The quantum of work done to move a 1 kg mass from the face of the earth to a point 10 ⁵ km from the centre of the earth is4.92 x 10 ⁸J.
The quantum of work done to move 1 kg mass from the face of the earth to a point 10 ⁵ km from the centre of the earth can be calculated using the gravitational implicit energy formula.
The gravitational implicit energy is the quantum of work done by an external force in bringing an object from perpetuity to a point in space where it can be told by graveness. When an object is moved from the face of the earth to a point 10 ⁵ km from the centre of the earth, the gravitational implicit energy of the object increases.
The formula for gravitational implicit energy is given by U = - GMm/ r where U is the gravitational implicit energy G is the universal gravitational constant M is the mass of the earth m is the mass of the object r is the distance between the object and the centre of the earth.
We know that the mass of the object is 1 kg, the mass of the earth is and the distance from the centre of the earth to a point 10 ⁵ km down is Plugging these values into the formula, we get thus, the quantum of work done to move a 1 kg mass from the face of the earth to a point 10 ⁵ km from the centre of the earth is 4.92 x 10 ⁸J.
the mass of the earth is [tex]5.97 * 10^2^4 kg[/tex],
and the distance from the centre of the earth to a point 10⁵ km away is:
[tex]= 6.38 * 10^6 + 10^5 km[/tex]
[tex]= 6.48 * 10^6 km[/tex]
[tex]= 6.48 * 10^9 m[/tex].
Plugging these values into the formula, we get
[tex]U = -6.67 * 10^-^1^1 * 5.97 * 10^2^4 * 1 / 6.48 * 10^9[/tex]
[tex]= -4.92 * 10^8 J[/tex]
Therefore, the amount of work done to move a 1 kg mass from the surface of the earth to a point 10⁵ km from the centre of the earth is 4.92 x 10⁸ J.
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i
need help
i attempted it but i dont thunk i did it right
Jason is on the cliff and dropped a rock, if the rock takes 10.1 seconds to reach the ground, how high is the cliff? Express your answer in meters (m)
The height of the cliff is approximately 495 meters. This was determined by using the equation of motion for free fall and the time it took for the rock to reach the ground.
To determine the height of the cliff, we can use the equation of motion for free fall:
[tex]h = \frac{1}{2} g t^2[/tex]
Where:
h is the height of the cliff
g is the acceleration due to gravity (approximately 9.8 m/s²)
t is the time it takes for the rock to reach the ground
Given:
t = 10.1 seconds
Substituting the values into the equation:
[tex]h = \frac{1}{2} \cdot 9.8 \,\text{m/s}^2 \cdot (10.1 \,\text{s})^2[/tex]
Calculating the expression:
h ≈ 494.99 meters
Therefore, the height of the cliff is approximately 494.99 meters.
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What angular acceleration would you expect from a rotating object with rotational inertia of 0.0655 kg.m² that was subjected to a net torque of 4.25 N·m? 2) If gravity is used to produce a torque applied to a rotating platform, do you expect the angular acceleration to be constant? Why or why not? 3) Would frictional errors affect this lab more or less if you had used a rotating system with a much larger rotational inertia? Explain your answer?
1) The formula for torque is given by τ = Iα where τ is the torque, I is the rotational inertia and α is the angular acceleration. The angular acceleration of a rotating object with a rotational inertia of 0.0655 kg.m² and subjected to a net torque of 4.25 N.
m is given by
α = τ/I
= 4.25/0.0655
= 64.885 m/s²2) No, the angular acceleration produced by gravity is not constant because the force acting on the rotating platform is not constant. As the platform rotates, the direction of the force due to gravity changes with the position of the platform. Therefore, the torque produced by gravity is not constant and hence the angular acceleration is not constant.3) If a rotating system with much larger rotational inertia is used, frictional errors will affect the lab less. This is because the larger the rotational inertia of a system, the less it is affected by external forces such as friction. This means that if the system has a larger rotational inertia, it will be less affected by frictional errors compared to a system with a smaller rotational inertia.
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the cable supports the three loads shown. determine the sags yb and yd of b and d. take p1 = 800 n, p2 = 500 n.
The cable supports the three loads shown, the sag of point B in the cable is 0.044 m and the sag of point D in the cable is 0.075 m.
For determining this, we can use the formula:
[tex]y=\frac{wL^2}{8T} \\\\[/tex]
At point A, the tension in the cable is:
[tex]T_A=P_1+P_2[/tex]
[tex]T_A=[/tex] 800N + 500N
=1300N
At point B:
[tex]T_B=T_A+P_2[/tex]
[tex]T_B=[/tex] 1300N + 500N
= 1800N
At point C:
[tex]T_C=T_B+P_3[/tex][tex]T_C=[/tex] 1800N + 600N
= 2400N.
Now,
[tex]y_B=\frac{wL^2_{AB}}{8T_B} \\\\y_D=\frac{wL^2_{CD}}{8T_D}[/tex]
Substituting the values:
[tex]y_B=\frac{(1)(4)^2}{8(1800)} =0.044m\\\\y_D=frac{(1)(6)^2}{8(2400)} =0.075m[/tex]
Thus, the sag of point B in the cable is 0.044 m and the sag of point D in the cable is 0.075 m.
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Your question seems incomplete, the probable complete question is:
the cable supports the three loads shown. determine the sags yb and yd of b and d. take p1 = 800 n, p2 = 500 n.
1) You drop a 2 kg brick and a 3 kg brick off of a bridge. They
fall to the bottom of the bridge, which brick has more kinetic
energy when it hits the bottom?
2) Which brick did gravity do more work o
1) The 3 kg brick has more mass, so it will have more kinetic energy.
2) Gravity does not do more work on one brick compared to the other.
1) The 3 kg brick has more kinetic energy than the 2 kg brick when it hits the bottom of the bridge because kinetic energy is directly proportional to mass. The formula for kinetic energy is KE=1/2mv², where m is the mass of the object and v is its velocity. Since both bricks are dropped from the same height and experience the same acceleration due to gravity, they will have the same velocity when they hit the bottom. However, the 3 kg brick has more mass, so it will have more kinetic energy.
2) Gravity does the same amount of work on both bricks because they both fall the same distance and experience the same force of gravity. Work is defined as force times distance, so in this case, the force of gravity is the same for both bricks and the distance they fall is also the same. Therefore, gravity does not do more work on one brick compared to the other.
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when an object is placed 32.5 cm in front of a convex spherical mirror, a virtual image forms 11.5 cm behind the mirror. determine the mirror's focal length in cm and the magnification.
The absolute value of the magnification (0.354) tells us that the image is 0.354 times the size of the object (reduced in size).
When an object is placed 32.5 cm in front of a convex spherical mirror, a virtual image forms 11.5 cm behind the mirror. The magnification and the mirror's focal length can be determined using the following formula:
1/f = 1/do + 1/di, Where, f = focal length, do = object distance, and di = image distance
Given: do = -32.5 cm (negative sign indicates object is placed in front of the mirror)di = -11.5 cm (negative sign indicates the image is virtual)Using the above formula:
1/f = 1/-32.5 + 1/-11.51/f = -0.0308f = -32.45 cm (the negative sign indicates that it is a convex mirror, which has a negative focal length)
Therefore, the mirror's focal length is 32.45 cm. The magnification can be determined using the formula:m = -di/do Where, m = magnification, do = object distance, and di = image distance
Given:do = -32.5 cmdi = -11.5 cm
Using the above formula:
m = -(-11.5)/(-32.5)m = 0.354If the magnification is positive, the image is upright, and if it is negative, the image is inverted. In this case, the magnification is negative, which means the image is inverted.
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the work function of a certain metal is ϕϕ = 3.65 ev. determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (planck's constant is: h = 4.1357×10-15 evs.)
The work function of a certain metal is ϕ = 3.65 eV. Planck's constant is h = 4.1357 × 10⁻¹⁵ eVs and we have to determine the minimum frequency of light f₀ for which photoelectrons are emitted from the metal.
What is work function?Work function is a measure of the energy required to remove an electron from a metal surface.What is photoelectric effect?Photoelectric effect is the process by which electrons are emitted from a metal surface when light falls on it.What is Planck's constant?Planck's constant is a physical constant that relates the energy of a photon to its frequency. The value of Planck's constant is
h = 6.626 × 10⁻³⁴ J s.
The minimum frequency of light required to remove an electron from the surface of a metal is given by the equation:
f₀ = (ϕ / h)
Where, f₀ is the minimum frequency of light, ϕ is the work function of the metal and h is Planck's constant.
f₀ = (3.65 / 4.1357 × 10⁻¹⁵)
= 8.82 × 10¹⁴ Hz
Therefore, the minimum frequency of light required to remove an electron from the surface of the metal is 8.82 × 10¹⁴ Hz.
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A penny has a mass of 2.50g and the Moon has a mass of ×7.351022kg . Use this information to answer the questions below. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of pennies? g Round your answer to 3 significant digits. How many moles of pennies have a mass equal to the mass of the Moon? Round your answer to 3 significant digits.
Therefore, there are 48.84 moles of pennies that have a mass equivalent to that of the Moon (rounded to 3 significant digits).
A mole of anything, whether it is pennies or anything else, is equivalent to Avogadro's number of atoms, molecules, or particles. Avogadro's number is given by 6.022 × 10²³.Using the mass of a single penny, we can calculate the mass of one mole of pennies by dividing the molar mass by Avogadro's number.
The molar mass is the mass of one mole of a substance, and it is equivalent to the atomic or molecular weight of a substance expressed in grams. It is the sum of all the atomic masses of an element's atoms. To begin, we must first convert the mass of a single penny from grams to kilograms: 2.50 g = 0.0025 kg .
The mass of one mole of pennies can now be calculated as follows: Molar mass = 0.0025 kg/mol = 0.0025 × 6.022 × 10²³= 15.055 × 10²⁰ g/mol or 1.506 × 10²¹ g/mol (rounded to 3 significant digits)Therefore, the mass of 1 mole of pennies is 1.506 × 10²¹ g/mol (rounded to 3 significant digits) .
To determine the number of moles of pennies required to equal the mass of the Moon, we will first convert the mass of the Moon from kilograms to grams.7.351 × 10²² g We'll then divide this mass by the mass of one mole of pennies:7.351 × 10²²g ÷ 15.055 × 10²⁰ g/mol= 48.84 moles of pennies . Therefore, there are 48.84 moles of pennies that have a mass equivalent to that of the Moon (rounded to 3 significant digits).
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In which one of the following situations does a car have an eastward acceleration? The car travels westward and slows down (a) and (d) The car travels westward at constant speed The car travels eastwa
The car has an eastward acceleration when it travels westward and slows down.
When a car travels westward at a constant speed, its velocity is directed to the west and remains constant. Acceleration is the rate of change of velocity, and if the car's velocity remains constant, then the acceleration is zero. Therefore, there is no eastward acceleration in this situation.
When the car travels eastward and speeds up, its velocity is directed to the east, and the rate of change of velocity is positive. This means that the car has an eastward acceleration.
When the car travels westward and slows down, its velocity is still directed to the west, but the rate of change of velocity is negative. In this case, the car experiences a westward deceleration, but there is no eastward acceleration.
Finally, when the car travels eastward and slows down, its velocity is directed to the east, but the rate of change of velocity is negative. This means that the car experiences a westward deceleration, and there is no eastward acceleration.
In summary, the car has an eastward acceleration when it travels eastward and speeds up.
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Complete Question:
In which one of the following situations does the car have a eastward acceleration. The car travels westward at constant speed. The car travels eastward and speeds up. The car travels westward and slows down The car travels eastward and slows down.
the wavelength of an electromagnetic wave is measured to be 0.067 m. (a) what is the frequency of the wave?
The frequency of the electromagnetic wave is 4.48 × 10⁹ Hz
The wavelength of an electromagnetic wave is measured to be 0.067 m. Therefore, we have to determine the frequency of the wave
The speed of light is constant in a vacuum, and it is represented by c.
The speed of light in a vacuum is 2.998 × 10⁸ m/s.
According to the formula for electromagnetic waves: v = fλwhere:v = the speed of lightf = frequencyλ = wavelength
Given that the wavelength of the electromagnetic wave is 0.067m, we can determine its frequency using the above formula.v = fλ⟹f = v/λ
Substitute the values into the above formula :f = 2.998 × 10⁸/0.067m = 4.48 × 10⁹ Hz
Therefore, the frequency of the wave is 4.48 × 10⁹ Hz.
In conclusion, the frequency of the electromagnetic wave is 4.48 × 10⁹ Hz
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what is the focal length of 2.0 d reading glasses found on the rack in a pharmacy
The focal length of 2.0 d reading glasses found on the rack in a pharmacy is 0.5 meters. Reading glasses are convex lenses that magnify objects up close, allowing those with presbyopia to read and perform close-up tasks.
The focal length of 2.0 d reading glasses found on the rack in a pharmacy is 0.5 meters. Focal length refers to the distance between the center of a lens and its focus.
Reading glasses are convex lenses that magnify objects up close, allowing those with presbyopia to read and perform close-up tasks. A lens that is 2.0 diopters has a power of +2.0. The formula for calculating the focal length of a lens is f = 1/d where f is the focal length and d is the power of the lens in diopters. Therefore, the focal length of 2.0 d reading glasses is f = 1/2 = 0.5 meters.
The focal length of 2.0 d reading glasses found on the rack in a pharmacy is 0.5 meters. Reading glasses are convex lenses that magnify objects up close, allowing those with presbyopia to read and perform close-up tasks.
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to calculate the radiation pressure on a highly polished metal surface, it would be best to use which approximation?
To calculate the radiation pressure on a highly polished metal surface, the most appropriate approximation to use is the small-angle approximation.
Radiation pressure refers to the pressure produced when electromagnetic radiation is absorbed or reflected by a surface. A highly polished metal surface is highly reflective, and therefore is expected to produce high radiation pressure.
The small-angle approximation is the assumption that the angle of incidence is small enough such that the sine of the angle is equal to the angle itself. This approximation is particularly useful in situations where the angle of incidence is small relative to 1 radian or less. This approximation can be used to calculate radiation pressure on highly polished metal surfaces because the angle of incidence is usually small (typically less than 1 radian), and therefore can be approximated using the small-angle approximation.
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how long (in nsns ) does it take light to travel 1.00 mm in vacuum?
The time taken by the light to travel 1.00 mm in vacuum is 3.33 × 10⁻⁹ nsns.
Light is an electromagnetic wave, which is a transverse wave that does not need a medium to travel through. In vacuum, light travels at a constant speed of 2.99792458 × 10⁸ m/s. It implies that if light travels for one second in vacuum, it will cover a distance of approximately 299,792,458 meters, that is 299,792,458,000,000 nanometers.
Therefore, the time taken by the light to travel 1.00 mm (1 × 10⁻³ m) in vacuum is;
Time = distance/speed of light in vacuum
= 1.00 × 10⁻³ m / 2.99792458 × 10⁸ m/s
= 3.33 × 10⁻⁹ s
= 3.33 × 10⁻⁹ nsns.
This calculation is done by dividing the distance light has to travel by its speed in vacuum.
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determine whether the vector field f(x,y) = (yex sin(y),ex xcos(y)) is conservative and, if it is, find a potential.
The vector field F(x, y) = (yex sin(y), ex xcos(y)) is not conservative,we calculate that after checking its components satisfy the condition of conservative vector fields.
conservative vector fields:
∂F/∂y = ∂(yex sin(y))/∂y = ex sin(y) + yex cos(y)
∂F/∂x = ∂(ex xcos(y))/∂x = ex cos(y)
Now, we need to check if ∂F/∂y = ∂F/∂x:
ex sin(y) + yex cos(y) = ex cos(y)
Since the two components of the vector field do not match, we conclude that the vector field F(x, y) is not conservative.
Therefore, there is no potential function associated with this vector field.
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Find the gauge pressure in the jar, in units of Pa. Assume the density of mercury is 13.5 g/cm³.
a. Gauge pressure
b. 13.5 g/cm³
The gauge pressure in the jar is -89945.95 Pa (approximated to two significant figures).
We can use the following formula to find the pressure, `p=hρg`,
where,
h is the height of the mercury column,ρ is the density of mercury, and
g is the acceleration due to gravity.
The pressure at the bottom of the jar is equal to the pressure due to the mercury column and the atmospheric pressure.`pabs = hρg + patm`
Substituting the values,
`pabs = 7.10 × 13.5 × 9.8 + 1.01 × 10⁵` = 10,754.05 Pa
Now, we can calculate the gauge pressure by using the formula;
`pgauge = pabs - patm``
pgauge = 10,754.05 - 1.01 × 10⁵``
pgauge = -89945.95 Pa`
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The gauge pressure in the jar is 19822.5 Pa. Given that density of mercury is 13.5 g/cm³. Gauge pressure in the jar is given by: P_gauge = hρg
Let's first find the absolute pressure in the jar. Pressure due to air = 1 atm. Pressure due to mercury = hρgThe total pressure P in the jar is the sum of the two pressures: P = P_air + P_mercuryP = 1 atm + hρg. Since gauge pressure is the difference between the absolute pressure and the atmospheric pressure, gauge pressure is: P_gauge = P - P_atmP_gauge = (1 atm + hρg) - 1 atmP_gauge = hρgwhere h is the height of mercury in the tube.
Using the given density of mercury, we can express it in kg/m³:ρ = 13.5 g/cm³ = 13500 kg/m³. Thus, gauge pressure in the jar is given by:P_gauge = hρg, Where, h = 15cm = 0.15m, ρ = 13500 kg/m³, g = 9.81 m/s². So,P_gauge = 0.15 m × 13500 kg/m³ × 9.81 m/s²= 19822.5 Pa. Hence, the gauge pressure in the jar is 19822.5 Pa.
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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of a gazelle assumes an acceleration of 4. 2 m/s2 for 6. 5 s , after which the gazelle continues at a steady speed
The gazelle travels 88.725 meters during the sprint. We use the following equation to calculate the distance traveled by the gazelle during the sprint: `d = vit + 0.5at²
A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of a gazelle assumes an acceleration of 4.2 m/s² for 6.5 s, after which the gazelle continues at a steady speed.
The gazelle's initial velocity is zero because it starts from rest. We can use the following equation to calculate the distance traveled by the gazelle during the sprint: `d = vit + 0.5at²`, where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time elapsed.
1. Substitute the given values into the equation.
`d = 0 + 0.5(4.2)(6.5)²`
2. Solve for d.
`d = 0 + 0.5(4.2)(42.25)`
`d = 88.725`
Therefore, the gazelle travels 88.725 meters during the sprint.
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the massless spring of a spring gun has a force constant k = 12n/cm
The spring potential energy (U) stored in the spring when compressed to 8 cm is 3.84 J.
The massless spring of a spring gun has a force constant k = 12 N/cm.
We need to determine the spring potential energy (U) stored in the spring when compressed to 8 cm.
The given variables are force constant k and displacement x of the massless spring.
Recall the formula for spring potential energy Spring potential energy (U) stored in the spring is given by:U = (1/2) k x²where:k is the force constant of the springx is the displacement of the spring from its equilibrium position
Substitute the given values in the formula
The displacement of the spring is 8 cm = 0.08 m
The force constant of the spring is k = 12 N/cm = 1200 N/m
Therefore, the spring potential energy (U) stored in the spring when compressed to 8 cm is:U = (1/2) k x²U = (1/2) × 1200 N/m × (0.08 m)²U = 3.84 J
Therefore, the spring potential energy (U) stored in the spring when compressed to 8 cm is 3.84 J.
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Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has a wavelength of 254 m? Energy = _____ kJ/photon
The energy of radio wave radiation with a wavelength of 254 meters is approximately 7.81 x 10^-26 kJ/photon.
To calculate the energy of a photon of radio wave radiation with a given wavelength, we can use the equation:
Energy = (Planck's constant × speed of light) / wavelength
Planck's constant (h) is approximately 6.626 x 10^-34 joule-seconds.
The speed of light (c) is approximately 3.00 x 10^8 meters per second.
The wavelength (λ) is given as 254 meters.
Plugging in the values into the equation:
Energy = (6.626 x 10^-34 J·s × 3.00 x 10^8 m/s) / 254 m
Calculating the value:
Energy ≈ 7.81 x 10^-23 Joules
To convert the energy from joules to kilojoules, we divide by 1000:
Energy = (7.81 x 10^-23 J) / 1000
Energy ≈ 7.81 x 10^-26 kilojoules
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Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. Energy = 4.9×10⁻²² kJ/photon. Radio waves are a type of electromagnetic radiation that travels through space at the speed of light.
Their wavelengths vary widely, ranging from 10⁻¹⁴ to 10⁴ meters. The energy of radio wave radiation with a wavelength of 254 m can be calculated using the formula: E = hc /λwhereE is energy, h is Planck's constant, which is 6.626 x 10^-34 joule-seconds, andλ is the wavelength of the radiation.
The speed of light, c, is 3 x 10⁸ meters per second. Substituting the values, we have: E = (6.626 × 10⁻³⁴ J·s) × (3 × 10⁸ m/s) / (254 m) = 7.82 × 10⁻²⁶ J/photon1 joule is equal to 1 x 10⁻³ kJ. Therefore, we can convert the energy of radio wave radiation to kJ/photon by dividing by 1000.7.82 × 10⁻²⁶ J/photon = 7.82 × 10⁻²⁹ kJ/photon
So, the energy of radio wave radiation with a wavelength of 254 m is 4.9 × 10⁻²² kJ/photon.
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The cholesterol content of large eggs of a particular brand is normally distributed with a mean of u = 195 mg and a standard deviation of o= 12 mg. Suppose we take a random sample of 50 eggs. What is
The sampling distribution of the mean of the cholesterol content of 50 eggs is normal with mean 195 and standard deviation 1.697 (rounded to 3 decimal places).
The central limit theorem states that the sampling distribution of the mean of a sufficiently large sample size from any population has a normal distribution, regardless of the population's distribution. As n >= 30, the sample size is large enough to use the central limit theorem in this case.The standard deviation of the sampling distribution, also known as the standard error of the mean (SEM), can be calculated using the formula: SEM = σ/√n, where σ is the population standard deviation and n is the sample size. Plugging in the given values, we get SEM = 12/√50 = 1.697 (rounded to 3 decimal places). Therefore, the sampling distribution of the mean of the cholesterol content of 50 eggs is normal with mean 195 and standard deviation 1.697 (rounded to 3 decimal places).
A measure of how dispersed the data are in relation to the mean is called the standard deviation (or ). Data with a low standard deviation are grouped around the mean, while data with a high standard deviation are more dispersed.
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A disk of radius 0.46 m and moment of inertia 2.1 kg·m2 is mounted on a nearly frictionless axle. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 34 N. What is the magnitude of the torque? After a short time the disk has reached an angular speed of 6 radians/s, rotating clockwise. What is the angular speed 0.85 seconds later? angular speed = ???? radians/s
The magnitude of the torque exerted on the disk when a constant force of 34 N is applied to the string wrapped around it is 15.64 N·m.
What is the magnitude of the torque exerted on the disk when a constant force of 34 N is applied to the string wrapped around it?The magnitude of the torque can be calculated using the formula: torque = force ˣ radius. Plugging in the given values, the torque is 34 N ˣ 0.46 m = 15.64 N·m.
What is the angular speed of the disk 0.85 seconds after reaching an initial angular speed of 6 radians/s?
The change in angular speed can be determined using the formula: change in angular speed = torque / moment of inertia ˣ time. Plugging in the values, the change in angular speed is (15.64 N·m) / (2.1 kg·m²) ˣ (0.85 s) = 0.118 rad/s.
To find the final angular speed, we add the change in angular speed to the initial angular speed: 6 rad/s + 0.118 rad/s = 6.118 rad/s.
Therefore, the angular speed of the disk 0.85 seconds later is 6.118 radians/s.
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4. Vo=50 m/s 10=53° des****** A projectile is fired at an angle 53° above the horizontal with 50 m/s initial velocity. a) Find its maximum height b) Find position and velocity 6 s later.
a) The maximum height reached by the projectile is approximately 67.35 meters.
b) 6 seconds later, the projectile will be at a horizontal position of approximately 155.33 meters and a vertical position of approximately 41.47 meters. The velocity at this time is approximately 19.98 m/s horizontally and -40.04 m/s vertically.
a) To find the maximum height reached by the projectile, we can use the kinematic equation for vertical motion. The formula to calculate the maximum height (h_max) is:
h_max = (V₀² * sin²θ) / (2 * g)
where V₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
Substituting the given values:
V₀ = 50 m/s
θ = 53° (converted to radians: 53° * π/180 ≈ 0.9273 rad)
g = 9.8 m/s²
h_max = (50² * sin²(0.9273)) / (2 * 9.8)
≈ 67.35 m
Therefore, the maximum height reached by the projectile is approximately 67.35 meters.
b) To find the position and velocity of the projectile 6 seconds later, we can analyze its horizontal and vertical motion separately.
For the horizontal motion, the projectile will continue to move at a constant velocity since there is no horizontal acceleration. Therefore, the horizontal position (x) will be:
x = V₀ * cosθ * t
Substituting the given values:
V₀ = 50 m/s
θ = 53° (converted to radians: 53° * π/180 ≈ 0.9273 rad)
t = 6 s
x = 50 * cos(0.9273) * 6
≈ 155.33 m
For the vertical motion, we can use the equation:
y = V₀ * sinθ * t - (1/2) * g * t²
Substituting the given values:
V₀ = 50 m/s
θ = 53° (converted to radians: 53° * π/180 ≈ 0.9273 rad)
g = 9.8 m/s²
t = 6 s
y = 50 * sin(0.9273) * 6 - (1/2) * 9.8 * 6²
≈ 41.47 m
Therefore, 6 seconds later, the projectile will be at a horizontal position of approximately 155.33 meters and a vertical position of approximately 41.47 meters.
The velocity at this time can be calculated by combining the horizontal and vertical components:
Vx = V₀ * cosθ
Vy = V₀ * sinθ - g * t
Substituting the given values:
V₀ = 50 m/s
θ = 53° (converted to radians: 53° * π/180 ≈ 0.9273 rad)
g = 9.8 m/s²
t = 6 s
Vx = 50 * cos(0.9273)
≈ 19.98 m/s
Vy = 50 * sin(0.9273) - 9.8 * 6
≈ -40.04 m/s
Therefore, the velocity of the projectile 6 seconds later is approximately 19.98 m/s in the horizontal direction and -40.04 m/s in the vertical direction.
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determine the ph of a solution prepared by adding 0.0500 mole of solid ammonium chloride to 100. millilitres of a 0.150 molar solution of ammonia. hint: use an ice box to help you solve this.
The pH of the solution is 8.79.
The pH of a solution prepared by adding 0.0500 mole of solid ammonium chloride to 100 millilitres of a 0.150 molar solution of ammonia can be determined as follows:
Step 1: Write the balanced equation for the dissociation of ammonium chloride (NH4Cl).NH4Cl (s) ⇌ NH4+ (aq) + Cl- (aq)
Step 2: Calculate the initial moles of NH3 (ammonia) in the solution.Initial moles of NH3 = 0.150 moles/L × 0.100 L = 0.015 moles
Step 3: Calculate the initial moles of NH4+ produced.Initial moles of NH4+ = 0.0500 moles
Step 4: Calculate the initial moles of Cl- produced.Initial moles of Cl- = 0.0500 moles
Step 5: Determine the change in moles of NH4+ and NH3 using the ICE table.Initially NH4+ (aq): 0.0500 mol (0 mol)NH3 (aq): 0.015 mol (0.015 mol)Change: -0.0500 mol + 0.0500 molEquilibrium: 0 mol (0.015 mol)Initially Cl- (aq): 0 mol (0 mol)Change: 0.0500 mol + 0 molEquilibrium: 0.0500 mol (0 mol)
Step 6: Calculate the concentration of NH3 and NH4+ at equilibrium.NH4+ (aq) = 0.0500 moles/0.100 L = 0.500 MNH3 (aq) = (0.015 moles + 0.0500 moles)/0.100 L = 0.650 M
Step 7: Calculate the value of Kb for ammonia.Kb for ammonia = Kw/Ka = 1.00 × 10-14/1.8 × 10-5 = 5.56 × 10-10
Step 8: Calculate the concentration of OH- at equilibrium.OH- (aq) = √(Kb [NH4+])/[NH3] = √(5.56 × 10-10 × 0.500)/0.650 = 6.21 × 10-6 M
Step 9: Calculate the pH of the solution.pOH = -log10(OH-) = -log10(6.21 × 10-6) = 5.207pH = 14.00 - pOH = 14.00 - 5.207 = 8.79
The pH of the solution prepared by adding 0.0500 mole of solid ammonium chloride to 100 millilitres of a 0.150 molar solution of ammonia was calculated using the ICE box method
. The initial concentration of NH3 was calculated to be 0.015 moles, and the initial concentration of NH4+ and Cl- was 0.0500 moles each.
After determining the equilibrium concentrations of NH3 and NH4+ ions, the value of Kb for ammonia was calculated to be 5.56 × 10-10.
The concentration of OH- at equilibrium was calculated to be 6.21 × 10-6 M, and the pH of the solution was determined to be 8.79.
In conclusion, the pH of the solution is 8.79.
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Listen Metallic chromium can be obtained from the mineral chromite (FeCr204). What is the mass percent of chromium in chromite? 46.46% 61.90% 23.23% 30.26% 41.99%
The mass percent of chromium in chromite is 46.46%.
How to find mass percent?Find the molar mass of Cr. It is 52 g/mol.
Find the molar mass of chromite. It is (52+2*56+4*16) g/mol. (FeCr2O4)
Find the mass of Cr in 1 mol of chromite. It is (52/120)*100%.
Calculate the mass percent of Cr in chromite using the below formula.
Mass percent of Cr = (mass of Cr/mass of chromite)×100%
Substitute the calculated values in the above formula.
Mass percent of Cr = (52/120) × 100% = 46.46%.
Hence, the correct option is 46.46%.
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Option (c), 46.46%. the mass percent of chromium in chromite is 30.26%.
This is the mass percent of chromium in chromite. Chromite, also known as FeCr2O4, is a mineral that contains both iron and chromium. To calculate the mass percent of chromium in chromite, we must first determine the molar mass of chromite. We can do this by adding up the molar masses of all the atoms in one formula unit of chromite:
Fe: 1 x 55.85 g/mol = 55.85 g/mol
Cr: 1 x 52.00 g/mol = 52.00 g/mol
O: 4 x 16.00 g/mol = 64.00 g/mol
Adding these together, we get a molar mass of 171.85 g/mol for chromite. Next, we need to determine the mass of chromium in one formula unit of chromite:
Cr: 1 x 52.00 g/mol = 52.00 g/mol
Finally, we can calculate the mass percent of chromium in chromite using the following formula:
mass percent of chromium = (mass of chromium / mass of chromite) x 100
mass percent of chromium = (52.00 g/mol / 171.85 g/mol) x 100
mass percent of chromium = 30.26%
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6. A jet plane is cruising at 340 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 3.4 km, the jet is moving with a speed of 400 m/s. What is the jet's acceleration,
The jet's acceleration is 6 m/s². This means that its velocity increases by 6 meters per second every second when the engines are at full throttle.
To find the jet's acceleration, we can use the equation:
acceleration = (final velocity - initial velocity) / time
First, let's convert the initial and final velocities to meters per second (m/s):
Initial velocity = 340 m/s
Final velocity = 400 m/s
Next, we need to calculate the time it took for the jet to increase its velocity from 340 m/s to 400 m/s. We can use the formula:
distance = velocity × time
Given that the jet traveled 3.4 km (or 3400 m) during this time, we can rearrange the formula to solve for time:
time = distance / velocity
time = 3400 m / 340 m/s
time = 10 seconds
Now we have all the values we need to calculate the acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (400 m/s - 340 m/s) / 10 s
acceleration = 60 m/s / 10 s
acceleration = 6 m/s²
The jet's acceleration is 6 m/s². This means that its velocity increases by 6 meters per second every second when the engines are at full throttle.
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A fast elevator starts from rest and is moving upward with a constant acceleration of a=4m/s. At 1-0 a bolt in the elevator ceiling h=3m above the elevator floor works loose and falls down.(a) How long does it take the bolt to reach the floor? (b)What is the velocity of the bolt relative to the elevator,as it hits the floor?cWhat is the velocity of'the bolt relative to the ground.as it hits the floor?d Relative to the ground,how far has the bolt traveled?
It takes the bolt (a) to reach the floor in 0.75 seconds. (b) The velocity of the bolt as it hits the floor, is 3 m/s downward. (c) The velocity of the bolt, is 3 m/s downward. (d) Relative to the ground, a distance of 1.125 meters.
Determine the time it takes for the bolt to reach the floor, we can use the equation of motion: h = (1/2)at², where h is the distance traveled, a is the acceleration, and t is the time. Plugging in the values, we find t = √(2h/a) = √(2(3 m)/(4 m/s²)) = 0.75 s.
The velocity of the bolt relative to the ground is the sum of the elevator's velocity (which is increasing at a constant rate) and the velocity of the bolt relative to the elevator.
Since the elevator starts from rest and has a constant acceleration, its velocity is given by v = at = 4 m/s² * 0.75 s = 3 m/s downward.
Therefore, the velocity of the bolt relative to the ground is also 3 m/s downward.
The distance traveled by the bolt relative to the ground, we can use the equation of motion: d = v₀t + (1/2)at², where v₀ is the initial velocity. Since the bolt starts from rest relative to the ground, v₀ = 0.
Plugging in the values, we find d = (1/2)at² = (1/2)(4 m/s²)(0.75 s)² = 1.125 meters.
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How long tg does it take for the balls to reach the ground? Use 10 m/s2 for the magnitude of the acceleration due to gravity. Express your answer in seconds to one significant figure. View Available Hint(s) Hint 1. How to approach the problem The balls are released from rest at a height of yo 5.0 m at time to 0 s Using these numbers and the kinematic equation y yo vot(1/2)at2 you can determine the amount of time it takes for the balls to reach the ground
The time taken for the ball to reach the ground is 1.4 seconds.
Using the given data and kinematic equation y = yo + vot (1/2)at² we have calculated the time taken by the ball to reach the ground. The initial velocity of the ball is zero. The initial height of the ball is 5.0m and using the given value of acceleration due to gravity g which is 10m/s², we can find out the time taken by the ball to reach the ground.
Using the given formula, y = yo + vot (1/2)at². Here, y = 0, yo = 5.0m, vo = 0, a = g = 10m/s²t = sqrt(2 * 5.0 / 10) = 1.4s. Therefore, the time taken for the ball to reach the ground is 1.4 seconds.
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The magnetic field at the center ofaolenoid L-55cm in length is »-046 T wheuanent of ,.65 Atm through the solenold wire. t Status re for vlew Status D A solenoid. The expression should be in terms of the given variables. 50% Part (a) Solve the formula fr the magnetic field near the center of a long, tightly wound solenoid for the number of turns.
A solenoid is a cylindrical coil of wire, usually made of copper or another electrically conductive material, used to produce a magnetic field when a current flows through it.
The formula for the magnetic field near the center of a long, tightly wound solenoid for the number of turns can be derived by using the formula of magnetic field on the axis of the solenoid given by;
B = μ0nI / L,
Given that
B = -0.046 T,
L = 55 cm = 0.55 m and
I = 0.65 A, and substituting the values in the formula above, we get;
-0.046 T = μ0n(0.65 A) / 0.55 mn(μ0 / 0.55)
= -0.046 T / (0.65 A) n
= 5000 / L.
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If the result of your calculation of a quantity has Si units of kg • m/(s2.C), that quantity could be Select one: A. an electric field strength. B. an electric potential difference. C. a dielectric constant. D. an electric potential energy. E. a capacitance
If the result of your calculation of a quantity has SI units of kg·m/(s²·C), that quantity could be an electric field strength. The electric field strength (E) is defined as the force per unit charge acting on an electric charge. Option (A) is correct.
It is a vector quantity with units of newtons per coulomb (N/C) or volts per meter (V/m). The formula to calculate electric field strength is given as E = F/q, where F is the force acting on the charge and q is the magnitude of the charge.The SI unit of force is the newton (N), and the SI unit of charge is the coulomb (C). Therefore, the units of electric field strength can be written as N/C or V/m. The given SI units of kg·m/(s²·C) can be rearranged to N/C. This confirms that the quantity being calculated is electric field strength.Other options such as electric potential difference, dielectric constant, electric potential energy, and capacitance have different SI units. Electric potential difference has SI units of volts (V), dielectric constant has no units, electric potential energy has SI units of joules (J), and capacitance has SI units of farads (F). Therefore, the answer to this question is option A.
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2. Suppose you set up a circuit with an AC power supply whose PEAK voltage is 8.00V. Suppose that the frequency is 30Hz. Based on your experience in lab, what will a voltmeter read if it is set on DC?
If an AC power supply with a peak voltage of 8.00V and a frequency of 30Hz is measured using a DC voltmeter, it will read approximately 5.65V, which is the average value of the AC waveform. The reading does not represent the peak or instantaneous values.
If an AC power supply with a peak voltage of 8.00V and a frequency of 30Hz is connected to a voltmeter set on DC, the voltmeter will read the average or RMS (root mean square) value of the AC voltage.
In this case, the voltmeter will read approximately 5.65V.
When an AC waveform is measured using a DC voltmeter, the meter will display the average value of the waveform.
The average value of an AC waveform is related to its peak value by a factor known as the form factor. For a sinusoidal waveform like the one described, the form factor is approximately 0.707.
To calculate the average value, we multiply the peak voltage by the form factor. In this case, 8.00V * 0.707 = 5.65V.
Therefore, if a voltmeter set on DC is used to measure the AC voltage with a peak value of 8.00V and a frequency of 30Hz, it will display an approximate reading of 5.65V.
It's important to note that the reading will only represent the average value of the AC waveform, and not the peak or instantaneous values.
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the magnetic field inside a 5.0-cm-diameter solenoid is 2.0 t and decreasing at 3.10 t/s.
solution shows that the magnetic flux through the solenoid is 3.926×10⁻³ Wb initially. As the magnetic field is decreasing at a rate of 3.10 T/s, the magnetic flux will decrease accordingly.
The induced emf is given by ɛ = -A(dB/dt) = -(1.963×10⁻³ m²)(-3.10 T/s)
= 6.08×10⁻⁶ V.
The question gives the magnetic field inside a solenoid with a diameter of 5.0 cm. The magnetic field is 2.0 T and decreasing at a rate of 3.10 T/s.A solenoid is a long wire wound into a coil. It is capable of creating a magnetic field inside it when a current is passed through it. The magnetic field strength is proportional to the number of turns in the solenoid per unit length, current flowing through it and the magnetic permeability of the medium.
The magnetic flux through the solenoid is given by φ = BA, where B is the magnetic field, and A is the cross-sectional area of the solenoid. The area of a circular cross-section is A = πr².
Therefore, A = π(5.0 cm/2)²
= 19.63 cm²
= 1.963×10⁻³ m²The initial magnetic flux through the solenoid is φ = (2.0 T)(1.963×10⁻³ m²) = 3.926×10⁻³ Wb
After time t, the magnetic flux through the solenoid will be
φ = (2.0 T - 3.10 T/s×t)(1.963×10⁻³ m²)The rate of change of magnetic flux is given by Faraday's law of electromagnetic induction as: ɛ = -dφ/dt
The negative sign indicates that the induced emf opposes the change in magnetic flux. The induced emf is given by ɛ = -A(dB/dt)Where A is the area of the solenoid, and dB/dt is the rate of change of the magnetic field inside the solenoid.
solution shows that the magnetic flux through the solenoid is 3.926×10⁻³ Wb initially. As the magnetic field is decreasing at a rate of 3.10 T/s, the magnetic flux will decrease accordingly.
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A ball with an initial velocity of 8.4 m/s rolls up a hill without slipping.
a) Treating the ball as a spherical shell, calculate the vertical height it reaches, in meters.
b) Repeat the calculation for the same ball if it slides up the hill without rolling.
a)Treating the ball as a spherical shell, the vertical height it reaches is 36.43 meters.
b) The vertical height it reaches is 8.68 times the distance traveled by the ball up the hill.
a) Assuming that the ball is a spherical shell and using the formula for potential energy and kinetic energy, we get:Initial Kinetic Energy (Ki) = 1/2 mu²
Potential Energy at maximum height (P) = mgh
Final Kinetic Energy (Kf) = 0
Total Mechanical Energy (E) = Ki + P = Kf
Applying this principle, we get:
mgh + 1/2 mu² = 0 + 1/2 mv² ⇒ gh + 1/2 u² = 1/2 v²
At the maximum height, the velocity of the ball will become zero (v = 0) and we can calculate the value of h using the above equation:
gh + 1/2 u² = 0h = u² / 2g = (8.4)² / 2 × 9.8 = 36.43 m
Therefore, the vertical height it reaches is 36.43 meters.
b)The formula can be represented as:
F × s = mgh - 1/2 mu²
Substituting the values, we get:
F × s = mgh - 1/2 mu²
F × s = mg(h - 1/2 u² / mg)
The maximum vertical height (h) can be calculated as:h = s + 1/2 u² / g + μk × s
The first two terms in the above equation represent the maximum height the ball can reach due to its initial velocity while the third term represents the extra height the ball can reach due to the frictional force acting on it.
h = s + 1/2 u² / g + μk × s = s + (8.4)² / 2 × 9.8 + 0.392s = 8.68s
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Treating the ball as a spherical shell, its maximum vertical height is 1.31 meters.
a) Treating the ball as a spherical shell, the vertical height it reaches can be calculated using the following equation:
mg = (2/5)Mv²
where,
m = 1.8 kg (mass of ball)
g = 9.8 m/s² (acceleration due to gravity)
h = ? (maximum vertical height)
M = 2/3mr² (moment of inertia of a spherical shell) = 1.2 mr²v = 8.4 m/s (initial velocity)
The equation can be simplified as follows:mgh = (2/5)Mv² ⇒ gh = (2/5) (v²/M) = (5/7) v² / r²
Hence, the maximum vertical height it reaches can be calculated as:h = v² / 2g * (5/7)r²h = (8.4)² / (2 × 9.8) × (5/7) × (0.3²)h = 1.31 meters
Therefore, treating the ball as a spherical shell, its maximum vertical height is 1.31 meters.
Given data:
Mass of ball, m = 1.8 kg
Initial velocity, v = 8.4 m/s
Radius of the ball, r = 0.3 m
Acceleration due to gravity, g = 9.8 m/s²
Calculating the maximum vertical height it reaches: Consider the ball a spherical shell.
Moment of inertia of a spherical shell, M = 2/3mr² = 1.2 mr²Now, the work done on the ball by the force of gravity (mgh) must be equal to its gain in kinetic energy (1/2mv²). By conservation of energy,mgh = (1/2)mv² ---(1)Also, by the work-energy principle, the total work done on the ball is equal to its change in kinetic energy. By treating the ball as a spherical shell, the total work done on the ball by the force of gravity can be found as shown below:
When the ball reaches the maximum height h, its speed becomes zero. Therefore, its kinetic energy becomes zero. Hence, the total work done by the force of gravity can be found by calculating the difference between the kinetic energy of the ball at the top and its kinetic energy at the bottom.
Total work done on the ball by gravity = Change in kinetic energy= 1/2m0² - 1/2mv²= - 1/2mv² --- (2) (Since the ball initially rolls without slipping, its velocity at the bottom of the hill is equal to the velocity at the top of the hill, which is zero)Now, equating equations (1) and (2), we get:
mgh = - 1/2mv²gh = (1/2)mv²/m --- (3)But, v = u + gt
where, u = 8.4 m/s (initial velocity)
t = Time taken by the ball to reach the maximum height
Let's find out t:
When the ball reaches the maximum height, its final velocity becomes zero. Hence, by the first equation of motion, we have:v = u + gt0 = 8.4 + (-9.8)t
Solving for t, we get:t = 0.857 seconds
Substituting the value of t in equation (3), we get:gh = (1/2)(8.4)² / (1.8) × (0.3)²gh = 1.31 meters
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