Answer:
Spring constant: approximately [tex]28\; {\rm N\cdot m^{-1}}[/tex].
Unloaded length: approximately [tex]7.0\; {\rm cm}[/tex].
(Assume that the weight of the spring is negligible, and that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)
Explanation:
Divide the tension [tex]F[/tex] on the spring by the displacement [tex]x[/tex] to find the spring constant. [tex]k[/tex]:
[tex]\begin{aligned}k &= \frac{F}{x}\end{aligned}[/tex].
Let [tex]L_{0}[/tex] denote the unloaded length of the spring (in meters.)
When the [tex]0.225\; {\rm kg}[/tex] mass is on the spring, the tension on the spring would be [tex]m\, g = (0.225)\, (9.81) \; {\rm N} \approx 2.207\; {\rm N}[/tex].
It is given that the length of the spring with this [tex]0.225\; {\rm kg}[/tex] mass attached is [tex]L = 0.15\; {\rm m}[/tex]. Subtract the initial length of the spring [tex]L_{0}[/tex] from the current length to find the displacement of the spring: [tex]x = L - L_{0} = 0.15 - L_{0}[/tex].
Divide the tension on the spring by the displacement to find an expression for the spring constant:
[tex]\begin{aligned}k &= \frac{F}{x} \approx \frac{2.207}{0.15 - L_{0}}\end{aligned}[/tex].
Similarly, when the [tex]1.85\; {\rm kg}[/tex] object is on the spring, the tension on the spring would be [tex]m\, g = (1.85)\, (9.81) \; {\rm N} \approx 18.15\; {\rm N}[/tex].
Subtract the initial length [tex]L_{0}[/tex] from the current length [tex]L = 0.725\; {\rm m}[/tex] to find the displacement: [tex]x = L - L_{0} = (0.725 - L_{0})\; {\rm m}[/tex].
Divide the tension on the spring by the displacement to find another expression for the spring constant:
[tex]\begin{aligned}k &= \frac{F}{x} \approx \frac{18.15}{0.725 - L_{0}}\end{aligned}[/tex].
Equate the two expressions for the spring constant [tex]k[/tex] and solve for the unloaded length [tex]L_{0}[/tex]:
[tex]\begin{aligned}\frac{18.15}{0.725 - L_{0}} &= \frac{2.207}{0.15 - L_{0}}\end{aligned}[/tex].
[tex](18.15)\, (0.15 - L_{0}) = (2.207)\, (0.725 - L_{0})[/tex].
[tex]\begin{aligned} L_{0} &\approx \frac{(18.15)\, (0.15) - (0.725) (2.207)}{18.15 - 2.207}\; {\rm m} \\ &\approx 0.070\; {\rm m}\end{aligned}[/tex].
Substitute [tex]L_{0}[/tex] back into one of the two expressions for the spring constant [tex]k[/tex] and evaluate:
[tex]\begin{aligned}k &\approx \frac{18.15}{0.725 - L_{0}} \\ &\approx \frac{18.15}{0.725 -0.070} \approx 28\; {\rm N\cdot m^{-1}} \end{aligned}[/tex].
Apply unit conversion:
[tex]0.070\; {\rm m} = (0.070 \times 10^{2})\; {\rm cm} = 7.0\; {\rm cm}[/tex].
In other words, the unloaded length of the spring would be approximately [tex]7.0\; {\rm cm}[/tex].
The correct answer to part a) is, the force constant of the spring can be found using Hooke's Law and it is 3.83 N/m, and part b) is, the unloaded length of the spring is the same as the length when the 1.85 kg mass is hanging from it, which is 0.725 m.
Part (a) To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is equal to the spring constant multiplied by the displacement from equilibrium.
The equation for this is F = kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium.
In this case, we can use the two given lengths and masses to find the displacement from equilibrium and the force exerted by the spring.
The displacement for the first mass is 0.15 m - 0.725 m = -0.575 m, and the force is 0.225 kg * 9.8 m/s^2 = 2.205 N. The displacement for the second mass is 0.725 m - 0.725 m = 0 m, and the force is 1.85 kg * 9.8 m/s^2 = 18.13 N.
We can then plug these values into the equation F = kx and solve for k:
2.205 N = k(-0.575 m)
k = -2.205 N / -0.575 m
k = 3.83 N/m
Therefore, the force constant of the spring is 3.83 N/m.
Part (b) To find the unloaded length of the spring, we can use the equation x = F/k, where x is the displacement from equilibrium, F is the force, and k is the spring constant.
Since the unloaded length of the spring is when the force is zero, we can plug in F = 0 and k = 3.83 N/m to solve for x:
x = 0 N / 3.83 N/m
x = 0 m
Since the displacement from equilibrium is zero, the unloaded length of the spring is the same as the length when the 1.85 kg mass is hanging from it, which is 0.725 m.
Therefore, the unloaded length of the spring is 0.725 m, or 72.5 cm.
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Which of the following is an example of acceleration?
O A. 15 m/s
OB. 25 m/s east
O C. 60 km/hr
OD. -2 m/s² west
Option D: -2 m/s² west is an example of acceleration.
What is the definition of acceleration?Acceleration is the rate of change of velocity with respect to time. It is a vector quantity and can be positive or negative depending on the direction of the change in velocity.
Why is option D an example of acceleration?Option D is an example of acceleration because it includes both the magnitude (-2 m/s²) and direction (west) of the change in velocity. The negative sign indicates that the velocity is decreasing, and the direction (west) indicates the direction of the change in velocity.
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o dramatize the loss of energy in an automobile, consider a car having a weight of 5000 lb that is traveling at 35 mi>h. if the car is brought to a stop, determine how long a 100-w light bulb must burn to expend the same amount of energy. (1 mi
The light bulb must burn for approximately 46.2 minutes to use the same amount of energy as the car lost during its deceleration.
Consider a 5,000-pound vehicle at 35 mph to illustrate energy loss in an automobile. When the automobile stops, its kinetic energy is transformed into heat and sound, resulting in a large energy loss. We use a 100-Watt light bulb to understand this loss.
The car's initial kinetic energy is [tex]KE = 0.5 * mass * velocity^2[/tex]. Next, we calculate 100-Watt light bulb energy usage per second. Finally, we divide the automobile's kinetic energy by the light bulb's energy consumption per second to determine the time needed for the light bulb to use the same amount of energy as the car.
The light bulb must burn for 46.2 minutes to utilise as much energy as the automobile lost during deceleration. This striking depiction illustrates the significant energy waste required in halting a moving automobile and emphasises energy conservation and efficient transportation systems.
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Respond to the following based on your reading.
How does the valence of an element relate to its chemical activity?
Identify the statements below as belonging to one of these laws: law of conservation of matter, law of conversation of energy, or law of definite proportions.
Energy can be neither created nor destroyed by chemical means but can be changed from one form to another.
The composition of a chemical compound never varies.
Matter can be neither created nor destroyed by chemical means but can be changed from one form to another.
How was the Periodic Table of Elements developed and how are the elements arranged on it?
Using the Periodic Table of Elements, fill in the chart below to identify and describe the properties of the noble gases.
Element Name Group Number Period Number Valence Properties/Reactivity Uses
Using the Periodic Table of Elements, fill in the chart below to identify and describe the properties of the halogens.
Element Name Group Number Period Number Valence Properties/Reactivity Uses
Using the Periodic Table of Elements, fill in the chart below to identify and describe the properties of the nonmetals.
Element Name Group Number Period Number Valence Properties/Reactivity Uses
Describe the unique properties of water. Be sure to include an explanation of the filtration systems we use to keep water safe for drinking.
What is the Earth’s atmosphere composed of, and why is it important?
The gases that make up the planet's atmosphere are mixed together.Nitrogen, oxygen, & carbon dioxide are among the atmospheric gases.The water cycle & weather depend on the atmosphere, which also sustains life.
Why is it crucial to us in Class 7?The atmosphere of the earth shields its people by absorbing dangerous sun rays and preserving a constant temperature.It protects Earth from many space hazards and enables life there.As a result, it is crucial for regulating climate and a source of life on planet.
What elements make up the atmosphere on Earth?Approximately 78% of the air of Earth's atmosphere is nitrogen, while 21% is oxygen.In trace proportions, additional gases like carbon dioxide, neon, & hydrogen are also present in air.
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Answer:
1. The number of valence electrons that an atom has dictates its ability to react chemically. The closer an atom is to achieving a full outermost shell, the more likely it will be to borrow electrons. Atoms are paired in a way that allows both to achieve a full stable outermost shell. For example, one atom of silicon has 14 electrons, four of which are valence electrons in the outermost shell. Silicon can lend, share, or gain four electrons to achieve a full outermost shell.
2.
a. Law of conservation of energy
b. Law of definite proportions
c. Law of conservation of matter
3. Mendeleev first published a table of elements arranged according to increasing atomic masses. He noticed that some elements near each other had differing properties, but elements in vertical columns had similar properties. Moseley then rearranged the table according to atomic numbers and this eliminated the discrepancies found in Mendeleev’s attempt. Today’s version of the periodic table displays elements in order based on their atomic number; the atomic number indicates the number of protons within the atoms of a particular element. Rows are called periods and columns are called groups. Elements in the same group have similar properties. Elements are grouped into nine categories: noble gases, halogens, nonmetals, alkali metals, alkaline earth metals, transition metals, other metals, metalloids, and rare earth elements.
( VIEW ATTACHMENT FOR 4, 5, AND 6.)
7. Water is colorless, odorless, and tasteless. It's the chief substance of living material. It has a specific gravity of 1.0 at 4° Celsius. It freezes at 0° Celsius and boils at 100° Celsius. Water is a powerful solvent. Distillation is needed to remove impurities. Purification methods include
Aeration (water is sprayed into the air so that light and oxygen can kill bacteria)
Filtration (water is passed through a filter to remove impurities)
Chlorination (chlorine is added to water to destroy bacteria)
Coagulation (chemicals are added to water to cause organic matter with bacteria to settle out)
8. Air contains nitrogen (78%), oxygen (21%), carbon dioxide (0.04%), and rare gases (less than 1%). It also contains water, dust, bacteria, and other materials.
The atmosphere supports life via two cycles. In the carbon dioxide cycle, carbon dioxide is taken in by plants and used in photosynthesis. From here, oxygen is released to the atmosphere and taken in by animals; it fuels combustion and is needed for fermentation and decay. These processes convert it back to carbon dioxide.
In the nitrogen cycle, nitrogen is fixed by bacteria in the soil into usable nitrogen compounds absorbed by plants. Plants use these to form proteins, which are used as food by animals. Decaying plants and animal wastes return the nitrogen to the soil and ammonia to air. Without atmospheric nitrogen, rust and decay of living material would happen more quickly.
Determine the total resistance for the current
The total flow of electricity from the source is equivalent to the sum of the currents through each path. To calculate the total impedance in a parallel circuit, use the method below : 1/Rt Equals 1/R1 + 1/R2 + 1/R3 +...
How do you find total resistance of a current?IT = VT/RT or I total = V total / R total or the total current = the total voltage / the total resistance.
Current still flows along the other routes even if one of the parallel ones is blocked.
R_total = R1 + R2 +... + Rn The sum of all individual resistances makes up the system's overall resistance. Take the example issue below as an
example. A resistor with a 100 ohm electrical resistance value is linked to a resistor with a 200 ohm electrical resistance value.
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vegetables donot burn if we use oil while frying it why?
Explanation:
This is because the oil is an unsaturated fat (liquid at room temperature) and when heated to such high temperature becomes saturated.
Which one of the following statements concerning electrostatic situations is false?A. No work is required to move a charge along an equipotential surface. B. If the electric potential with a region of space is zero volts, the electric field within that region must also be zero V/m. C. The electric field is never perpendicular to equipotential surfaces. D. The electric field is zero V/m everywhere inside a conductor
The correct option is C, the following statements concerning electrostatic situations is false of The electric field is never perpendicular to equipotential surfaces.
Equipotential surfaces are imaginary surfaces in space where every point on the surface has the same electrical potential or voltage. In other words, these surfaces represent points in space where the electric potential is constant.
Equipotential surfaces are important in the study of electric fields because the direction of the electric field is always perpendicular to the equipotential surfaces. This means that the electric field lines must cross the equipotential surfaces at right angles. The spacing between equipotential surfaces also gives an indication of the strength of the electric field - the closer the surfaces, the stronger the field. The electric field created by the charge will have a certain strength and direction at every point in space.
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A battery of EMF 1. 5V has terminal pd of 1. 25V when a resistor of 25 ohm is joined to it. Calculate the current flowing, internal resistance and terminal pd when resistance of 25 ohm is replaced by 10 ohm
Using Ohm's law, we can find the current flowing in the circuit: I = V / R where I is the current, V is the voltage, and R is the resistance.
When a resistor of 25 ohms is connected to the battery, the terminal pd is 1.25 V. This means that the voltage drop across the internal resistance of the battery is: 1.5 V - 1.25 V = 0.25 V. Using Ohm's law, we can find the internal resistance of the battery: 0.25 V / I = R_internal. Substituting the given values, we get: R_internal = 0.25 V / (1.5 V / 25 ohm) = 4.17 ohm. The current flowing in the circuit is: I = 1.25 V / 25 ohm = 0.05 A. When the resistance is replaced by 10 ohms, the current flowing in the circuit is: I = 1.5 V / 35 ohm = 0.043 A. Using the same approach as before, the terminal pd can be calculated as: 1.5 V - 0.043 A * (R_internal + 10 ohm) = V_terminal. Substituting the value of R_internal, we get: V_terminal = 1.5 V - 0.043 A * (4.17 ohm + 10 ohm) = 1.34 V. Therefore, when the resistance is replaced by 10 ohms, the current flowing in the circuit is 0.043 A, the internal resistance of the battery is 4.17 ohm, and the terminal pd is 1.34 V.
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Princess Leia once fell out of the spaceship. She came back with the Force. From the farthest spot (zero speed) back to the spaceship was 10 meters, and it took 5 second. She weighed 50 kg. Assume constant force during this trip. Find the force needed. (It is not much, but to get a force while floating in space is magic.)
The force needed to propel Princess Leia from a stationary position 10 meters back to the spaceship in 5 seconds would be 100 Newtons. This force can be determined by using the formula F = m x a, where m is the mass of the object and a is the acceleration.
The force required to propel Princess Leia from a stationary position 10 meters back to the spaceship in 5 seconds would be equal to the work done divided by the time taken, or 50 kg x 10 m/5 s = 100 Newtons. This is the force that is needed to move her 10 meters in 5 seconds, taking into account her 50 kg mass.
To explain this in a little more detail, the force required is determined by the formula F = ma, where F is force, m is mass and a is acceleration. Since the acceleration is constant in this scenario, the force applied is equal to the mass multiplied by the acceleration, or F = m x a. This means that to move the 50 kg mass of Princess Leia from zero speed to 10 meters in 5 seconds, a force of 100 Newtons is required.
To illustrate this with an example, if you were to pick up a 50 kg weight and move it 10 meters in 5 seconds, you would need to exert 100 Newtons of force. This is the same force required to move Princess Leia in the same way in space, taking into account her 50 kg mass.
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What is the condition of the atmosphere at a certain place and time?
The condition of the atmosphere at any given place and time depends on a variety of factors, such as air temperature, air pressure, wind speed and direction, humidity, precipitation, and cloud cover.
Air temperature can influence wind speed and direction, humidity, and cloud cover. Air pressure can impact wind speed and direction, and can create large-scale weather patterns. Wind speed and direction can determine the location of rain or snow, and also affect the temperature. Humidity influences air temperature, as well as precipitation and cloud cover. Precipitation can determine the amount of moisture in the atmosphere, and can be affected by the temperature and humidity of the air. Cloud cover affects the amount of solar radiation that reaches the surface, and can be determined by the temperature and humidity of the atmosphere.
In conclusion, the condition of the atmosphere at any given place and time depends on the interaction of several different factors. Air temperature, air pressure, wind speed and direction, humidity, precipitation, and cloud cover all interact to form the atmospheric conditions in any particular place and time.
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a transformer on a pole near a factory steps the voltage down from 3600 v to 120 v. the transformer is to deliver 1000 kw to the factory at 90% efficiency. find (a) the power delivered to the primary, (b) the current in the primary, and (c) the current in the secondary.
A) The power delivered to the primary is 1111.1 kWB) The current in the primary is 308.6 A.C) The current in the secondary is 9258 A.
We can use the formula for power to solve this problem: Power = Voltage x Current x Efficiency, where efficiency is a decimal between 0 and 1.
(a) The power delivered to the primary can be found by dividing the power delivered to the secondary by the efficiency of the transformer:
Power_primary = Power_secondary / Efficiency
Power_secondary is the power delivered to the factory, which is 1000 kW.
Efficiency is given as 90%, or 0.9.
Power_primary = 1000 kW / 0.9 = 1111.1 kW
Therefore, the power delivered to the primary is 1111.1 kW.
(b) To find the current in the primary, we can use the formula for power again:
Power_primary = Voltage_primary x Current_primary
We know that the voltage on the primary side is 3600 V. Rearranging the formula, we get:
Current_primary = Power_primary / Voltage_primary
Current_primary = 1111.1 kW / 3600 V = 308.6 A
Therefore, the current in the primary is 308.6 A.
(c) To find the current in the secondary, we can use the turns ratio of the transformer:
Turns_ratio = Voltage_primary / Voltage_secondary
We know that the voltage on the secondary side is 120 V, and the turns ratio is:
Turns_ratio = 3600 V / 120 V = 30
The turns ratio tells us that the secondary voltage is 1/30th of the primary voltage, and the current on the secondary side is 30 times greater than the current on the primary side.
Therefore, the current in the secondary is:
Current_secondary = Current_primary x Turns_ratio
Current_secondary = 308.6 A x 30 = 9258 A
Therefore, the current in the secondary is 9258 A.
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The moment magnitude scale is a scale that rates earthquakes by estimating the total energy released by an earthquake . Estimating the total amount of energy released, enables comparison of earthquakes more accurately. It measures only small earthquakes.
It uses data collected by a seismograph.
It shows only the strength of seismic waves that were produced.
It determines the amount of damage caused by an earthquake.
The total amount of energy released, enables comparison of earthquakes more accurately. Thus, b. It makes use of seismograph data.
How the moment magnitude scale rates earthquakes?
The moment magnitude scale rates earthquakes by estimating the total energy released by an earthquake, based on data collected by a seismograph. It takes into account several factors, including the size of the fault that ruptured, the amount of movement that occurred, and the strength of the rocks involved.
The resulting magnitude value indicates the relative size or strength of the earthquake, and allows for more accurate comparison of earthquakes of different sizes. It does not directly determine the amount of damage caused by an earthquake, but a larger magnitude earthquake is generally capable of causing more damage than a smaller magnitude earthquake.
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I runner makes one lap around a 1,765m
track an average speed of 441.25 meters a day
How long did it take him?
Answer:
To find the time, we can use the formula:
time = distance ÷ speed
In this case, the distance is 1,765 meters and the speed is 441.25 meters per minute (not per day, as stated in the question). We need to convert the speed to minutes per lap:
441.25 meters/min = 7.3542 meters/sec
1 lap = 1765 meters
Now we can calculate the time:
time = 1765 meters ÷ 7.3542 meters/sec
time ≈ 240.11 seconds
Therefore, it took the runner approximately 240.11 seconds, or 4 minutes and 0.11 seconds, to complete one lap around the track at an average speed of 441.25 meters per minute.
Explanation:
A spacecraft is orbiting a planet at a distance of 10,000 km from the planet's center. The spacecraft's speed is constant at 20,000 km/hr. Suddenly, the spacecraft's engines fail and it begins to free fall towards the planet's surface. Assuming that the planet has a uniform density and a radius of 5,000 km:
1. What is the gravitational acceleration experienced by the spacecraft as it falls towards the planet's surface?
2. What is the maximum velocity that the spacecraft can reach just before impacting the planet's surface?
3. What is the kinetic energy of the spacecraft just before impacting the planet's surface?
The gravitational acceleration experienced by the spacecraft would be 13.33 m/s^2.
The maximum velocity the spacecraft can reach would be 1633.5 m/s.
The kinetic energy of the spacecraft would be KE = 1/2mv^2
Gravitational acceleration problemThe gravitational acceleration experienced by the spacecraft as it falls toward the planet's surface can be calculated using the formula
g = G*M/r^2Thus:
g = (6.67 x 10^-11 N m^2/kg^2) * (4/3 * pi * (5000 km)^3 * 5500 kg/m^3) / (10000 km)^2 = 13.33 m/s^2.
The maximum velocity that the spacecraft can reach just before impacting the planet's surface can be calculated using the formula
v = sqrt(2gh)The height from which the spacecraft falls is equal to the distance from the planet's surface to the spacecraft's initial position, which is 10,000 km - 5000 km = 5000 km.
Thus, v = sqrt(2 * 13.33 m/s^2 * 5,000,000 m) = 1633.5 m/s.
The kinetic energy of the spacecraft just before impacting the planet's surface can be calculated using the formula
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2 bodies of equal masses are placed at heights h/2 & 2h respectively. Find the-ratio of their gravitational potential energy
The-ratio of their gravitational potential energy is [tex]1:2[/tex]. The unit of measurement for any and all types of energy, which would include kinetic and potential energy, is kilograms per square second, or kg*m2/s2 (J).
Height [tex]H_{1} = h[/tex]
Height [tex]F_{2} = 2h[/tex]
Mass of body 1 = m
Mass of body 2 = m
Gravitational potential energy of body [tex]1 = mgH 1 =mgh[/tex]
Gravitational potential energy of Body 2 [tex]= mgH 2 =mg(2h)[/tex].
Ratio of gravitational potential energies
[tex]=\frac{mgh}{mg (2h)} =\frac{mgh}{2mgh} =\frac{1}{2} = 1:2[/tex]
What Unit of Measurement Is Potential Energy in?
Due to the fact that energy and work quantify identical types of force, the Joule is indeed the proven scientific measurement device for either of those.
How do you calculate kinetic energy?
K E = 1 2 m v 2 is the formula for kinetic energy. When m denotes the body's weight as well as v denotes its velocity, KE stands for kinetic energy.
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work of joules is done in stretching a spring from its natural length to 14 beyond its natural length. what is the force (in newtons) that holds the spring stretched at the same distance (14 )?
The force that holds the spring stretched at the same distance (14) is 25.76 N.
The work done in stretching a spring from its natural length to 14 beyond its natural length = W = 180 J
The stretch distance is = x = 14
Work done in stretching a spring = (1/2) k x² Where,
W = work done in stretching a spring
k = spring constant
x = distance stretched from its natural length
Therefore, the formula for finding the spring constant k is: k = (2W) / x²
From the given data, Work done = W = 180 J
Stretch distance = x = 14
Therefore, the spring constant k is: k = (2W) / x²= (2 × 180 J) / (14)²= (2 × 180) / 196= 360 / 196= 1.84 N/m
The force that holds the spring stretched at the same distance (14) is given by the formula: F = k x Where,
k = spring constant
x = stretch distance
Therefore, F = k x= 1.84 N/m × 14= 25.76 N
Therefore, the force that holds the spring stretched at the same distance (14) is 25.76 N.
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A small squishy ball is thrown against the side of a large cube of concrete. While the objects are in contact with each other, which experiences the larger force?
the correct option is b. The large cube of concrete exerts a larger force on the small squishy ball.
It tends to be described as a connection that influences an article's movement on the off chance that there is no resistance. The straightforward meaning of force, then again, is the push or pull that any article gets. Force is a vector amount, consequently, it has a magnitude and a bearing.
Nonetheless, gravity is a force in a more broad sense since it represents the connection that happens when two masses are near each other. Gravitational effects are on a very basic level brought about by the extension of spacetime and the movement of things through the extended spacetime. The impact, by and by, is by all accounts the consequence of utilizing force.
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the complete question is:
1) A small squishy ball is thrown against the side of a large cube of concrete. While the objects are in contact with each other, which experiences the larger force? a. The small squishy ball exerts a larger force on the large cube of concrete b. The large cube of concrete exerts a larger force on the small squishy ball C. Neither object exerts a force on the other while they are in contact. d. The small squishy ball and the large cube of concrete exert equal force on each other.
a. Two charged parallel plates are 4.0 cm apart and have a uniform electric field of 750 V /m between them. What is the potential difference between the plates?
b. How much work is necessary to move a +7.0 mC charge from the positive plate to the negative plate!
Answer:
a. The potential difference between two parallel charged plates can be calculated using the formula V = Ed, where V is the potential difference, E is the electric field strength between the plates, and d is the distance between the plates. In this case, the distance between the plates is 4.0 cm or 0.04 m and the electric field strength is 750 V/m. Plugging these values into the formula gives V = 750 V/m * 0.04 m = 30 V. So, the potential difference between the two parallel charged plates is 30 V.
b. The work done to move a charge between two points in an electric field can be calculated using the formula W = qV, where W is the work done, q is the charge being moved, and V is the potential difference between the two points. In this case, the charge being moved is +7.0 mC or 0.007 C and the potential difference between the positive and negative plates is 30 V (calculated in part a). Plugging these values into the formula gives W = 0.007 C * 30 V = 0.21 J. So, it would take 0.21 joules of work to move a +7.0 mC charge from the positive plate to the negative plate.
What would be the weight of a 60-kg astronaut on a newly discovered planet that is found to have density 2/3 ρᴇ and radius 2Rᴇ , where ρᴇ and Rᴇ are the density and radius of Earth, respectively?
A) 200N
B) 600N
C) 800N
D)1800N
The weight of the astronaut on the newly discovered planet is approximately 1800N, which is option D.
How is acceleration due to gravity calculated?The weight of an object is given by the formula W=mg, where g is the gravitational acceleration and m is the mass of the object. The gravitational acceleration on the surface of a planet can be calculated using the formula [tex]g = GM/R^2[/tex], where M is the mass of the planet, R is its radius, and G is the gravitational constant.
In this case, we are given that the planet has density 2/3 ρᴇ and radius 2Rᴇ. The mass of the planet can be calculated using the formula M = (4/3)π[tex]R^3[/tex]ρ, where ρ is the density of the planet. Substituting the given values, we get:
M = (4/3)π(2Rᴇ[tex])^3[/tex](2/3ρᴇ) = (32/27)πρᴇRᴇ[tex])^3[/tex]
The gravitational acceleration on the surface of the planet can be calculated as:
g = GM/R^2 = G[(32/27)πρᴇRᴇ^3]/(2Rᴇ[tex])^2[/tex] = (4/3)Gπρᴇ
The weight of the astronaut on the planet can be calculated as:
W = mg = 60[(4/3)Gπρᴇ] = (80/3)Gπρᴇ
Substituting the value of G and ρᴇ, we get:
W = (80/3)([tex]6.67 * 10^-^1^1[/tex])([tex]5.5 * 10^3[/tex])π ≈ 1818.9 N
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A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wrapped around the drum of radius 1. 39 m exerts a force of 3. 34 N to the right on the cylinder. A rope wrapped around the core of radius 0. 48 m exerts a force of 8. 52 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinder about the rotation axis? Answer in units of N ∗ m
The magnitude of the net torque acting on the cylinder about the rotation axis is 8.73 Nm.
τ1 = F1 * r1
where F1 is the force on the drum, and r1 is the radius of the drum. Substituting the given values:
τ1 = 3.34 N * 1.39 m = 4.64 Nm
The torque due to the force on the core is given by:
τ2 = F2 * r2
where F2 is the force on the core, and r2 is the radius of the core. Substituting the given values:
τ2 = 8.52 N * 0.48 m = 4.09 Nm
The net torque is the sum of the two torques:
τnet = τ1 + τ2
τnet = 4.64 Nm + 4.09 Nm = 8.73 Nm
Magnitude is a term used in physics to describe the size or amount of a physical quantity, such as length, mass, force, or energy. It is often represented by a numerical value and a unit of measurement. In physics, the magnitude of a vector is the length of the vector, and it is a scalar quantity. For example, the magnitude of the force acting on an object is the amount of force, irrespective of its direction.
Magnitude is also used to describe the intensity of phenomena such as earthquakes, sound waves, and light waves. The Richter scale, for example, is a measure of the magnitude of an earthquake, while the decibel scale is used to measure the magnitude of sound waves. In summary, magnitude is an important concept in physics that allows us to quantify the size or amount of physical quantities and phenomena.
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halfway through her swing, when the golf club is parallel to the ground, does the acceleration vector of the club head point (a) straight down, (b) parallel to the ground, approximately toward the golfer's shoulders, (c) approximately toward the golfer's feet, or (d) toward a point above the golfer's head? explain.
The acceleration vector of the club head point straight down(a)
Acceleration is a vector, and it may have a direction other than the acceleration of gravity.
The club head accelerates due to the forces exerted by the golfer, which include a torque applied by the golfer's hands, which generates angular acceleration that causes the club to rotate.
The acceleration vector points down because the club head is still at the beginning of the downswing.
As the club head approaches the ball, it begins to move away from its initial position and pick up velocity, resulting in a more complicated acceleration profile.
To summarise, during the downswing, halfway through the swing, the acceleration vector of the golf club head points straight down.
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During a tug of war game, both teams apply a force of 75 N. What is the net force?
The net force is 0 N. This is because the forces from each team are equal but opposite in direction, so the net force is the difference between the two forces, which is 0 N.
What is force ?Force is an interaction between two or more objects that results in a change in the motion of one or more of the objects. It is a vector quantity, meaning it has both magnitude and direction. Force can be divided into two categories: contact forces, which occur when two objects physically touch one another, and non-contact forces, which occur when objects interact without direct physical contact. Examples of contact forces include friction, tension, normal force, and air resistance. Examples of non-contact forces include gravitational force, electric force, and magnetic force.
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4. You throw a m = 128 g ball up into the air at an initial velocity v₁ = 4.8 m/s. How high, h, does the ball reach before slowing down to vi = 2.4 m/s?
The ball will reach a height of 1.17 meters before slowing down to Vi = 2.4 m/s
Kinematic motion problemFirst, we can use the equation for the final velocity of an object under constant acceleration:
v_f^2 = v_i^2 + 2ah
where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, h is the height reached by the ball, and we assume that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts downwards).When the ball is thrown upwards, it experiences a constant downward acceleration due to gravity, so we can take a = -9.8 m/s^2.
At the maximum height, the final velocity of the ball will be zero, so we can set v_f = 0 and solve for h:
0 = (4.8 m/s)^2 + 2(-9.8 m/s^2)h
h = [(4.8 m/s)^2]/[2(9.8 m/s^2)] = 1.17 m
So the ball reaches a maximum height of 1.17 meters.
Next, we can use the same equation with v_i = 2.4 m/s and v_f = 0 to find the height where the ball reaches a velocity of 2.4 m/s on its way down:
0 = (2.4 m/s)^2 + 2(-9.8 m/s^2)h
h = [(2.4 m/s)^2]/[2(9.8 m/s^2)] = 0.29 m
Therefore, the ball reaches a maximum height of 1.17 meters and then falls to a height of 0.29 meters before reaching a velocity of 2.4 m/s.
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algol is a binary system consist of a 3.7 msun main-sequence star and a 0.8 msun subgiant (a sub giant is just a small red giant). why is this surprising? 1 the two stars in a binary system should both be at the same stage of their lives. 2 a star with a mass of 3.7 msun is too big to be a main-sequence star. 3. it doesn't make sense to find a subgiant in a binary star system. 4 in a binary system, the more massive star should be more evolved than its companion.
The correct answer is 4. In a binary system, the more massive star should be more evolved than its companion.
In the case of Algol, the 0.8 m sun subgiant is more evolved than the 3.7 m sun main-sequence star, which is surprising and not what we would expect based on our understanding of stellar evolution. This discrepancy is believed to be due to the fact that the two stars did not form together as a binary system, but rather the less massive star was captured by the more massive star after it had already evolved off the main sequence. The more massive star evolves more quickly and will reach later stages of evolution (such as a red giant phase) before the less massive star
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explain how a sand blaster can be used to clean surfaces such as concrete even though every grain of sand is very small and light
The combination of high velocity, kinetic energy, hardness, and irregularity of the sand particles make sand blasting an effective method for cleaning and preparing surfaces, including concrete.
How does the collective force of thousands of small and light grains of sand hitting a surface at high velocity create a powerful abrasive action that effectively removes dirt, grime, paint, rust, and other unwanted materials from the surface when using a sand blaster?
A sand blaster uses compressed air to force a stream of abrasive particles, such as sand, onto a surface to clean or prepare it for painting or other treatments.
Although each grain of sand is small and light, the collective force of thousands of grains of sand hitting the surface at high velocity creates a powerful abrasive action that can effectively remove dirt, grime, paint, rust, and other unwanted materials from the surface.
The sand particles have kinetic energy due to their high velocity, which allows them to physically abrade and remove the unwanted materials from the surface. Additionally, the sand particles are typically harder than the surface being cleaned, which allows them to scratch away the surface layer of the material and expose a fresh, clean layer underneath. This process is known as mechanical abrasion.
Moreover, the sand particles used in sand blasting are often of irregular shapes and sizes, which allows them to reach crevices and corners that may be difficult to clean with other methods.
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2. A disc speeds up from rest at a constant rate of 2. 5 rad/s^2.
a. What is the final angular speed of the disc after 18 secs. ? (45 rad/s)
b. How many revolutions will the disc go through during this time period?
(64. 5 revs)
C. WHat is the linear speed of a point at the end of the disc after 18 secs. (The radius
of the disc is 50 cm. (22. 5 m/s)
I
a. The final angular speed of the disc after 18 seconds is 45 rad/s.
b. The disc goes through approximately 64.5 revolutions during this time period.
c. The linear speed of a point at the end of the disc after 18 seconds is 22.5 m/s.
a. The angular acceleration of the disc is constant at 2.5 [tex]rad/s^2[/tex].
Using the formula for angular velocity with constant angular acceleration, we have:
ωf = ωi + αt
Where:
ωi = initial angular velocity (0, since the disc starts from rest)
α = angular acceleration (2.5 [tex]rad/s^2[/tex])
t = time (18 s)
ωf = 0 + 2.5 × 18
ωf = 45 rad/s
Therefore, the final angular speed of the disc after 18 seconds is 45 rad/s.
b. To find the number of revolutions the disc goes through during this time period, we need to first find the total angle rotated by the disc. Using the formula for angular displacement with constant angular acceleration, we have:
θ = ωit + 1/2 α[tex]t^2[/tex]
Where:
ωi = initial angular velocity (0, since the disc starts from rest)
α = angular acceleration (2.5 [tex]rad/s^2[/tex])
t = time (18 s)
θ = 0 + 1/2 × 2.5 × [tex]18^2[/tex]
θ = 405 rad
One revolution is equal to 2π radians,
so the number of revolutions is given by:
N = θ / 2π
N = 405 / 2π
N ≈ 64.5 revolutions
Therefore, the disc goes through approximately 64.5 revolutions during this time period.
c. The linear speed of a point at the end of the disc is given by:
v = rω
Where:
r = radius of the disc (50 cm = 0.5 m)
ω = angular velocity (45 rad/s)
v = 0.5 × 45
v = 22.5 m/s
Therefore, the linear speed of a point at the end of the disc after 18 seconds is 22.5 m/s.
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if the sun were to collapse into a black hole, the point of no return for an investigator would be approximately 3 km from the center singularity. would the investigator be able to survive visiting even 300 km from the center? answer this by finding the difference in the gravitational attraction the black holes exerts on a 1.0-kg mass at the head and at the feet of the investigator.
No, the investigator would not be able to survive visiting 300 km from the center of the collapsed sun. According to Newton's Law of Gravitation, the gravitational force experienced by a 1.0-kg mass is given by the equation F = GMm by r2, where G is the gravitational constant, M is the mass of the black hole, m is the mass of the investigator, and r is the distance between them.
If the investigator is at a distance of 3 km from the center, the force experienced by the 1.0-kg mass at their head is GMm by 32. If the investigator is at a distance of 300 km from the center, the force experienced by the 1.0-kg mass at their feet is GMm by 3002, which is much greater than the force experienced at the head. Thus, the difference in gravitational attraction experienced by the investigator is too great for them to survive at a distance of 300 km from the center.
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What is physical significance of dimension of physical quantity
The dimension of a physical quantity is the power to which the fundamental units must be raised to, to represent it. Mass, length, time, temperature, electric current, luminous intensity and amount of substance are the fundamental quantities
Temperature is a measure of the degree of hotness or coldness of an object or substance. It is a fundamental concept in physics, chemistry, and engineering and is one of the most commonly measured physical parameters in scientific and industrial applications. The temperature of an object or substance is related to the average kinetic energy of its particles. In other words, the faster the particles are moving, the higher the temperature.
Temperature is measured using a variety of devices, including thermometers, thermocouples, and pyrometers. The most common unit of temperature is the Celsius (°C) scale, which is based on the freezing and boiling points of water. Another commonly used scale is the Fahrenheit (°F) scale, which is used primarily in the United States.
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Krystal throws a rappelling rope at a speed of 10 m/s down a 50 m cliff. when will the rope hit the ground? use the drop-down to put the correct order to solve for when the rope will hit the ground. question 2 options: the rope will hit the ground in about 4.37 s.
The rope will hit the ground in about 4.37 seconds.
With the help of the kinematic equation, we can resolve this issue:
[tex]d = vit + 1/2at^2[/tex]
where:
d = 50 m (distance or height of the cliff)
vi = 10 m/s (initial velocity)
a = [tex]9.8 m/s^2[/tex] (acceleration due to gravity, pointing downwards)
t = ? (time to hit the ground)
Solving for t:
[tex]50 = 10t + 1/2(9.8)t^2[/tex]
[tex]0 = 4.9t^2 + 10t - 50[/tex]
[tex]t = (-10+ \sqrt{(10^2 - 4(4.9)(-50))) / (2(4.9)} )[/tex]
[tex]t = 4.37 s[/tex]
Therefore, the rope will hit the ground in about 4.37 seconds.
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a hoop with a moment of inertia of 0.1 kg m2 spins about a frictionless axle with an angular velocity of 5 rad/s. at what radius from the center of the hoop should a force of 2 n be applied for 3 seconds in order to accelerate the hoop to angular speed of 10 rad/s?
Tthe radius from the center of the hoop at which a force of 2 N should be applied for 3 seconds in order to accelerate the hoop to angular speed of 10 rad/s is 0.25 m.
Given data:
Initial angular velocity ω1 = 5 rad/s
Final angular velocity ω2 = 10 rad/s
Force applied F = 2 N
Time taken t = 3 seconds
Moment of inertia I = 0.1 kg m²
Formula used: τ = Iα Where,τ = torque
I = moment of inertiaα = angular acceleration
Let's calculate the torque required to change the angular speed: Initial torque τ1 = Iω1
Final torque τ2 = Iω2
Change in torque Δτ = τ2 - τ1 = I(ω2 - ω1)
We know that τ = F × r where,
F is the force applied
r is the distance from the center of mass where the force is applied
Let's calculate the distance r: Δτ = τ = F × rr = Δτ / F = I(ω2 - ω1) / F
Substitute the values, r = 0.1 × (10 - 5) / 2 = 0.25 m
Therefore, the radius from the center of the hoop at which a force of 2 N should be applied for 3 seconds in order to accelerate the hoop to angular speed of 10 rad/s is 0.25 m.
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Drag each tile to the correct box.
Arrange the images in order to show how lake-effect snow occurs.
The images below demonstrate how lake effect snow occurs:
Cold Arctic air is passing over warm lake waters.
The surface of the lake adds heat and water vapour to the cold air mass.
Water vapour condenses in rising air to form clouds.
Snow blankets the lake and the downwind shore.
Where does lake effect snow most commonly occur?Lake effect snow most commonly occurs in the Great Lakes region of North America, which includes parts of the United States and Canada. The Great Lakes generate significant amounts of moisture, and when cold air passes over the relatively warm water, it picks up moisture and heat, leading to the formation of clouds and precipitation. As the air moves over land, it can release large amounts of snowfall in areas downwind of the lakes. This phenomenon is most common during the winter months and can produce heavy snowfalls that can impact transportation, infrastructure, and daily life in affected areas.
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