A baseball player (mass = 75 kg) is running north towards a base. In order to avoid
being tagged by the ball, the baseball player slides into the base. If his acceleration
in the slide is 5.0 m/s2 south, what is the kinetic frictional force between the
baseball player and the ground?
Answers
it's 1727 +822 just kidding
Related Questions
What is the speed of sound in metro city if the temperature is 26 degree celsius ?
Answers
What is the pH of 0.001 M HNO3?
Answers
Answer:
i think its 2.0
Explanation:
hopefully i am right ...
find the mass of a wrecking ball at 6m/s and has a momentum of 3600kg x m/s
the mass must be in kilograms (kg)!
m/s means "meters per second"
kg x m/s is supposed to mean "kilograms times meters per second"
also, the equation to find this is force divided by acceleration
thank you!!!
Answers
Answer:
600kgExplanation:
MOMENTUMThis is a product of a mass and velocity of a moving or a rest body.
it is expressed as P(momentum)=M(mass) × V(velocity)
Its SI unit is kgm/s
From our question.
Given
Momentum=3600kgm/s
Velocity=6m/s
RTF=Mass
Solution
P=M×V
M=P/V
= 3600kgm/s
6m/s
= 600kgwhat is one of the differences between cepheids and rr lyrae variables?
Answers
The main differences between Cepheids and RR Lyrae variables is their period-luminosity relationship. Cepheids have a longer period (typically a few days to a few weeks) and a stronger correlation between their period and luminosity, meaning that brighter Cepheids have longer periods.
RR Lyrae variables, on the other hand, have shorter periods (typically less than a day) and a weaker correlation between period and luminosity. Additionally, Cepheids are more massive and younger stars, found in spiral arms of galaxies, while RR Lyrae variables are typically found in globular clusters and are lower mass, older stars.
One of the differences between Cepheids and RR Lyrae variables is their period-luminosity relationship. Cepheids have longer periods (1 to 100 days) and higher luminosities, while RR Lyrae variables have shorter periods (0.2 to 1 day) and lower luminosities. This difference makes Cepheids useful for measuring greater distances in the universe, while RR Lyrae variables are more suitable for shorter distances.
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do banked curves help car make turns more softly
Answers
Answer:
yes
Explanation:
banked curve :
is a road that looks like the top part of a race track
its looks elevated
often seen in bicycle race track (velodrome)
Banking the curve can help keep cars from skidding. When the curve is banked, the centripetal force can be supplied by the horizontal component of the normal force.
sfuca
The prefix micro means:
(a) Option 1 5.64
(b) Option 2 7.14
(c) Option 3 7.16
(d) Option 4 7.64
Answers
Answer:
7.64mm is the answer I guess
The electric field strength at one point near a point charge is 1000 n/c. what is the field strength in n/c if the distance from the point charge is doubled?
Answers
The electric field strength near a point charge is inversely proportional to the square of the distance. Doubling the distance reduces the electric field strength by a factor of four.
The electric field strength at a point near a point charge is directly proportional to the inverse square of the distance from the charge. So, if the distance from the point charge is doubled, the electric field strength will be reduced by a factor of four.
Let's say the initial electric field strength is 1000 N/C at a certain distance from the point charge. When the distance is doubled, the new distance becomes twice the initial distance. Using the inverse square relationship, the new electric field strength can be calculated as follows:
The inverse square relationship states that if the distance is doubled, the electric field strength is reduced by a factor of four. Mathematically, this can be represented as:
(new electric field strength) = (initial electric field strength) / (2²)
Substituting the given values:
(new electric field strength) = 1000 N/C / (2²)
= 1000 N/C / 4
= 250 N/C
Therefore, if the distance from the point charge is doubled, the electric field strength will be 250 N/C.
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A ferry boat is 2.0 m wide and 4.0 m long. When a truck pulls onto it, the boat sinks 2.00 cm in the water. What is the weight of the truck?DENSITY OF WATER IS 1000 Kg/m^3
Answers
Answer:
156,800 N
Explanation:
Given: Width = 2.0 m Length = 4.0 m Depth = 2.00 cm. Density of water = 1000 Kg/m³
To find: What is the weight of the truck?
Formula: \(F = m\) × \(9.8 m/s^2\)
Solution: Weight is a measure of the force of gravity pulling down on an object. It depends on the object's mass and the acceleration due to gravity, which is 9.8 m/s² on Earth. F is the object's weight in Newtons (N) and m is the object's mass in kilograms.
Firstly, multiply the length, width, and depth.
\(2.0\) × \(4.0\) × \(2.00\) \(= 16\)
Next, the weight of the truck is found by multiplying the density of water, the result found by multiplying the length, width, and depth, and the formula;
\(1000 kg/m^3\) × \(16\) × \(9.8m/s^2\)\(= 156,800\)
Therefore, the weight of the truck is 156,800 N
A wave oscillates with a period of 4.25 X 10-10 s. How many waves will strike a signal detector in 3 seconds?
Answers
Answer:
\(n=7.05\times 10^9\) waves
Explanation:
Given that,
A wave oscillates with a period of \(4.25\times 10^{-10}\ s\)
We need to find the number of waves that will strike a signal detector in 3 seconds.
We know that,
Frequency = (no. of oscillations)/time
Time period = time/(no. of oscillations)
Let there are n waves that will strike
So,
\(n=\dfrac{3}{4.25\times 10^{-10}}\\\\=7.05\times 10^9\)
Hence, \(7.05\times 10^9\) waves will strike in 3 seconds.
FAILURE OF THE PRODUCT Instructions 1. Select THREE from everyday below items from the list and discuss the way this item can potentially fail (list minimum THREE failures). Justify your answer by considering Load Strength graph and what can be done to prevent those failures. -Ball Pen -Room Key - Blender
Answers
The three product which can be potentially fail considering Load Strength graph and precautionary measure to prevent failure are as below;
Ball Pen:
1. Ink Leakage: One potential failure of a ball pen is ink leakage. This can occur due to poor sealing between the ink reservoir and the ballpoint mechanism. Ink leakage can result in messy hands, stained documents, and reduced functionality of the pen. To prevent this failure, manufacturers can improve the quality control process to ensure proper sealing and use high-quality materials for the pen's components.
2. Ballpoint Jamming: Another failure is ballpoint jamming, where the ball gets stuck and prevents smooth writing. This can be caused by a buildup of dried ink or debris inside the pen's mechanism. To prevent ballpoint jamming, regular cleaning and maintenance of the pen can be recommended. Additionally, manufacturers can design the pen with features that facilitate easy cleaning or provide instructions on how to clear any blockages.
3. Weak Barrel Construction: The barrel of the pen may also be prone to failure if it is weak or brittle. Excessive pressure or rough handling can lead to cracks or breakage, rendering the pen unusable. To prevent this, manufacturers can use durable materials for the pen barrel, such as sturdy plastics or reinforced metal, and perform quality checks to ensure structural integrity.
Room Key:
1. Keycard Malfunction: A potential failure of a room key is a malfunction in its electronic components. This can result in the keycard being unreadable by the door lock system, preventing access to the room. To prevent this failure, regular maintenance and replacement of keycard readers can be implemented. Additionally, guests should be advised to keep their keycards away from magnets and electronic devices that can interfere with the card's functionality.
2. Magnetic Strip Damage: Another failure can occur if the magnetic strip on the keycard gets damaged or demagnetized. This can happen due to exposure to magnetic fields or physical damage. To prevent this failure, keycards can be made more durable with protective coatings or alternative technologies such as RFID. Guests should also be educated on proper handling and storage of keycards to avoid damage.
3. Battery Drain: Some room keys use batteries to power their electronic components. A failure can occur if the battery drains, leading to an inactive keycard. To prevent this, low-power consumption designs can be implemented, and regular battery checks or replacements can be carried out by hotel staff. Guests should be informed about the importance of returning the keycard to the front desk for recycling or proper disposal to ensure the battery is replaced as needed.
Blender:
1. Motor Burnout: One potential failure of a blender is motor burnout due to prolonged use or overloading. Continuous operation at high speeds or attempting to blend hard or frozen ingredients beyond the blender's capacity can cause the motor to overheat and fail. To prevent motor burnout, manufacturers can provide clear guidelines on the maximum load capacity and recommended usage durations. Automatic thermal protection mechanisms can also be incorporated to shut off the blender if it detects excessive heat.
2. Blade Jamming: Another failure can occur if food particles or ingredients get jammed between the blender's blades, preventing them from spinning freely. This can happen if the blender is not properly cleaned or if ingredients are not adequately prepared before blending. To prevent blade jamming, users should be advised to clean the blender thoroughly after each use and ensure that ingredients are cut into manageable sizes. Manufacturers can also design blades with accessible mechanisms for easy cleaning or provide cleaning tools.
3. Leakage: A failure in a blender can also manifest as leakage. This can happen if the blender jar or its sealing components are damaged or improperly assembled. Liquid or food can leak out during blending, resulting in a messy and potentially unsafe situation. To prevent leakage, manufacturers should ensure proper sealing mechanisms and use high-quality materials for the blender jar and lid. Regular inspection of the sealing components can be advised,
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What are 5 examples of potential energy?
Answers
Potential energy is the energy an object possesses because of its position or state. It is the energy that an object has the capacity to do work, but has not yet done so. Here are five examples of potential energy:
Gravitational potential energy: This is the energy an object possesses due to its position in a gravitational field. For example, a ball held above the ground has gravitational potential energy because it has the potential to fall and do work. The formula for gravitational potential energy is mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the ground.Elastic potential energy: This is the energy stored in an object when it is stretched or compressed. For example, a stretched rubber band has elastic potential energy. The more it is stretched, the more potential energy it has. The formula for elastic potential energy is 1/2 kx^2, where k is the spring constant and x is the displacement from the equilibrium position.Chemical potential energy: This is the energy stored in the bonds between atoms and molecules. For example, the chemical potential energy in a stick of dynamite is released when the chemical bonds are broken, creating an explosion.Nuclear potential energy: This is the energy stored in the nucleus of an atom. For example, the nuclear potential energy in a uranium atom is released when the nucleus is split, creating a nuclear reaction.Thermal potential energy: This is the energy stored in the motion of the atoms and molecules in a substance. For example, the thermal potential energy in a cup of hot coffee is released when it cools down to room temperature.It's important to note that potential energy can be converted into kinetic energy, which is the energy of motion. This can happen spontaneously if an object is able to move due to a force acting on it, such as gravity.
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Suppose we have one metal plate at 200 volts and another plate parallel to and facing the first plate at 0 volts. What is the magnitude of the electric field vector between the plates and in which direction does it point
Answers
The magnitude of the electric field between the plates would be approximately 200 volts per small distance. The direction of the electric field vector is from the positively charged plate to the negatively charged plate.
The magnitude of the electric field vector between the plates is given by the formula E = (V1 - V2) / d, where V1 is the voltage of the first plate (200 volts), V2 is the voltage of the second plate (0 volts), and d is the distance between the plates. Assuming that the plates are close to each other, we can take d to be very small. Therefore, the magnitude of the electric field between the plates would be very large, approximately 200 volts per small distance. The direction of the electric field vector is from the positively charged plate (200 volts) to the negatively charged plate (0 volts), which is from the first plate towards the second plate.
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When an ambulance drives towards you, the pitch of the siren is higher. After itdrives past, the pitch of the siren is lower.The reason it is lower is that the frequency of the wavelengthA. increasesB. decreasesC. Stays the same
Answers
Doppler's effect:
The Doppler effect states that the frequency of sound increases as it source approaches you and it decreases as the source turns away from you.
When the ambulance drives past you, the pitch of the siren is lower. This is due to the fact that the frequency of the wavelength decreases.
A higher pitch means frequency increases and a lower pitch means frequency decreases.
Therefore, the correct answer is option B. decreases
how do i multiply 9.6 by 3/2
Answers
Solve the given problem as
\(\begin{gathered} x=9.6\times\frac{3}{2} \\ =9.6\times1.5 \\ =14.4 \end{gathered}\)what causes condensation on ac vents
Answers
Answer:
Condensation on AC vents occurs when warm, humid air comes into contact with the cool surfaces of the vents. This phenomenon is similar to what happens when water droplets form on the outside of a cold glass on a hot day. The main causes of condensation on AC vents are:
Explanation:
Temperature difference: Air conditioning systems lower the temperature of indoor spaces, creating a significant temperature difference between the cold AC vents and the surrounding air. When warm, humid air from the room comes into contact with the cold surface of the vents, the moisture in the air condenses into water droplets.
High humidity: Humidity refers to the amount of moisture present in the air. Higher humidity levels mean that the air is holding more moisture. When the indoor humidity is high, and the AC vents are cooler than the dew point temperature (the temperature at which air becomes saturated and condensation occurs), condensation is more likely to form on the vents.
Inadequate insulation: Poor insulation or improper installation of air conditioning ductwork can lead to condensation issues. If the cool air from the ducts escapes into unconditioned spaces, such as attics or crawl spaces, the temperature difference can cause condensation to form on the vents.
Vent blockage: Blocked or restricted vents can disrupt the airflow from the AC system, resulting in lower temperatures near the vents. This can increase the likelihood of condensation forming on the vent surfaces.
Improper AC sizing: If the air conditioning system is oversized for the space it is cooling, it may cool the air too quickly, leading to shorter cycles and less dehumidification. This can result in higher humidity levels and increased condensation on the vents.
Calculate the first and second velocities of the car with four washers attached to the pulley, using the formulas
v1 = 0.25 m / t1 , and v2 = 0.25 m / (t2 – t1)
where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. Record these velocities, to two decimal places, in Table E
Answers
The first and second velocities of the car, with four washers attached to the pulley, are both 0.125 m/s.
We must apply the following calculations to determine the car's first and second velocities with the four washers attached to the pulley:
v1 = 0.25 m / t1
v2 = 0.25 m / (t2 - t1)
Here, t1 denotes the typical time needed for the car to go 0.25 metres, and t2 is the typical time needed to travel 0.50 metres.Let's say that t1 and t2 are the typical times, each lasting two seconds on average.The first formula allows us to determine v1:
v1 = 0.25 m / 2 s = 0.125 m/s
Using the second formula, we can calculate v2:
v2 = 0.25 m / (4 s - 2 s) = 0.125 m/s
Table E can be used to record these values as the velocities at the relevant distances.
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The displacement of a car moving with constant velocity 9.5 m/s in time interval between 3 seconds to 5 seconds is given by odt. What is the displacement of the car during that interval in meters?
Answers
The displacement of a car moving with a constant velocity of 9.5 m/s in a time interval between 3 seconds to 5 seconds is 19 meters.
It given by the formula: Δx = vΔt where Δx = displacement v = velocity Δt = time interval Substituting the given values, we get:Δx = 9.5 m/s × (5 s - 3 s)Δx = 9.5 m/s × 2 sΔx = 19 m, the displacement of the car during the given interval is 19 meters.
The given formula is derived from the definition of velocity which is the change in displacement per unit time. Since the velocity of the car is constant, we can assume that its acceleration is zero. Therefore, the car is not changing its velocity, which means that the displacement during that interval is equal to the product of velocity and time.In this case, we are given the initial and final times, and we need to find the displacement during that time interval.
The difference between the two times is 2 seconds. Multiplying the velocity with the time interval, we get the displacement of the car. The unit of displacement is meter, which is the same as the unit of distance.
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which of these is NOT an example of a lever?
a. shovel
b. pushup
c. ramp
d. seesaw
Answers
Answer:
c
Explanation:
Out of the given options, shovel, pushup, and seesaw are all examples of levers. Hence, the one that is not an example is :
rampA transformer is used to light a lamp rated 40w, 240v from a 400v A.C supply. Calculate:
A. The ratio of the number of turns of the primary to the secondary coil.
B. Current drawn from the main circuit if the efficiency of the transformer is 90%.
Answers
Answer:1.81
(a) Explanation:the turn ratio= input voltage÷output voltage.
400÷220=1.81.
Don't know how to solve b part...
hydrogen molecules, with a molar mass of 2.016 g/mol, in a certain gas have an rms speed of 395 m/s. What is the temperature of this gas, in kelvins?
Answers
The ideal gas law is,
PV = nRT
Substituting the values,
Vrms = √(3RT/M)
The root-mean-square speed of the hydrogen molecule (H2) is given as 395 m/s.
The molar mass of H2 is 2.016 g/mol.
Converting grams to kilograms:2.016 g/mol = 0.002016 kg/mol
Substituting the values into the formula;
395 m/s = √((3 × 8.314 J/mol-K × T) / (0.002016 kg/mol))
Square both sides;
(395 m/s)² = (3 × 8.314 J/mol-K × T) / (0.002016 kg/mol)
T = (395 m/s)² × 0.002016 kg/mol / (3 × 8.314 J/mol-K)
T = 373.95 K ≈ 374 K
Therefore, the temperature of the gas containing hydrogen molecules, with a molar mass of 2.016 g/mol, having an rms speed of 395 m/s is approximately 374 K.
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the sputnik 1 satellite orbited earth (mass=5.98 x 10^24 kg) in a circle of radius 6.96 x 10^6 m. what was its orbital velocity?
Answers
Answer:
v = 4.79 10³ m / s
Explanation:
For this exercise we will use the universal law of gravitation and Newton's second law
F = G M m / r²
where G is the gravitational constant, m the mass of the satellite, M the mass of the Earth, r the distance from the center of the planet to the satellite
F = m a
since the satellite has a circular path, the acceleration is centripetal
a = v² / r
we substitute
G M m / r² = m v² / r
G M / r = v²
We calculate
v² = 6.67 10⁻¹¹ 5.98 10²⁴ / 6.96 10⁶
v = √ (22.94 10⁶)
v = 4.79 10³ m / s
Answer:
7570 m/s
Explanation:
take the square root of: 6.67E-11× 5.98E24/6.96E6
100 POINTS
Plutonium-235
Given: energy released = about 200 MeV per individual reaction (mass = 239 amu)
Part A
List the balanced nuclear reaction.
I NEED THIS ASAP
Answers
Answer:
One gram atom of any substance contains one Avogadro number of atoms. Therefore, the number of atoms of U-235 present in 235 g of the substance = Avogadro Number = 6.02×10^23 atoms
The number of U-235 atoms present in 1 g of the substance = (6.02×10^23/235) atoms=2.56×10^21 atoms
Energy released in one atom fissioning = 200 MeV= 200 MeV × 1.6 × 10^-6 erg/MeV= 3.2 ×10^-4 erg/atom.
Energy released in the fission of all the atoms contained in 1g of Uranium-235 fissioning = (2.56×10^21 atom)×(3.2×10^-4 erg/atom)=8.19 ×10^17 erg
Now
1 kWh = 1000 Watt hour = 1000×( 10^7 erg/s)×3600s = 3.6 ×10^13 erg/kWh
So energy in kWh released in fission of 1 g of U-235 = (8.19× 10^17 erg)÷(3.6×10^13 erg/kWh) = 2.55 ×10⁴ kWh.
Added on 19th February, 2019.
The energy consumption of an average home in India in a month is between 300–500 kWh in the metropolitan cities. Let us take it at 400 kWh. So one gram of nuclear fuel (uranium) would be able to meet the energy requirements of a town of about 6000 for a month.
The energy in a fissile material is highly concentrated, compared to other types of fuels like wood, coal, diesel etc. So nuclear powered submarines can remain below water for very long periods of time, as they do not need to surface for getting fuel.
Explanation:
Energy in kWh released in fission of 1 g of U-235 is 2.55 ×10⁴ kWh.
How do find the energy released in fission?One gram atom of any substance contains one Avogadro number of atoms. Therefore, the number of atoms of U-235 present in 235 g of the substance = Avogadro Number = 6.02×10^23 atoms
The number of U-235 atoms present in 1 g of the substance = (6.02×10^23/235) atoms=2.56×10^21 atoms
Energy released in one atom fissioning = 200 MeV= 200 MeV × 1.6 × \(10^{-6}\) erg/MeV= 3.2 ×\(10^{-4}\) erg/atom.
The energy released in the fission of all the atoms contained in 1g of Uranium-235 fissioning = (2.56×\(10^{21}\) atom)×(3.2×\(10^{-4}\) erg/atom) = 8.19 ×\(10^{17}\) erg
1000 Watt hour = 1000×( \(10^{7}\) erg/s)×3600s = 3.6 ×10^13 erg/kWh
So the energy in kWh released in fission of 1 g of U-235 = 2.55 ×10⁴ kWh.
Fission takes place when a neutron slams into a bigger atom, forcing it to excite and break up into smaller atoms—additionally referred to as fission products. Extra neutrons are also launched that may initiate a sequence reaction. whilst each atom splits, a high-quality amount of electricity is released.
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experiments and modeling of the viscoelastic behavior of polymeric gels. journal of the mechanics and physics of solids
Answers
The Journal of the Mechanics and Physics of Solids has published numerous studies on the experiments and modelling of the viscoelastic behaviour of polymeric gels. These studies aim to understand and predict the mechanical properties of polymeric gels, which are essential in various applications such as biomaterials, soft robotics, and drug delivery systems.
Polymeric gels exhibit a viscoelastic behaviour, meaning they possess both viscous (fluid-like) and elastic (solid-like) properties. The Journal of the Mechanics and Physics of Solids has been a prominent platform for researchers to share their findings on the experimental characterization and theoretical modelling of these materials. Experimental studies on polymeric gels often involve rheological tests, which measure the gel's response to external forces and deformation. These experiments help in understanding the time-dependent mechanical behaviour of the gels, including their stress relaxation and creep properties.
Modelling the viscoelastic behaviour of polymeric gels is crucial for predicting their mechanical response under different conditions. Researchers have developed various constitutive models, such as the Maxwell model, the Kelvin-Voigt model, and the Burgers model, to describe the viscoelastic properties of gels. These models incorporate parameters like elastic modulus, viscous damping coefficient, and relaxation time to capture the gel's behaviour accurately.
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The Journal of the Mechanics and Physics of Solids has published numerous studies on the experiments and modelling of the viscoelastic behaviour of polymeric gels. These studies aim to understand and predict the mechanical properties of polymeric gels, which are essential in various applications such as biomaterials, soft robotics, and drug delivery systems.
Polymeric gels exhibit a viscoelastic behaviour, meaning they possess both viscous (fluid-like) and elastic (solid-like) properties. The Journal of the Mechanics and Physics of Solids has been a prominent platform for researchers to share their findings on the experimental characterization and theoretical modelling of these materials. Experimental studies on polymeric gels often involve rheological tests, which measure the gel's response to external forces and deformation. These experiments help in understanding the time-dependent mechanical behaviour of the gels, including their stress relaxation and creep properties.
Modelling the viscoelastic behaviour of polymeric gels is crucial for predicting their mechanical response under different conditions. Researchers have developed various constitutive models, such as the Maxwell model, the Kelvin-Voigt model, and the Burgers model, to describe the viscoelastic properties of gels. These models incorporate parameters like elastic modulus, viscous damping coefficient, and relaxation time to capture the gel's behaviour accurately.
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Hola cómo están todos
Answers
a) A cell of dry air is moved vertically from its original position under adiabatic conditions. Depending on the temperature profile of the surrounding atmosphere, this gas cell can keep on moving in the same direction, or it may come back to its original position. Considering the temperature profile of the atmosphere, change of the air cell temperature as it moves up and down in the surrounding atmosphere, as well as relative densities of the air cell and atmosphere, explain why and when the atmosphere is considered to be convectively stable and convectively unstable. In answering this question, use diagrams of temperature change with altitude. (13 marks) b) Explain why the adiabatic lapse rate of dry air is different from the adiabatic lapse rate of wet saturated air. Show them both in a diagram. (5 marks) c) Wet unsaturated air rises from the ocean surface. The ambient lapse rate is higher than the adiabatic lapse rate for dry air. There is a temperature inversion layer at higher altitudes. Show in a schematic diagram how the temperature of the wet air changes with altitude, in comparison with the ambient temperature. Explain at what altitudes the cumulus clouds are formed and why. (7 marks)
Answers
The question addresses the stability of the atmosphere and the factors that determine convective stability or instability. It also explains the difference between the adiabatic lapse rate of dry air and wet saturated air.
a) The stability of the atmosphere is determined by the temperature profile and relative densities of the air cell and atmosphere. If the temperature of the surrounding atmosphere decreases with altitude at a rate greater than the adiabatic lapse rate of the air cell, the atmosphere is considered convectively stable.
In this case, the air cell will return to its original position. Conversely, if the temperature of the surrounding atmosphere decreases slower than the adiabatic lapse rate of the air cell, the atmosphere is convectively unstable. The air cell will continue moving in the same direction.
b) The adiabatic lapse rate refers to the rate at which temperature decreases with altitude for a parcel of air lifted or descending adiabatically (without exchanging heat with its surroundings). The adiabatic lapse rate of dry air is higher (around \(9.8^0C\) per kilometer) compared to the adiabatic lapse rate of wet saturated air (around 5°C per kilometer).
This difference arises because when water vapor condenses during the ascent of saturated air, latent heat is released, reducing the rate of temperature decrease. A diagram can illustrate the difference between the two lapse rates, showcasing their respective slopes.
c) When wet unsaturated air rises from the ocean surface, its temperature decreases at a rate equal to the dry adiabatic lapse rate. However, if the ambient lapse rate (temperature decrease with altitude) is higher than the adiabatic lapse rate for dry air, a temperature inversion layer forms at higher altitudes.
In this inversion layer, the temperature increases with altitude instead of decreasing. A schematic diagram can depict the temperature changes of the wet air in comparison to the ambient temperature, showing the inversion layer.
Cumulus clouds form at the altitude where the rising moist air reaches the level of the temperature inversion layer. These clouds are formed due to the condensation of water vapor as the air parcel cools to its dew point temperature.
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A boy weighing 50 kg is running at 2m/s. What is its kinetic energy? *
10 points
Answers
The formula is KE = 1/2mv^2
Then you plug in 50 to the m
And 2 to the v . That’s how I got 100 .
Hope this helps
6. Explain a change that you can make in your diet to further your overall health.
Answers
Answer:
“Eat more delicious whole-food plants at every meal and snack. Fruits, vegetables, whole grains, nuts and seeds are packed with nutrients and contain satisfying fiber that is good for digestion, disease prevention and sustained energy.
Explanation:
Hope this helped
A 22 g block of ice is cooled to −70 ◦C. It is added to 589 g of water in an 83 g copper calorimeter at a temperature of 21◦C. Find the final temperature. The specific heat of copper is 387 J/kg · ◦C and of ice is 2090 J/kg ·◦C . The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg · ◦C . Answer in units of ◦C
Answers
I
Given the densities of five objects: 2.00 grams/cm^3, 3.00 grams/cm^3, 4.00 grams/cm^3, 5.00 grams/cm^3, and 6.00 grams/cm^3. How many of these objects would float in water? A)1 B)2 C)3 D)4 E)0
Answers
Givne densities are,
\(\begin{gathered} \rho_1=2\text{ g/}cm^3 \\ \rho_2=3\text{ g/}cm^3 \\ \rho_3=4\text{ g/}cm^3 \\ \rho_4=5\text{ g/}cm^3 \end{gathered}\)The density of water is ρ=1 g/cm³.
Any object will float if the denisty of the object is less than the density of water, if the density of object is greater than the dnesity of water, then the object sink in the water.
In the given case, all the objects have density greater than the density of water. Therefore, all the objects will sink in the water.
Thus, option E is corret, no object will flot on the water.
Dizzy is speeding along at 22.8 / as she approaches the level section of track near the loading dock of the Whizzer roller coaster ride. A braking system abruptly brings the 328 car (rider mass included) to a speed of 2.9 / over a distance of 5.55 . Determine the braking force applied to Dizzy's car
Answers
The braking force applied to Dizzy's car is approximately 50,600 Newtons in the opposite direction to the car's motion.
What is braking force?Braking force is the force applied by a vehicle's brakes to slow down or stop the vehicle. To determine the braking force applied to Dizzy's car, we can use the equation:
force = mass x acceleration
where the acceleration is the rate at which the car's velocity changes during braking.
First, we need to calculate the initial velocity of the car before braking:
v0 = 22.8 m/s
Next, we need to calculate the final velocity of the car after braking:
vf = 2.9 m/s
The change in velocity, or delta-v, is:
delta-v = vf - v0 = 2.9 m/s - 22.8 m/s = -19.9 m/s
Note that the negative sign indicates a decrease in velocity.
We also need to calculate the time it takes for the car to come to a stop. We can use the equation:
delta-x = (v0 + vf) / 2 x t
where delta-x is the distance traveled during braking, which is given as 5.55 m.
Rearranging the equation, we get:
t = 2 x delta-x / (v0 + vf)
Substituting the given values, we get:
t = 2 x 5.55 m / (22.8 m/s + 2.9 m/s) = 0.393 s
Now we can calculate the acceleration:
acceleration = delta-v / t = (-19.9 m/s) / (0.393 s) = -50.6 m/s^2
Again, the negative sign indicates a deceleration or braking acceleration.
Finally, we can calculate the force applied to the car:
force = mass x acceleration
The mass of the car is not given, but we can assume it is approximately the same as the average mass of a roller coaster car, which is around 1000 kg.
Substituting the values, we get:
force = 1000 kg x (-50.6 m/s^2) = -50,600 N
Therefore, the braking force applied to Dizzy's car is approximately 50,600 Newtons in the opposite direction to the car's motion.
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