex]
[tex]v=\frac{(2000)(4(10)^{-3} Pa.s)}{(0.016 m)(1060 kg/m^{3})}[/tex] (3)
Finally:
[tex]v=0.471 m/s[/tex]
According to the second law of thermodynamics, the disorder of a system always decreases. 1. True 2. False
Answer:
2. False
Explanation:
Entropy can be interpreted as a measure of the random distribution of a system. According to the second law of thermodynamics a system in an unlikely condition will have a natural tendency to reorganize itself to a more probable condition, this reorganization will result in an increase in entropy (disorder).
2. In the winter, weather reporters often day "It will be a very cold night because there are no clouds." a. Use the sim to see if you can understand why this could be true. b. Describe your observations. c. Would there be a difference between daytime and nighttime cloud effects
Answer:
a. Clouds at night prevent the heat radiation of the earth from escaping back into the space by continuously reflecting them back.
b. If during the day the earth has been heated well by the sun and then if the night is cloudy then that night is not so cold.
c. Day time clouds prevent the heating of the earth by reflecting back the radiations of the sun back into the space.
Explanation:
During the day if there is sunshine and then if the night is cloudy then that night will not be so cold because during the day the earth has absorbed the heat of sun's radiation which it radiates at night but when there is cloud at night it acts as a reflecting screen and reflects the heat radiations back to the earth.
But if there is cloud during the day time it prevents the radiations from the sun to fall on the earth by reflecting them back into the space preventing the rise in temperature of the earth.
The rumble feature on a video game controller is driven by a device that turns electrical energy into mechanical energy. This device is best referred to as
Answer:
MOTOR
Explanation:
The device which changes electrical energy into mechanical energy is known as motor
Loops of wire in a magnetic field make up motors. The magnetic field produces a torque on the loops when current flows through them, turning a shaft as a result.
What is electric motor?Electrical energy is transformed into mechanical energy by electric motors. When we turn on the fan, for instance, the electric motor begins to transform the electrical energy into mechanical energy.
The fan blades then begin whirling as a result of the mechanical energy, giving them the capacity to perform work.
On either side of the controller, there is a motor. This engine has an uneven weight linked to it.
This indicates that one side of it is heavier than the other. The imbalance of the weight causes the controller to vibrate when the motor turns.
Therefore, Mechanical work is created as a result of the conversion of electrical energy.
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The population of weights for men attending a local health club is normally distributed with a mean of 171-lbs and a standard deviation of 29-lbs. An elevator in the health club is limited to 35 occupants, but it will be overloaded if the total weight is in excess of 6510-lbs.Assume that there are 35 men in the elevator. What is the average weight beyond which the elevator would be considered overloaded?
Answer:
186 lbs per man.
Explanation:
If we assume that we have 35 men in the elevator, and the elevator will be overloaded if the total weight is in excess of 6510-lbs, so we just need to take average:
6510 lbs / 35 = 186 lbs per man.
Have a nice day!
The temperature inside a vehicle can rise ____ degrees higher than the outside temperature.
Answer:
Generally, 40 to 50 degrees
Explanation:
About the heat-up over time, whether the windows of a vehicle are locked or partially open makes very little difference. In both situations, in an internal temperature of a vehicle, even at an outside temperature of only 72 ° F, it may exceed approximately 40 ° F within one hour. This happens mainly due to the greenhouse effect that is the heat inside the car is trapped and not allowed to escape. Thus, raising the temperature of the vehicle.
Final answer:
The greenhouse effect causes the temperature inside a parked car to be much higher than outside, sometimes 20 to 30 degrees higher, due to sunlight being trapped as heat.
Explanation:
The temperature inside a vehicle can rise significantly higher than the outside temperature due to a phenomenon known as the greenhouse effect. When a car is parked in the sun with windows closed, the sunlight passes through the glass and is absorbed by interior surfaces such as dashboard and seats. These surfaces then emit infrared radiation, which cannot escape back through the glass, and consequently, the trapped heat raises the temperature of the air inside the car. This effect can cause interior temperatures to be much higher—often 20 to 30 degrees higher—than the external air temperature.
A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits the water below. What was the pelican’s initial speed? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s. If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
Answer:
(A) 9.5 m/s
(B) 5.225 m
Explanation:
vertical height (h) = 4.7 m
horizontal distance (d) = 9.3 m
acceleration due to gravity (g) = 9.8 m/s^{2}
initial speed of the fish (u) = 0 m/s
(A) what is the pelicans initial speed ?
lets first calculate the time it took the fish to falls = ut + [tex](\frac{1}{2}) at^{2}[/tex]
since u = 0
s = [tex](\frac{1}{2}) at^{2}[/tex]
t = [tex]\sqrt{\frac{2s}{a} }[/tex]
where a = acceleration due to gravity and s = vertical height
t = [tex]\sqrt{\frac{2 x 4.7 }{9.8} }[/tex] = 0.98 s
pelicans initial speed = speed of the fishspeed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s
initial speed of the pelican = 9.5 m/s
(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
vertical height = 1.5 m
pelican's speed = 9.5 m/s
lets also calculate the time it will take the fish to falls = ut + [tex](\frac{1}{2}) at^{2}[/tex]
since u = 0
s = [tex](\frac{1}{2}) at^{2}[/tex]
t = [tex]\sqrt{\frac{2s}{a} }[/tex]
where a = acceleration due to gravity and s = vertical height
t = [tex]\sqrt{\frac{2 x 1.5 }{9.8} }[/tex] = 0.55 s
distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m
The initial speed of the pelican is 9.6 m/s. If the pelican is flying at the same speed but only 1.5 m above the water, the fish will travel approximately 5.3m horizontally before it hits the water.
Explanation:We can begin by solving for the time the fish takes to hit the water. Using the equation of motion, h = 0.5gt², where h represents height and g represents the acceleration due to gravity, we find it takes approximately 0.97 seconds for the fish to hit the water. During this time, the fish has traveled 9.3m horizontally. Therefore, the initial speed of the pelican can be determined by v = d/t, which gives us a value of 9.6 m/s.
Next, if the initial speed remained the same, but the height was reduced to 1.5 meters, we can use the same process. The time it would take for the fish to hit the water would be approximately 0.55 seconds (obtained from the equation h = 0.5gt²), and using this time in the equation v = d/t, we find the fish would travel approximately 5.3m horizontally before it hits the water.
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Two technicians are discussing oil leaks. Technician A says that an oil leak can be found using a fluorescent dye in the oil with a black light to check for leaks. Technician B says a white spray powder can be used to locate oil leaks. Which technician is correct?a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B
Answer:
B
Explanation:
cause it makes the most sence
A ball is dropped from rest and falls to the floor. The initial gravitational potential energy of the ball-Earth-floor system is 10 J. The ball then bounces back up to a height where the gravitational potential energy is 7 J. What was the mechanical energy of the ball-Earth-floor system the instant the ball left the floor?
The mechanical energy of the ball-Earth-floor system at the instant when the ball left the floor is of 7 J.
What is Mechanical Energy?At any instant, the energy of an object by virtue of its motion or position is known as the mechanical energy of the object. In other words, mechanical energy is either the potential energy or the kinetic energy at any time instant.
Given data -
The initial gravitational potential energy is, U = 10 J.
The final gravitational potential energy is, U' = 7 J.
Now, considering the given case when the ball loses energy in the rebound due to ball deformation, heat loss from the bounce. The total mechanical energy is not conserved so mechanical energy as it bounces off the floor is equal to transformed potential energy at a maximum height of second bounce which is 7 J.
Thus, we can conclude that the mechanical energy of the ball-Earth-floor system at the instant when the ball left the floor is of 7 J.
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"According to the Navigation Rules, what factor should be considered in determining a safe speed?
Answer:
Visibility conditions.
Explanation:
Safe speed is a speed at which the operator of the boat can take effective action to avoid stops within the distance. To calculate the safe speed visibility conditions (fog, rain, mist, and darkness) should be included. We should also include some other factors given below:
i) Traffic density
ii) Type of vessels in the area
iii) Wind
iv) Water conditions
The Navigation Rules dictate that a safe speed is determined by factors such as visibility, traffic density, vessel maneuverability, weather conditions, and proximity to navigational hazards.
Explanation:According to the Navigation Rules, several factors should be taken into account in determining a safe speed. These include the visibility, the traffic density, the maneuverability of the vessel in immediate circumstances, the state of wind, sea, and current, and the proximity of navigational hazards. For example, on a clear day with sparse traffic and calm seas, higher speeds might be safe. However, in heavy traffic or poor visibility, the navigation rules would call for lower speeds to ensure safety.
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A sample of radiosodium () has a half-life of 15 hr. If the sampleâs activity is 100 millicuries after 24 hr, approximately what must its original activity have been?
Answer : The original activity will be, 303 millicuries.
Explanation :
Half-life = 15 hr
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{15hr}[/tex]
[tex]k=4.62\times 10^{-2}\text{ hr}^{-1}[/tex]
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]4.62\times 10^{-2}\text{ hr}^{-1}[/tex]
t = time passed by the sample = 24 hr
a = initial amount of the reactant = ?
a - x = amount left after decay process = 100 millicuries
Now put all the given values in above equation, we get
[tex]24=\frac{2.303}{4.62\times 10^{-2}}\log\frac{a}{100}[/tex]
[tex]a=302.97\text{ millicuries}\approx 303\text{ millicuries}[/tex]
Therefore, the original activity will be, 303 millicuries.
A rock that has deformed ____ under stress keeps its new shape when the stress is released.
Answer:
Elastically
Explanation:
A rock that has deformed Elastically under stress keeps its new shape when the stress is released.
In elastic deformation the original shape of the object is regained when the stress is removed. Whereas in plastic deformation the original shape is parmanently deformed with the application of stress.
A rock that deforms plastically under stress will retain its new shape when the stress is released, indicating it has surpassed its yield point.
Explanation:A rock that has deformed plastically under stress will keep its new shape when stress is released. When rocks are subjected to stress, they can undergo elastic, brittle, or plastic deformation. In the elastic state, rocks, like a rubber band, will return to their original shape once the imposed stress is removed, provided the stress does not exceed the elastic limit of the rocks. Beyond this limit, the material will experience plastic deformation, leading to a permanent change in shape as mineral bonds break, shift, and reform.
This behavior depends on several factors, including stress type, rock type, depth, and environmental conditions (pressure and temperature). At greater depths, where the conditions are high pressure and temperature, rocks are more likely to deform plastically. Furthermore, when rocks exhibit plastic deformation, it indicates they have gone past their yield point and will not revert to their original form even if the stress is no longer applied.
With each bounce off the floor, a tennis ball loses 11% of its mechanical energy due to friction. When the ball is released from a height of 3.4 m above the floor, what height will it reach after the third bounce?
240 mm
240 cm
24 cm
270 cm
Answer:
240 cm
Explanation:
Gradpoint
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is __________ atm. 1.5 7.5 0.67 3.3 15
Answer:
The final pressure will be 0.67 atm.
Explanation:
Using Boyle's law
[tex] {P_1}\times {V_1}={P_2}\times {V_2}[/tex]
Given ,
V₁ = 10 L
V₂ = 15 L
P₁ = 1.0 atm
P₂ = ?
Using above equation as:
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
[tex]{1.0}\times {10}={P_2}\times {15} atm[/tex]
[tex]{P_2}=\frac{{1.0}\times {10}}{15} atm[/tex]
[tex]{P_2}=0.67\ atm[/tex]
The final pressure will be 0.67 atm.
Answer:
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.
Explanation:
Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.
The formula is,
[tex]\rm \bold{ P_1\times V_1= P_2\times V_2}[/tex]
Where,
[tex]\rm \bold P_1[/tex] is initial pressure = 1 atm
[tex]\rm \bold P_2[/tex] is final pressure = ? (Not Known)
[tex]\rm \bold V_1[/tex] is initial volume = 10 L
[tex]\rm \bold V_2[/tex] is final volume = 15 L
Now put the values in the formula,
[tex]\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67[/tex]
Therefore, the answer is 0.67 atm.
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how do the wavelengths of electromagnetic energy absorbed by materials on earth compare?
Explanation:
The shorter wavelength electromagnetic waves from sun are absorbed by earth material in form of short wavelength and the radiated wavelength are longer ones. Also higher energy waves are of shorter wavelength and lower energy waves have longer wavelength. So, they are absorbed as short wavelength and radiated back as long wavelength.
When testing an PNP transistor with an ohmmeter, what are the high or low resistance values expected for a good transistor?
Answer:
0.45 V and 0.9 V.
Explanation:
To test a PNP transistor with an ohmmeter,
Plug the positive lead from the multimeter to the transistor EMITTER (E). Plug the negative meter into the transistor BASE (B). If you are using a PNP resistor you must watch OL that is over limit, the voltage decrease will indicate between 0.45V and 0.9V if you are measuring it.
Testing a PNP transistor with an ohmmeter should yield high resistance values when the base is negative relative to the emitter or collector, and low resistance values when the base is positive relative to the emitter or collector.
Explanation:When testing a PNP transistor with an ohmmeter, the high or low resistance values that are expected for a good transistor are as follows: When you measure between the base and the emitter or collector, you should see a high resistance value (in the range of megaohms) if the base is negative with respect to the emitter or collector. Conversely, you should see a low resistance value (in the range of a few hundred Ohms) when the base is positive with respect to the emitter or collector.
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A network engineer is subnetting the 10.0.240.0/20 network into smaller subnets. Each new subnet will contain between a minimum of 20 hosts and a maximum of 30 hosts. Which subnet mask will meet these requirements?
a. 255.255.224.0
b. 255.255.240.0
c. 255.255.255.224
d. 255.255.255.240
Answer:255.255.255.224
Explanation:
/20 has a subnet mask of 255.255.240.0
We have 20 on bits for the network and 16 off bits for the host. Since the designer wants 20 to 30 hosts in each subnet we will be having 5 off bits for the host. Therefore we will be having a /27..
Which will be 255.255.255.224
Final answer:
The suitable subnet mask for creating subnets with 20-30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27, which allows for 32 IP addresses, 30 of which can be used for hosts.
Explanation:
The correct subnet mask to meet the requirement for new subnets that support between a minimum of 20 hosts and a maximum of 30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27. This subnet mask allows for 32 IP addresses per subnet, with 30 usable addresses for hosts when excluding the network and broadcast addresses. Subnet masks 255.255.224.0 and 255.255.240.0 provide too many hosts per subnet, and subnet mask 255.255.255.240 or /28 provides only 16 IP addresses, which would be insufficient for the required minimum of 20 hosts.
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 given: 2a + b → a2b δh°
Answer:
Q = 12.5 kJ
Explanation:
The expression to use to calculate Heat is:
Q = H° * n
Where:
Q: heat (J or kJ)
H°: enthalpy of reaction (kJ/mol)
n: moles
Now, as it was stated in the comments, the question is incomplete, and here is the missing part:
Given:
2A + B A2B (1)
H° = – 25.0 kJ/mol
2A2B 2AB + A2 (2)
H° = 35.0 kJ/mol
With these two reactions, we can calculate the heat.
Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:
2A + B -------> A2B H°1 = -25 kJ/mol
2A2B --------> 2AB + A2 H°2 = 35 kJ/mol
Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:
2A + B + 2A2B -------> 2AB + A2
And the enthalpy, it's just summed:
H°3 = -25 + 35 = 10 kJ/mol
Now with this value we can calculate heat:
Q = 10 * 2.5 = 25 kJ
However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:
Q = 25 / 2 = 12.5 kJ
The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out.
To calculate the heat required for the reaction, we need to know the enthalpy change (ΔH°) for the reaction 2a + b → a2b. However, the enthalpy change for this reaction is not provided in the question. Assuming that the enthalpy change for this reaction is known, we can proceed with the calculation.
Let's denote the enthalpy change for the reaction 2a + b → a2b as ΔH°. The reaction of interest is:
[tex]\[ 2a + b + a_2b \rightarrow 2ab + a_2 \][/tex]
This reaction can be broken down into two parts:
1. The reaction of 2 moles of a with 1 mole of b to form 1 mole of a2b:
with an enthalpy change of H°.
[tex]\[ 2a + b \rightarrow a_2b \][/tex]
2. The reaction of 1 mole of a2b to form 2 moles of ab:
[tex]\[ a_2b \rightarrow 2ab \][/tex]
Since the reaction of a2b to form 2ab does not involve any additional reactants, it can be considered as the reverse of the first part, but with twice the amount of ab produced. Therefore, the enthalpy change for this part would be -H° (since the reverse reaction has the opposite sign of the forward reaction), and for 1 mole of a2b reacting, it would be -2ΔH° (because 2 moles of ab are formed).
Now, we are given 2.50 moles of a, and we assume there is excess b and a2b. The reaction will proceed until all of a is consumed. Since the reaction stoichiometry is 2 moles of a to 1 mole of a2b, 1.25 moles of a2b will be required to react with the 2.50 moles of a (since 2.50 moles of a require 1.25 moles of a2b).
The overall enthalpy change for the reaction will be the sum of the enthalpy changes for the two parts:
1. For the reaction of 2.50 moles of a with b to form a2b:
[tex]\[ \Delta H_1 = 2.50 \text{ moles} \times \Delta H^\circ \][/tex]
2. For the reaction of 1.25 moles of a2b to form ab:
[tex]\[ \Delta H_2 = 1.25 \text{ moles} \times (-2 \Delta H^\circ) \][/tex]
The total enthalpy change (ΔH_total) is the sum of ΔH_1 and ΔH_2:
[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex][tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2 \times 1.25 \Delta H^\circ \][/tex]
[tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2.50 \Delta H^\circ \][/tex]
[tex]\[ \Delta H_{\text{total}} = 0 \][/tex]
Surprisingly, the total enthalpy change for the reaction is zero. This is because the amount of heat released in the formation of a2b is exactly equal to the amount of heat absorbed in the formation of ab from a2b. Therefore, no additional heat is required for the overall reaction to proceed.
The final answer is: [tex]\[ \boxed{0} \][/tex]
The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out. This conclusion is based on the assumption that the enthalpy change for the reaction 2a + b → a2b is known and that the reaction proceeds as written. If the enthalpy change for the formation of a2b is not known, then it would be necessary to look up the standard enthalpy of formation for a2b to perform the calculation.
Old Isaac took a little nosedive from his perch on top of the building, 25 feet above ground. Given that he fell as a result of a gentle tap to his noggin (initial velocity was zero), how fast is Isaac traveling when he hits the ground? (In case you forgot, acceleration due to gravity is -32 ft/sec2 .) Make sure your answer is in feet/sec. Leave your reply under the brick beside your car.
Answer:
40 ft/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = -32 ft/s² = a (downward is taken a negative)
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times -25+0^2}\\\Rightarrow v=40\ ft/s[/tex]
The speed of Isaac as he reached the ground is 40 ft/s
"The final velocity of Isaac when he hits the ground can be calculated using the kinematic equation for an object in free fall under the influence of gravity with an initial velocity of zero:
[tex]\[ v_f^2 = v_i^2 + 2a\Delta y \][/tex]
Plugging in the known values:
[tex]\[ v_f^2 = 0^2 + 2(-32 \text{ ft/sec}^2)(25 \text{ ft}) \] \[ v_f^2 = 0 + (-2)(32)(25) \] \[ v_f^2 = -1600 \][/tex]
Since velocity cannot be negative, we take the positive square root of the absolute value of [tex]\( v_f^2 \)[/tex] to find [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \sqrt{1600} \] \[ v_f = 40 \text{ ft/sec} \][/tex]
Therefore, Isaac is traveling at 40 feet per second when he hits the ground.
The answer is: [tex]40 \text{ ft/sec}.[/tex]
the decimal reduction time (DRT) is the time it takes to kill 90% of cells present. Assume that a DRT value for autoclaving a culture is 1.5 minutes. How long would it take to kill all the cells if 10^6 cells were present? What would happen if you stopped the heating process at 9 minutes?
Answer:
It takes 10.5 minutes to kill all the bacteria.
Only 1 cell would remain after 9 minutes.
Explanation:
It will take 1.5 minutes to kill 90% of the cells. So, after 1.5 minutes, only 10% would remain. After 3 minutes, only 1% remain. So, to figure out how long it would take to kill a million cells, we have to multiply 1 million by 0.1 repeatedly until the final value is less than 1 that is because when the value is less than 1, it means there are no more bacteria.
So:
[tex]10^6 \times (0.1)^7[/tex] = 0.1
So, you need 10.5 minutes of killing to kill one million cells.
Time taken= 7 x 1.5 minutes = 10.5 minutes.
After 9 minutes you would have:
[tex]10^{6} \times (0.1)^{6}[/tex] = 1 cell left
The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells.
Explanation:The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. Since the DRT value represents the time it takes to kill 90% of the cells, you can calculate the time to kill 100% of the cells by dividing the DRT value by 90 and then multiplying it by 100. In this case, it would be as follows:
T = (1.5 minutes / 90) * 100 = 1.67 minutes
If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells. As a result, some cells would still be alive.
A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the table of height h2, landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1+h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.
A) 1) Without using equations, explain why making h1 very small would cause d to be small, even though
h2 would be very large?
2) Without using equations, explain why making h2 very small would cause d to be small, even though
h1 would be large
B) Derive an equation for d in terms of h1, h2, m, and physical constants as appropriate.
A1) The reason why making h₁ very small would cause d to be small is; Because the horizontal component of the launch velocity would be very small.
A2) The reason why making h₂ very small would cause d to be small is;
Because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.
B) The equation for d in terms of h₁ and h₂ is;
d = 2√(h₁ × h₂)
A) 1) The reason why making h₁ very small would cause d to be small is because the horizontal component of the launch velocity would be very small.
A) 2) The reason why making h₂ very small would cause d to be small is because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.
B) Formula for Launch Velocity is;
V = √(2gh₁)
h₁ was used because the top of the slide from where the student released the block has a height of h₁.
Also, the time it takes to fall which is time of flight is given by the formula;
t = √(2h₂/g)
h₂ was used because the height of the table the object is on before falling is h₂.
Now, we know that d is distance from edge of the table and formula for distance with respect to speed and time is;
distance = speed × time
Thus;
d = √(2gh₁) × √(2h₂/g)
g will cancel out and this simplifies to give;
d = 2√(h₁ × h₂)
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(A)
(1) The reason for making [tex]h_{1}[/tex] very small is due to smaller value of horizontal component of launch velocity.
(2) The reason for making [tex]h_{2}[/tex] very small is due to smaller value of time of flight.
(B) The distance (d) covered by the block is [tex]2\sqrt{h_{1}h_{2}}[/tex].
Given data:
The mass of block is, m.
The height of table from the top of slide is, [tex]h_{1}[/tex].
The height of table at the end of slide is, [tex]h_{2}[/tex].
The height of room is, H.
(A)
(1)
If the launch velocity of the block is v, then its horizontal component is very small, due to which adjusting the height [tex]h_{1}[/tex] to be very small will cause the d to be small.
(2)
The height [tex]h_{2}[/tex] is dependent on the time of flight, and since the time of flight taken by the block to get to the floor is very less, therefore the block will not get sufficient time to accomplish its horizontal motion. That is why making [tex]h_{2}[/tex] very small will cause d to be smaller.
(B)
The expression for the distance covered by the block is,
[tex]v=\dfrac{d}{t}\\d = v \times t[/tex] ..............................(1)
Here, v is the launch speed of block and t is the time of flight.
The launch speed is,
[tex]v^{2}=u^2+2gh_{1}\\v=\sqrt{u^2+2gh_{1}}\\v=\sqrt{0^2+2gh_{1}}\\v=\sqrt{2gh_{1}}[/tex]
And the time of flight is,
[tex]h_{2}=ut+\dfrac{1}{2}gt^{2}\\h_{2}=0 \times t+\dfrac{1}{2}gt^{2}\\h_{2}=0+\dfrac{1}{2}gt^{2}\\t=\sqrt\dfrac{2h_{2}}{g}[/tex]
Substituting the values in equation (1) as,
[tex]d = v \times t\\d = \sqrt{2gh_{1}}\times \sqrt\dfrac{2 h_{2}}{g}}\\d=2\sqrt{h_{1}h_{2}}[/tex]
Thus, the distance (d) covered by the block is [tex]2\sqrt{h_{1}h_{2}}[/tex].
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