Jacob reports that as he varies the charge on the left particle, he observes that the force between the two particles changes.
Specifically, when the charge on the left particle is positive, the force between the two particles is attractive, and when the charge on the left particle is negative, the force between the two particles is repulsive. This is because opposite charges attract and like charges repel. As the magnitude of the charge on the left particle increases, the force between the two particles increases as well. The force of attraction is strongest when the charges on both particles are the same, and the force of repulsion is strongest when the charges on the particles are opposite.
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a charge of 21 nc is uniformly distributed along a straight rod of length 3.6 m that is bent into a circular arc with a radius of 2.5 m. what is the magnitude of the electric field at the center of curvature of the arc?
A straight rod of length 3.6 m is bent into a circular arc with a radius of 2.5 m, dispersing a charge of 21 nc evenly throughout its length. The strength of the electric field at the arc's curvature center is 11.868 N/C.
l=rθ
3.6=2.5θ
θ=82.5°
E=[tex]\frac{\lambda }{4\pi E_{o}r }(sin\frac{\theta }{2} +sin\frac{\theta }{2} )[/tex]
E=[tex]\frac{a/l }{4\pi E_{o}r }(2 sin\frac{\theta }{2} )[/tex]
E=[tex]\frac{9*10^{9} }{2.5}*\frac{21*10^{-9}}{3.6} (2 sin\frac{82.5 }{2} )[/tex]
E=11.868 N/C
In physics, an electric field is a fundamental concept that describes the effect of electric charges on the space around them. Electric fields arise due to the presence of charged particles or electrically charged objects. The electric field is a vector field that describes the force experienced by a charged particle at any given point in space. The strength of the electric field is measured in units of volts per meter (V/m) and is directly proportional to the amount of charge and inversely proportional to the distance between the charges. The electric field plays a critical role in a variety of physical phenomena, including electric motors, electronics, and the behavior of charged particles in particle accelerators.
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describe the relationship between frequency and energy as you move through all parts of the electromagnetic spectrum.
In electromagnetic spectrum, the frequency and energy of electromagnetic waves increase.
The electromagnetic spectrum includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. At the lowest end of the spectrum, radio waves have low frequency and low energy, while at the highest end, gamma rays have high frequency and high energy.
The relationship between frequency and energy can be described by the equation E = hf, where E is energy, h is Planck's constant, and f is frequency. As frequency increases, energy increases proportionally, and this relationship is consistent throughout the electromagnetic spectrum.
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the fundamental frequency of a string fixed at both ends is 331 hz. how long does it take for a wave to travel the length of this string?
The fundamental frequency of a string that is fixed at both ends is given by
f = v /2sL
where v and L are the speed of sound and string length, respectively.
The amount of time it takes for the wave to travel the length of the string, on the other hand, is given by
t = L/v
As a result, we have
t = L/v
t = v /2f /v
t = 1 /2f
t = 1 /2f
t= 1/2(255hz)
t=1.96 *10⁻³s
The fundamental frequency is computed as f = v/2*L, where v is the speed of the sound wave and L is the length of a tube or device through which the wave is moving. Wavelength multiplied by frequency equals speed. The wave length in this equation is given in metres, while the frequency is measured in hertz (Hz), or the number of waves per second. As a result, wave speed is expressed in metres per second, the SI unit of speed. If the frequencies are all integers or precise multiples of a fundamental frequency, the greatest common divisor of the frequencies can be used. If the frequencies are 1760, 2200, and 3080, then the fundamental frequency is 440 since it is the greatest common divisor.
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Water is pumped from a large reservoir to a point 20 m higher than the reservoir.
Calculate the pump head if 0.01 m3/s flows through a 0.15 m diameter pipe and the
total frictional head loss is given to be 35^2/2g?
The pump head is -42.065 m
Since the pump head is negative, the pump is unable to push the water to the height of 20 m, and additional work must be done to overcome the frictional losses in the pipe.
How The Pump Head Calculation Was DoneThe pump head is equal to the total head minus the frictional head loss. The total head can be calculated as follows:
h_total = h_static + h_dynamic
where h_static is the static head, which is equal to the difference in height between the reservoir and the point 20 m higher, and h_dynamic is the dynamic head, which is a measure of the pressure generated by the flow of water.
h_static = 20 m
To calculate the dynamic head, we need to first calculate the velocity of the water in the pipe. We can use the equation of continuity, which states that the flow rate through a pipe must remain constant along its length.
Q = A * v
where Q is the flow rate (0.01 m^3/s), A is the cross-sectional area of the pipe (pi * (d/2)^2), and v is the velocity of the water.
A = pi * (0.15/2)^2 = 0.00176 m^2
v = Q / A = 0.01 / 0.00176 = 5.68 m/s
Next, we can calculate the dynamic head using the Bernoulli equation:
h_dynamic = (v^2) / (2g)
where g is the acceleration due to gravity (9.8 m/s^2).
h_dynamic = (5.68^2) / (2 * 9.8) = 16.06 m
Finally, we can calculate the total head:
h_total = h_static + h_dynamic
= 20 + 16.06
= 36.06 m
The pump head is then:
h_pump = h_total - h_friction
= 36.06 - (35^2) / (2 * 9.8)
= 36.06 - 78.125
= -42.065 m
Since the pump head is negative, it means that the pump is unable to lift the water to the desired height of 20 m, and additional work must be done to overcome the frictional losses in the pipe.
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you're flying a cessna 182 skylane, n935fa, on a cross-country vfr flight with flight following. during a handoff, the ideal call sign to use when contacting the new controller is .
You're flying a Cessna 182 Skylane, N935FA, on a cross-country VFR flight with flight following. During a handoff, the ideal call sign to use when contacting the new controller is Skylane 935FA.
Flight following is a relatively simple concept as it’s an aircraft flying under VFR that is taking use of Air Traffic Control (ATC) services. Practically, it indicates that several advisories may be available from the controller and the flight is radar identified by ATC.
The service is given on a workload-permitting basis and involves multiple layers of service, there’s not entirely uniform delivery among Center or Approach control facilities. The fusion of these realities has resulted in misinformation, misunderstandings, and even no awareness of its existence.
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a smooth metal plate with a surface area of 380 cm2 is traveling in water at a velocity of 6.3 m/s, with the flow parallel to the long dimension of the surface. if the reynolds number at the trailing edge of the plate is 950,000, what is the friction drag acting on the surface of the plate?
The friction drag acting on the surface of the plate is approximately 0.074 N.
The friction drag acting on the surface of the plate can be calculated using the following formula:
[tex]$F_D = \frac{1}{2} \rho V^2C_D A$[/tex]
where \rho is the density of water,
V is the velocity of the plate relative to the water,
C_D is the drag coefficient, and
A is the surface area of the plate.
To determine the drag coefficient C_D.
We need to know the Reynolds number of the flow.
For a flat plate, the drag coefficient can be estimated using the following formula for laminar flow:
[tex]$C_D = \frac{1.328}{\sqrt{Re_x}}$[/tex]
where Re_x is the Reynolds number based on the distance from the leading edge of the plate to the point of interest (in this case, the trailing edge).
For turbulent flow, we can use the following formula for the drag coefficient:
[tex]$C_D = \frac{0.074}{Rex^{1/5}}[/tex]
The critical Reynolds number for a flat plate is around 5,000, above which the flow is typically turbulent.
In this case, the Reynolds number of 950,000 indicates that the flow is definitely turbulent.
To calculate the drag coefficient, we need to use the turbulent flow formula:
[tex]$C_D = \frac{0.074}{Rex^{1/5}}[/tex]
[tex]= \frac{0.074}{(380 \times 6.3 / \nu)^{1/5}}[/tex]
where nu is the kinematic viscosity of water at the given temperature.
Assuming a temperature of 25°C, the kinematic viscosity of water is approximate nu = 8.85* [tex]10^{-7}[/tex] [tex]m^2\\[/tex]/s.
Plugging in the values, we get:
[tex]C_D = \frac{0.074}{(380 \times 6.3 / 8.85 \times )^{1/5}}[/tex] approx 0.0093
Finally, we can calculate the friction drag:
[tex]F_D = \frac{1}{2} \rho V^2 C_D A\\ = \frac{1}{2} \times 1000 \times 6.3^2 \times 0.0093 \times 0.0038\\ \approx. 0.074N[/tex]
Therefore, the friction drag acting on the surface of the plate is approximately 0.074 N.
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A pipe 1.00 m long is closed at one end. A guitar is placed near the open end of the tube and the string is plucked. The guitar string is 0.50 meters long and has a mass of 0.0010 kg. The string vibrates in its fundamental mode (1st harmonic) and produces a 1st overtone (3rd harmonic) standing wave in the closed pipe. If the speed of sound is 340 m/s, determine the tension in the guitar string.
Hint: the frequency of the vibrating guitar string is the same as the frequency produced in the closed pipe.
The tension in the guitar string is approximately 165 N.
How to determined the fundamental frequency (1st harmonic) of the closed pipeThe fundamental frequency (1st harmonic) of the closed pipe
can be determined using the formula:
f = v/2L
where
f is the frequencyv is the speed of sound L is the length of the pipeSubstituting the given values, we get:
f = 340 m/s / 2(1.00 m) = 170 Hz
The frequency of the 1st overtone (3rd harmonic) is three times the fundamental frequency, so:
f_overtone = 3f = 3(170 Hz) = 510 Hz
The tension in the guitar string can be determined using the formula:
f = 1/2L√(T/μ)
where T is the tension in the string and μ is the linear mass density (mass per unit length) of the string. The linear mass density can be determined from the given mass and length of the string:
μ = m/L = 0.0010 kg / 0.50 m = 0.0020 kg/m
Substituting the given and calculated values, we get:
510 Hz = 1/2(0.50 m)√(T/0.0020 kg/m)
Simplifying and solving for T, we get:
T = (4π^2)(0.0020 kg/m)(510 Hz)^2(0.50 m)^2 = 165 N
Therefore, the tension in the guitar string is approximately 165 N.
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(c) Determine the total work done on the ball from A to B
V = sqrt (KE * 2 / m) = 4.913 m/s, where KE = 0.5 * 0.58 * 1.6*1.6 = 0.7424 J. (c) Work done = change in KE = 7 - 0.7424 = 6.2576 J.
Work is the energy used by one thing to move another object across a distance by applying a force. The equation W = F x d calculates the work done on an item with a given force, F, and a certain distance, d. It should be noted that this equation presupposes that the force is applied in a direction parallel to the object's direction of motion. How to manage non-parallel circumstances will be covered in a later lecture.
When we apply force "F" to a block, the body moves with some acceleration or, moreover, its speed increases or decreases depending on the direction of the force. The system's kinetic energy changes as speed increases or decreases.
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The complete question is:
A 0.58 kg rubber ball has a speed of 1.60 m/s at point A and kinetic energy of 7.0 J at point B. (a) Determine the ball's kinetic energy at A (b) Determine the ball's speed at B (c) Determine the total work done on the ball from A to B
one light-minute is the distance that light travels in one minute. how far is this, in kilometers? (recall that the speed of light is 300,000 km/s.)
One light-minute is approximately 18 million kilometers.
One light-minute is the distance that light travels in one minute, at the speed of light, which is approximately 300,000 km/s. we can simply multiply the speed of light by the number of seconds in one minute:
1 light-minute = 60 seconds x 300,000 km/s
1 light-minute = 18,000,000 km
Therefore, one light-minute is approximately 18 million kilometers.
Light is an electromagnetic wave that travels through space at a constant speed of approximately 300,000 km/s. This means that in one second, light can travel a distance of 300,000 kilometers.
300,000 km/s x 60 seconds = 18,000,000 km
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(ii) One possible path of the spacecraft as it approaches the planet is shown in fig.1.1 The spacecraft enters the orbit at point A with speed 3.7x10^3 ms^-1.At point B, a distance of 5.00x10^7 from the centre of the planet, the spacecraft has a speed of 4.1x10^3ms^-1. The mass of the spacecraft is 650kg. For the spacecraft moving from point B to point A, show that the change in gravitational potential energy of the spacecraft is8.3x10^9
The change in gravitational potential energy of the spacecraft as it moves from point B to point A is 8.3x10^9 J.
The change in gravitational potential energy of the spacecraft as it moves from point B to point A can be calculated using the formula:
[tex]ΔPE = -GMm(1/rA - 1/rB)[/tex] , where
ΔPE is the change in gravitational potential energy,
G is the gravitational constant,
M is the mass of the planet,
m is the mass of the spacecraft,
rA is the distance from the planet's center at point A, and
rB is the distance from the planet's center at point B.
Using the given values, we have:
[tex]ΔPE = -(6.67x10^-11 Nm^2/kg^2)(1.50x10^24 kg)(650 kg)(1/5.00x10^7 m - 1/0)[/tex][tex]ΔPE = -(6.67x10^-11 Nm^2/kg^2)(1.50x10^24 kg)(650 kg)(1/5.00x10^7 m - 1/0)[/tex]
Simplifying the expression, we get:
[tex]ΔPE = -8.3x10^9 J[/tex]
Therefore, the change in gravitational potential energy of the spacecraft as it moves from point B to point A is [tex]8.3x10^9[/tex]J, which is a negative value because the spacecraft is moving closer to the planet and its gravitational potential energy is decreasing.
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(ii) Atmospheric pressure on that day is 1.0 x 10^5Pa
Calculate the total pressure at the bottom of the pond
Answer:
Explanation:Force applied by atmosphere = atmospheric pressure × ares of table
F=P×A
F=1.013×[tex]10^{5}[/tex] ×(2×1)=2.026×[tex]10^{5}[/tex]N
. two blocks are connected by a massless rope as shown below. the mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. the table and the pulley are frictionless. (a) find the acceleration of the system. (b) find the tension in the rope. (c) find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.
A- The acceleration of the system is 1.96 [tex]m/s^{2}[/tex], B- the tension in the rope is 7.84N, C- the velocity at which the hanging mass strikes the ground if it begins at rest and is positioned 1.0 meters off the ground is 1.98 m/s.
A- both the masses will have same magnitude of acceleration a ,
m2g-T = m2a
T=m1a
from both the equations a = m2g/(m1 + m2) = 1.96[tex]m/s^{2}[/tex]. Therefore, The acceleration of the system is 1.96 [tex]m/s^{2}[/tex]
B- T = m1a = 7.84 N is the tension in the rope
C- [tex]V_{f^{2} }[/tex] = [tex]V_{i^{2} }[/tex] + 2as
= 0 + 2am
= 1.98 m/s
Hence , the velocity at which the hanging mass strikes the ground if it begins at rest and is positioned 1.0 meters off the ground is 1.98 m/s.
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The complete question is :
As seen in the illustration below, a massless rope links two blocks. The hanging mass is 1.0 kg, whereas the block on the table weighs 4.0 kg. No friction exists between the table and the pulley. Find the system's acceleration in (a). (b) Determine the rope's tension. (c) determine the speed at which the hanging mass will strike the ground if it begins at rest and is 1.0 m from the ground.
the work done to compress a gas is 74 j. as a result, 26 j of heat is given by the system to the surroundings. what is the change in the internal energy of the gas?
The change in the internal energy of the gas is 100 J and this calculation assumes that the system is closed and no other forms of energy are involved.
The change in the internal energy of a system can be calculated using the First Law of Thermodynamics: ΔU = Q - W where ΔU is the change in internal energy of the system, Q is the heat added to the system, and W is the work done by the system. In this case, the work is done on the system (compressing the gas), so W is negative.
Given that the work done to compress the gas is 74 J and 26 J of heat is given by the system to the surroundings, we can substitute these values into the equation to get:
ΔU = Q - W
ΔU = 26 J - (-74 J)
ΔU = 26 J + 74 J
ΔU = 100 J
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Rolling railroad car a collides inelastically with railroad car b of the same mass, which is initially at rest. If the two cars stick together after the collision, how does their speed after the collision compare with the initial speed of car a ?
The final velocity of the two cars after they stick together is half the initial velocity of car a. In other words, their speed after the collision is half the initial speed of car a.
In an inelastic collision, the two objects stick together after the collision and move together with a common final velocity. In this case, the rolling railroad car a collides inelastically with railroad car b of the same mass, which is initially at rest.
Let's assume that the initial velocity of car a is v and the mass of each car is m. Since car b is initially at rest, its initial velocity is 0.
Using the law of conservation of momentum, we can write:
(momentum before collision) = (momentum after collision)
mv + 0 = (m + m)vf
where vf is the final velocity of the two cars after they stick together.
Simplifying the equation, we get:
vf = v/2
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the range of possible frequencies of radiation is called the ______ spectrum.
The range of possible frequencies of radiation is called the electromagnetic spectrum.
What is the electromagnetic spectrum and how is it classified?The electromagnetic spectrum is the variety of all possible frequencies of electromagnetic radiation. This consists of radio waves, microwaves, infrared radiation, seen mild, ultraviolet radiation, X-rays, and gamma rays. The spectrum is classed based totally at the frequency and wavelength of the radiation, with each sort of radiation occupying a specific area of the spectrum. The class is primarily based on their capability to tour thru one of a kind substances and their interactions with matter. The distinct areas of the spectrum have an expansion of realistic applications, from conversation technology to medical imaging.
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Can a vector with zero magnitude have one or more components that are zero? Explain.
Yes, a vector with zero magnitude can have one or more components that are zero. This is because a vector's magnitude is determined by the length of its components, which can be zero even if the magnitude of the vector is zero.
What is magnitude?Magnitude is a measure of the size, intensity, or strength of something. It is most commonly used to describe the size of an earthquake, magnitude being a measure of its energy release. It can also refer to the size of an astronomical object, such as a star or planet, or the strength of a magnetic field. Magnitude is expressed using a logarithmic scale, with each increase in magnitude representing a ten-fold increase in the strength or size of an object. For example, an earthquake of magnitude 5.0 is ten times as powerful as one of magnitude 4.0. Magnitude is an important concept in physics, astronomy, and other sciences.
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uppose you have two metal cubes, one made of iron and one made of aluminum. You transfer the same amount of heat Q to each of them. Which cube will have the higher final temperature, given they have the same masses and initial temperatures?a. Iron Cubeb. Aluminum Cube
Answer:
Q = C M ΔT where C is specific heat in cal / gm*deg C
C (Fe) = .11
C (Al) = .22
obviously ΔT has to be twice as great for Iron (Fe) as for (Al) for the same amount of heat to be transferred
ΔT = Q /(C * M) where ΔT is the change in temperature
a) iron would have the higher final temperature
three motions are listed: an object changes direction an object changes mass an object changes position which of the three is considered acceleration? a only b only a and c b and c
Acceleration is a measure of how quickly an object changes its velocity, and is given by the rate of change of its velocity over time.
Of the three motions listed, changing direction and changing position are considered acceleration. This is because changing direction involves a change in velocity, even if the speed remains constant. Similarly, changing position involves a change in velocity, as the object is accelerating in a particular direction. On the other hand, changing mass is not considered acceleration because it does not affect the object's velocity. While it may affect other properties of the object's motion, such as its momentum or kinetic energy, it does not result in a change in velocity, and therefore is not considered acceleration. Acceleration is a measure of how quickly an object changes its velocity, and is given by the rate of change of its velocity over time.
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Answer:
A and B
Explanation:
I took the exam
If two objects have different temperatures, in which direction will heat move? When will the heat stop moving?
Answer: When there are two objects of different temps, the heat will always move from the higher temp to the lower temp. The energy will stop moving when there is equilibrium, when both objects are at the same temperatures.
Help now please! I don't really understand the question and would be very grateful for some clarification.
El Toro is the third tallest wooden roller coaster in the world. It has a drop of 55 meters. One empty cart has a mass of 958 kg and 511,000 J of kinetic energy when it reaches the base level of the track. How much energy is converted into heat when the cart is at the base level of the track? Show your work.
The amount of energy converted into heat when the cart is at the base level of the track is approximately 7490 J.
What is kinetic energy?The energy an object has as a result of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it.
Kinetic energy:
E = mgh
E = (958 kg) x (9.81 m/[tex]s^2[/tex]) x (55 m) = 5.26 x [tex]10^5[/tex] J
E = KE + Q
KE is the kinetic energy of the cart and Q is the energy converted into heat. Substituting the given value for the kinetic energy, we get:
5.26 x [tex]10^5[/tex] J = 511,000 J + Q
Solving for Q, we get:
Q = 5.26 x [tex]10^5[/tex] J - 511,000 J = 7490 J
Thus, 7490 J energy is converted into heat.
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Steady precipitation preceding a front is an indication of:a) stratiform clouds with moderate turbulence
b) cumuliform clouds with little or no turbulence
c) stratiform clouds with little or no turbulence
Steady precipitation preceding a front is most likely an indication of stratiform clouds with little or no turbulence.
The steady precipitation preceding a front is an indication of stratiform clouds with little or no turbulence. Stratiform clouds are layered and cover a large area, producing a wide, steady, and uniform precipitation that can last for many hours. These clouds form when moist air is lifted and cooled, resulting in a broad cloud layer with a relatively uniform base and top.
The steady precipitation ahead of a front is often caused by the uplift of warm air over cooler air, which creates a wide, relatively stable front. This stable front tends to produce stratiform clouds that extend over a large area and produce steady precipitation. On the other hand, cumuliform clouds are characterized by vertical development, producing showers and thunderstorms with a lot of turbulence.
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light from a slit passes through a transmission diffraction grating of 325 lines/mm, which is located 2.6 m from a screen. what are the distances on the screen (from the unscattered slit image) of the three brightest visible (first-order) hydrogen lines? (enter your answers in cm.)
The distance on the screen from the unscattered slit image to the first-order violet line is 0.00514 cm.
What is Transmission?
In physics, transmission refers to the passage of waves, particles, or energy through a medium or barrier, without any change in the shape or form of the wave. This can occur in various contexts, such as the transmission of electromagnetic waves (such as light) through a material, the transmission of sound waves through the air, or the transmission of particles (such as electrons or ions) through a solid-state material.
The distance d between the central bright fringe and the first-order bright fringe for a transmission diffraction grating is given by:
dλ = mD
where λ is the wavelength of light, m is the order of the bright fringe, D is the distance between the grating and the screen, and d is the distance between adjacent slits on the grating.
For a hydrogen spectrum, the wavelengths of the first three visible (first-order) lines are:
λ_1 = 656.3 nm (red line)
λ_2 = 486.1 nm (blue-green line)
λ_3 = 434.0 nm (violet line)
The distance between adjacent slits on the grating is:
d = 1/325 mm = 0.00308 cm
The distance between the grating and the screen is D = 2.6 m = 260 cm.
For the first-order red line (m = 1, λ = 656.3 nm), we have:
dλ = mD
0.00308 cm * 656.3 nm = 1 * 260 cm * x
Solving for x, we get:
x = (0.00308 cm * 656.3 nm) / 260 cm = 0.00776 cm
So, the distance on the screen from the unscattered slit image to the first-order red line is 0.00776 cm.
For the first-order blue-green line (m = 1, λ = 486.1 nm), we have:
dλ = mD
0.00308 cm * 486.1 nm = 1 * 260 cm * x
Solving for x, we get:
x = (0.00308 cm * 486.1 nm) / 260 cm = 0.00579 cm
So, the distance on the screen from the unscattered slit image to the first-order blue-green line is 0.00579 cm.
For the first-order violet line (m = 1, λ = 434.0 nm), we have:
dλ = mD
0.00308 cm * 434.0 nm = 1 * 260 cm * x
Solving for x, we get:
x = (0.00308 cm * 434.0 nm) / 260 cm = 0.00514 cm
So, the distance on the screen from the unscattered slit image to the first-order violet line is 0.00514 cm.
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ramon has contact lenses with the prescription 2.0 d. a. what eye condition does ramon have? b. what is his near point without the lenses?
The closest object that the eye can concentrate on is referred to as the near point. Typically, vision at a near point of 25 cm is regarded as “normal.”
What is the near point without the lenses?So, because the power is negative, he won't be able to see the nearby objects quickly. This kind of victim is known as the metro pia, and it starts out right. Far-sightedness, or hyper metro pia, affects Raymond.
Sightedness for fire. The person in question is unable to stand on two feet as a result of this consequence. The farthest thing an average eye can image onto the retina is called the far point of a human eye.
Therefore, choose a prominent form behind the retina of the eye rather than focusing on the close items clearly. And to correct this flaw, a convex lens is utilized.
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a child holds a ball of mass m a distance h above the ground. in which system(s) is the force of gravity on the ball an internal force to the system?
The system in which the force of gravity on the ball an internal force to the system is Option B. system of the earth and the ball together.
Every object that has mass exerts a gravitational pull or force on every other mass. The strength of this pull depends on the millions of objects at play. graveness keeps the globes in route around the sun and the moon around the Earth. Hence, we define graveness as graveness is a force that attracts a body towards the centre of the earth or any other physical body having mass.
Originally, the direct instigation of the" ball earth" system is zero. So, according to the conservation of direct instigation, final direct instigation of the system must also be zero. therefore, if the ball moves overhead with some haste, the earth moves in downcast direction so as to conserve the instigation. Hence, the ball and the earth moves down from each other.
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Complete question:
A child holds a ball of mass m a distance h above the ground. In which system(s) is the force of gravity on the ball an internal force to the system? The system of just the ball.
The system of the earth and the ball together.
The system of the earth, the ball, and the child's hand.
The system of the earth, the ball, and the entire child.
your roommate is working on his bicycle and has the bike upside down. he spins the 60.0 cm -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. what is the acceleration of the pebble
The acceleration of the pebble stuck in the tread goes by three times every second is 106.592 m/s².
Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.
The pebble in the tread goes by 3 times every second.
This is the same as 3 times per second.
Recall the unit of frequency is Hertz or per second, s⁻¹
So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz
For angular motion:
Angular speed, ω = 2πf
= 2*π*3
= 6π rad/s
Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s
Linear acceleration, a = v² / r
a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²
Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²
Angular speed = 6π rad/s ≈ 18.840 rad/s
The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s
The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²
The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²
The rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.
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a small object of mass 2.6 g and charge 14 uc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what is the magnitude and direction of the electric field?
Uniform electric field of magnitude [tex]1.81 * 10^5 N/C[/tex]pointing upward, away from the ground.
The small object is suspended motionless in a uniform electric field, which means that the electric force on the object is balanced by the force of gravity. We can use the equation for the electric force on a point charge in an electric field to find the magnitude of the electric field:
F = qE
where F is the electric force on the charge q, and E is the electric field. Since the object is motionless, the electric force on it must be equal and opposite to the gravitational force:
[tex]F_E = F_g[/tex]
qE = mg
m: mass of the object
g: acceleration due to gravity.
Substituting values:
E = [tex](mg) / q = [(2.6 g) * (9.8 m/s^2)] / (14 microC)[/tex]
μC: microcoulombs.
Use conversion factor:
[tex]1 microC = 10^(-6) C[/tex]
[tex]E = 1.81 * 10^5 N/C[/tex]
Electric field magnitude is [tex]1.81 * 10^5 N/C[/tex], and its direction is perpendicular to the ground, which is the direction of the force on the charge. The charge on the object is negative, so the direction of the electric field is opposite to the direction of the force on a positive charge. Therefore, the electric field points upward, away from the ground.
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two point charges, a and b, are separated by a distance of 17.0 cm . the magnitude of the charge on a is twice that of the charge on b. if each charge exerts a force of magnitude 47.0 n on the other, find the magnitudes of the charges.
The charge on plate a is [tex]54.8 * 10^{-3}C[/tex] while that on plate b is [tex]27.4 * 10^{-3}C[/tex] that are separated by a distance of 17.0 cm.
Let the magnitude of the charge on charge a be qa and that of the charge on charge b be qb.
The magnitude of the force on each charge due to the other charge can be found using Coulomb's law:
[tex]F = k(qa*qb)/r^2,[/tex]
where k is the Coulomb constant, qa is the magnitude of the charge on charge a, qb is the magnitude of the charge on charge b, and r is the distance between the two charges.
Given the force of magnitude = 47.0N
The distance between the charges are = 17cm
Since the magnitude of the charge on charge a is twice that of the charge on charge b, we can write:
qa = 2qb
Substituting the given values into the equation, we get:
[tex]47.0 N = (8.99 * 10^9 Nm^2/C^2)(qa*qb)/(17.0cm)^2[/tex]
Rearranging, we get:
[tex]qaqb = (47.0 N)*(17.0 cm)^2/(8.99 * 10^9 Nm^2/C^2)[/tex]
[tex]qb^2 = 754.6 * 10^{-9}[/tex]
[tex]qb = 27.4 * 10^{-3}C[/tex]
then [tex]qa = 2 * 27.4 * 10^{-3} = 54.8 *10^{-3}C[/tex]
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An airplane flies with a constant speed of
540 miles per hour. How long will it take to travel a distance of 2160 miles?
Answer:
4 hours
Explanation:
2160 miles ÷ 540 miles/hr = 4
Which type of satellite is useless for meteorologists at 3AM?
Select one:
a. visible satellite imagery
b. infrared satellite imagery
c. water vapor imagery
what if? if the woman still pulls on the strap with a 34.5 n force but wishes to accelerate the suitcase at a rate of 0.500 m/s2, at what angle (in degrees) must she pull on the strap? assume that the rolling friction is independent of the angle of the strap.
The angle at which the woman pull on the strap with a 34.5N force but wishes to accelerate the suitcase at a rate of 0.500 m/s2 is 0°
Given:
F = 34.5 N
Friction force = 20.0 N
m = 16.5 kg
The net force, Fnet, must be equal to the mass of the suitcase, 16.5 kg, times the acceleration, 0.500 m/s^2, or 8.25 N.
Fnet = ma
8.25 N = 16.5 kg * 0.500 m/s^2
The net force, Fnet, will be equal to the force of the strap, F, minus the friction force, Ffriction, or 34.5 N - 20.0 N = 14.5 N.
Fnet = F - Ffriction
14.5 N = 34.5 N - 20.0 N
To find the angle, θ, at which the strap must be pulled, we can use the following equation:
Fnet = Fcosθ - Ffriction
14.5 N = 34.5 Ncosθ - 20.0 N
Solving for θ, we get:
θ = 0°
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complete question: A woman at an airport is towing her 16.5 kg suitcase at a constant speed by pulling on a strap at an angle θ above the horizontal (see figure). She pulls on the strap with a 34.5N force, and the friction force on the suitcase is 20.0 N. A woman holds the strap of a suitcase while pulling it to the right. The strap makes an angle θ measured counterclockwise from the horizontal. What If? If the woman still pulls on the strap with a 34.5 N force but wishes to accelerate the suitcase at a rate of 0.500 m/s2, at what angle (in degrees) must she pull on the strap? Assume that the rolling friction is independent of the angle of the strap
