To create an upright image that is half the height of the object and located 10 meters from the object using a spherical mirror, we need to determine the focal length (f) of the mirror.
In this scenario, we can use the mirror equation to find the focal length. The mirror equation relates the object distance (dₒ), image distance (dᵢ), and the focal length of the mirror (f) using the formula: 1/f = 1/dₒ + 1/dᵢ.
Given that the image is located 10 meters from the object and has half the height of the object, we know that the magnification (m) is -1/2. The magnification is given by the formula: m = -dᵢ/dₒ.
Since the magnification is negative, it indicates that the image is upright. By substituting the known values into the magnification formula, we can solve for the object distance (dₒ).
With the object distance known, we can then substitute the object distance and image distance into the mirror equation. Rearranging the equation, we can solve for the focal length (f).
By substituting the values into the equation, we can calculate the focal length (f) of the spherical mirror that would create the desired image.
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Ball A is thrown horizontally, and ball B is dropped from the same height at the same moment,
Select one:
A Bat B has the greater speed when it reaches the ground
B. Ball A reaches the ground ist
Ball Beaches the ground first
D. D. Balt A has the greater speed when it reaches the ground,
Answer:
a....................
The answer is:
D. Ball A has the greater speed when it reaches the ground
10) A wall moving in the positive x-direction with velocity v hits a stationary ball and keeps moving with the same velocity. The ball's velocity after the collision is:
a) v
b) -v
c) 0
d) 2v
e) -2v
11) For a potential U(x)=-4x³+2, find the force F at x = 2m:
a) F= ON
b) F = 12N
c) F=-12N
d) F= 48N
e) F= -48N f) F = 36N
The ball's velocity after the collision is v
For a potential U(x)=-4x³+2, the force F at x = 2m is 36N
Define force
A force is an effect that changes, or accelerates, the motion of a mass-containing object. It is a vector quantity since it can be a push or a pull and always has magnitude and direction.
A force is a physical quantity that alters the shape or size of an item, affects the direction of motion of an object in motion, or tends to create a motion in an object at rest.
P2 will be -m(2V)=-2mV
P₁ = m(V) = mV
Pi = -mV
Pf=mv1+mv2
P₁ = Pf
-mV = mv2-mv1
V2 + V₁ = -V
e=1(elastic collision)
V2 - V1 /2V + V will be equal to e
V2+V₁ = 3V
V₁ = −2V and V₂ = V
For a potential U(x)=-4x³+2, find the force F at x = 2m:
U(2)=-4*2³+2
U(2)= 36N
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'If you increase the frequency of a sound wave four times, what will happen to its speed?
ОА.
The speed will increase four times.
OB.
The speed will decrease four times.
O c.
The speed will remain the same.
OD
The speed will increase twice.
O E.
The speed will decrease twice.
the use of an electronic throttle control (etc) system allows the elimination of all of these except
The electronic throttle control (ETC) system offers the ability to eliminate certain components and mechanisms.
The electronic throttle control (ETC) system is a technology that replaces traditional mechanical linkages between the accelerator pedal and the engine throttle with an electronic sensor and actuator. By doing so, it provides several advantages in terms of efficiency, control, and safety. One significant benefit is the elimination of various components and mechanisms found in conventional throttle systems.
These include the throttle cable, throttle position sensor, idle air control valve, and cruise control module, among others. With the ETC system, these components are no longer needed, simplifying the overall design and reducing maintenance requirements.
Instead, the ETC system relies on electronic signals and actuators to precisely control the engine throttle opening, allowing for improved responsiveness and fuel efficiency. Furthermore, the elimination of mechanical linkages reduces the risk of failures or malfunctions associated with wear and tear. Overall, the ETC system streamlines the throttle control process while enhancing performance and reliability.
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A 5kg particle moving at a speed of 10m/s to the right makes an elastic collision with a wall and rebounds backward calculate the magnitude of the impulse of the body
Answer:
The magnitude of the impulse experienced by the particle is 100 kg.m/s.
Explanation:
Given;
mass of the particle, m = 5 kg
initial velocity of the particle, v₁ = 10 m/s
assuming the particle rebounds with same velocity backwards, v₂ = - 10 m/s
The impulse experienced by the particle is the change in linear momentum;
J = ΔP = mv₁ - mv₂
J = m(v₁ - v₂)
J = 5 (10 - (-10))
J = 5 (10 + 10)
J = 5(20)
J = 100 kg.m/s
Therefore, the magnitude of the impulse experienced by the particle is 100 kg.m/s.
Al changes save
3. The graph shows the magnitude of the force exerted by a given spring as a function of the distance x that the spring is stretched. How much work is needed to stretch this spring a distance of 5 cm,
starting with it unstretched?
350
300
250-
200-
F(N)
150
100
50
1
2
7 8
x (cm)
The work needed to stretch the spring a distance of 5 cm is 1100 N·cm.
To determine the work needed to stretch the spring a distance of 5 cm, we need to calculate the area under the force vs. distance graph within that range. Looking at the graph, we can see that the force initially increases linearly as the distance increases and then levels off.
To calculate the work, we need to find the area of the triangle formed by the initial linear part of the graph and the rectangle representing the constant force. The height of the triangle is the force at 5 cm, which appears to be around 200 N. The base of the triangle is 5 cm. The area of the triangle is given by 0.5 * base * height, which is 0.5 * 5 cm * 200 N = 500 N·cm .The rectangle representing the constant force has a height of 200 N and a base of 3 cm (since it starts at 2 cm and ends at 5 cm). The area of the rectangle is base * height, which is 3 cm * 200 N = 600 N·cm.
Adding the areas of the triangle and the rectangle, we get a total work of 500 N·cm + 600 N·cm = 1100 N·cm.
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a tall, open container is full of glycerine. at what depth ℎ below the surface of the glycerine is the pressure 2830 pa greater than atmospheric pressure? the density of glycerine is 1.26×103 kg/m3 .
The pressure in the glycerin at a depth of 0.2306 meters below the surface where it is 2830 Pa higher than atmospheric pressure.
To determine the depth below the surface of the glycerin at which the pressure is 2830 Pa greater than atmospheric pressure, we can use the concept of pressure in a fluid.
The pressure at a certain depth in a fluid is given by the equation:
P = P₀ + ρgh
Where:
P is the pressure at the depth h,
P₀ is the atmospheric pressure (assumed to be the reference pressure),
ρ is the density of the fluid (glycerin in this case),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
h is the depth below the surface.
We can rearrange the equation to solve for h:
[tex]h = \frac{P - P_0}{\rho g}[/tex]
Given that the pressure difference is 2830 Pa and the density of glycerine is 1.26×10³ kg/m³, we can substitute the values into the equation:
[tex]h = \frac{2830\text{ Pa}}{1.26\times 10^3\text{ kg/m}^3 \times 9.8\text{ m/s}^2} = 0.24\text{ m}[/tex]
Calculating the value, we find:
h ≈ 0.2306 meters
Therefore, the depth below the surface of the glycerin at which the pressure is 2830 Pa greater than atmospheric pressure is approximately 0.2306 meters.
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The image formed by the eye’s lens is normally
a) real and inverted
b) virtual and inverted
c) real and upright
d) virtual and inverted
Option (a) Real and inverted , is the correct answer .
The image formed by the eye's lens is normally a) real and inverted.The image formed by the eye's lens is normally real and inverted. This is due to the way light is refracted by the lens and focused onto the retina.
The human eye works similarly to a camera, with a lens that focuses light onto the retina at the back of the eye. The lens in the eye refracts (bends) light to form an image on the retina. This process creates a real and inverted image.
Light rays from an object pass through the cornea and the lens of the eye. The lens adjusts its shape to focus the incoming light onto the retina. The retina contains light-sensitive cells called photoreceptors that convert light into electrical signals, which are then transmitted to the brain for interpretation.
When light rays converge on the retina, an inverted image is formed. This means that the top of the object is projected onto the bottom of the retina, and the bottom of the object is projected onto the top of the retina. This inverted image is the initial visual information that is sent to the brain.
The image formed by the eye's lens is normally real and inverted. This is due to the way light is refracted by the lens and focused onto the retina. The inverted image is then processed by the brain to perceive the object in its correct orientation.
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in one hour, a machine can dig a hole with diameter 10 cm through a 1.75 m depth of consistently hard rock. the machine is required to dig to a total depth of 112 m. if the machine is used for a total of 7 hours per day, how many days will it take for the machine to complete the task?
If the machine is used for a total of 7 hours per day. Hence, the machine will take approximately 9.15 days to complete the task.
Let's solve the problem step by step: Volume of the cylindrical hole: V = πr²h where, r is the radius of the hole, and h is the depth of the hole.
Diameter = 10 cm ⇒ radius, r = 5 cm = 0.05 m Depth = 1.75 m
∴ Volume of the cylindrical hole dug by the machine in 1 hour =
V = πr²h= π × (0.05 m)² × (1.75 m)= 0.004326 m³
We need to find the time required to dig a hole of total depth 112 m.
Total number of such cylindrical holes dug by the machine:
Total number of holes = 112 / 1.75= 64
∴ Total volume of all the 64 holes = 64 × 0.004326 m³= 0.27744 m³
Total time required to dig this volume of rock:
Let t be the time required. In one day, the machine works for 7 hours.
Thus, Volume of rock dug in 1 day = 7 × 0.004326 m³= 0.030282 m³
∴ Total number of days required to dig the required volume of rock = (0.27744 / 0.030282) days
= 9.1504 days (approx.)
∴ The machine will take approximately 9.15 days to complete the task.
Answer: In one hour, the machine can dig a hole with diameter 10 cm through a 1.75 m depth of consistently hard rock.
The volume of the cylindrical hole dug by the machine in 1 hour is
V = πr²h
where r is the radius of the hole, and h is the depth of the hole. The diameter of the hole is 10 cm, and therefore, the radius is 5 cm or 0.05 m. The depth of the hole is 1.75 m.
Hence, the volume of the cylindrical hole dug by the machine in 1 hour is 0.004326 m³.
We need to find the time required to dig a hole of total depth 112 m.
The total number of such cylindrical holes dug by the machine is 112 / 1.75 or 64.
The total volume of all the 64 holes is
64 × 0.004326 m³ = 0.27744 m³.
Let t be the time required. In one day, the machine works for 7 hours.
Thus, the volume of rock dug in 1 day is
7 × 0.004326 m³ = 0.030282 m³.
Therefore, the total number of days required to dig the required volume of rock is
0.27744 / 0.030282 days or approximately 9.15 days.
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A rope is used to pull a box 15.0 m across a floor. The rope is held at an angle of 46.0˚ and a force of 628 N is used along the rope. What is the work done? Your answer should be rounded to the tenths place, and include the correct units. View question
Answer:
6544.07 J
Explanation:
From the question given above, the following data were obtained:
Distance (d) = 15 m
Force (F) = 628 N
Angle (θ) = 46°
Workdone (Wd) =?
The work done can be obtained by using the following formula:
Wd = Fd × Cos θ
Wd = 628 × 15 × Cos 46
Wd = 9420 × 0.6947
Wd = 6544.07 J
Therefore, the workdone is 6544.07 J
Every second the sun gives out 400 million joules of energy, but how much of that actually reaches the earth?
Answer:
about 50 million
Explanation:
100% guessing lol
Answer:
the best thing about this place was to have the kids to do it and
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distance of 1.4 cm. What is the electric field strength?
The electric field strength is approximately -3.34 × 10^29 newtons per coulomb (N/C).
How to solve for the fieldTo determine the electric field strength, we can use the following equation that relates the change in speed of a charged particle to the electric field strength:
Δv = a * Δt
Where:
Δv is the change in velocity (speed) of the electron
a is the acceleration of the electron
Δt is the time taken
Given:
Initial velocity (v1) = 2.0 × 10^7 m/s
Final velocity (v2) = 4.0 × 10^7 m/s
Distance (d) = 1.4 cm = 0.014 m
The change in velocity can be calculated as:
Δv = v2 - v1
Δv = (4.0 × 10^7 m/s) - (2.0 × 10^7 m/s)
Δv = 2.0 × 10^7 m/s
We can rearrange the equation to solve for acceleration (a):
a = Δv / Δt
To find the time (Δt), we can use the equation:
d = (1/2) * a * Δt^2
Rearranging this equation to solve for Δt:
Δt = sqrt((2 * d) / a)
Substituting the given values:
Δt = sqrt((2 * 0.014 m) / (2.0 × 10^7 m/s))
Δt = sqrt(1.4 × 10^(-8) s^2 / m^2)
Δt = 3.74 × 10^(-4) s
Now we can calculate the acceleration (a):
a = Δv / Δt
a = (2.0 × 10^7 m/s) / (3.74 × 10^(-4) s)
a = 5.35 × 10^10 m/s^2
Finally, we can find the electric field strength (E) using the equation:
E = a / q
Where q is the charge of the electron. The charge of an electron is approximately -1.6 × 10^(-19) coulombs.
E = (5.35 × 10^10 m/s^2) / (-1.6 × 10^(-19) C)
E ≈ -3.34 × 10^29 N/C
The electric field strength is approximately -3.34 × 10^29 newtons per coulomb (N/C).
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how does the frequency of a particular spectral line of the sun compare with the frequency of that line observed from a source on earth?
The frequency of a particular spectral line of the sun compare with the frequency of that line observed from a source on earth may differ slightly due to the Doppler Effect, the difference is typically small and can be accurately measured.
When comparing the spectral lines of the sun with those observed from Earth, it's important to consider the Doppler Effect, which causes the wavelengths of light to appear shifted when an object is moving relative to an observer. The frequency of a particular spectral line of the sun will appear slightly different than the frequency of that line observed from a source on Earth due to the Doppler Effect. This effect causes the light from the sun to appear slightly redshifted or blueshifted depending on the relative motion of the sun and Earth.
However, the difference in frequency is typically small and can be accurately measured by modern telescopes. By comparing the frequencies of the spectral lines observed from the sun and Earth, astronomers can study the motion of celestial bodies and determine their chemical compositions. In conclusion, while the frequencies of particular spectral lines of the sun and those observed from a source on Earth may differ slightly due to the Doppler Effect, the difference is typically small and can be accurately measured.
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calculate the energy released in the following fusion reaction. the masses of the isotopes are: 14n (14.00307 u), 32s (31.97207 u), 12c (12.00000 u), and 6li (6.01512 u). $$
The energy released in the fusion reaction is approximately 3.598 × 10¹⁶ Joules.
To calculate the energy released in a fusion reaction, we need to determine the mass defect and then apply Einstein's mass-energy equivalence equation, E = mc².
The mass defect (Δm) is the difference in mass between the reactants and the products. It is given by the sum of the masses of the reactants minus the sum of the masses of the products.
Reactants:
14n (14.00307 u)
32s (31.97207 u)
6li (6.01512 u)
Products:
12c (12.00000 u)
Δm = (mass of reactants) - (mass of products)
Δm = (14.00307 u + 31.97207 u + 6.01512 u) - (12.00000 u)
Δm = 51.99026 u - 12.00000 u
Δm = 39.99026 u
Now, we can calculate the energy released (E) using Einstein's equation, E = Δmc².
E = (39.99026 u) * (c²)
E = (39.99026 u) * (2.998 × 10⁸ m/s)²
E ≈ 3.598 × 10¹⁶ Joules
Therefore, the energy released in the fusion reaction is approximately 3.598 × 10¹⁶ Joules.
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I'm giving 40 points for this
Answer:
1 is true 4 is false 5 is c i think 6 is a 7 is c 3 is a 2 is c 8 is c 9 is true 10 is c
Explanation:
Please help, I do NOT need any links.
Answer:
P since without a host the parasite won't be able to survive they would start decreasing as well but if the hosts were no more then they would go extinct. But since it is just decreasing then it should be P
A 7.55 × 1014 Hz electromagnetic wave travels in carbon tetrachloride with a speed of 2.05 ×108 m/s. What is the wavelength of the wave in this material?
361 nm
301 nm
272 nm
397 nm
338 nm
The wavelength of the electromagnetic wave in carbon tetrachloride is approximately 361 nm.
The speed (v) of a wave is related to its frequency (f) and wavelength (λ) by the equation:
v = f × λ
We are given the frequency (f) of the electromagnetic wave as 7.55 × 10^14 Hz and the speed (v) in carbon tetrachloride as 2.05 × 10^8 m/s.
Rearranging the equation, we can solve for the wavelength (λ):
λ = v / f
Substituting the given values:
λ = (2.05 × 10^8 m/s) / (7.55 × 10^14 Hz)
λ ≈ 2.71 × 10^-7 m
To convert the wavelength to nanometers (nm), we multiply by 10^9:
λ ≈ 2.71 × 10^-7 m × 10^9 nm/m
λ ≈ 271 nm
Therefore, the wavelength of the electromagnetic wave in carbon tetrachloride is approximately 361 nm.
The wavelength of the electromagnetic wave in carbon tetrachloride is approximately 361 nm. This calculation is based on the given frequency and speed of the wave, using the equation relating wavelength, frequency, and speed of the wave.
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A ball of mass ¼(kg) is dropped vertically towards a surface and its velocity at the moment of its arrival is (10m/s), and it bounces back at a speed of (10m/s), so the change in its momentum after the ball bounces in unit (NS) is:
a) 5
b)-5
c)¼
d)zero
The change in the momentum of the ball after the ball bounces back at a speed of 10 m/s, given that its initial speed is 10 m/s is 5 Ns (option A)
How do i determine the change in the momentum of the ball?First, we shall list out the given parameters from the question. Details below:
Mass of ball (m) = ¼ Kg = 0.25 KgInitial velocity (u) = 10 m/sFinal velocity (v) = 10 m/sChange in momentum =?The change in the momentum of the ball can be obtained as follow:
Change in momentum = m(v + u) (since there is a rebound)
Change in momentum = 0.25 × (10 + 10)
Change in momentum = 0.25 × 20
Change in momentum = 5 Ns
Thus, we can conclude that the change in the the momentum of the ball is 5 Ns (option A)
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a projectile is fired horizontally with an initial speed of 57 m/s. what are the horizontal and vertical components of its displacement 3.0 s after it is fired?
After 3.0 seconds, the projectile will have a horizontal displacement of 171 meters and a vertical displacement of 44.1 meters.
When a projectile is fired horizontally, its initial vertical velocity is zero. However, the horizontal velocity remains constant throughout its motion. We can calculate the horizontal and vertical components of displacement after 3.0 seconds using the following equations:
Horizontal component of displacement:
d_horizontal = v_horizontal * t
Vertical component of displacement:
d_vertical = v_vertical * t + (1/2) * g * t^2
Since the projectile is fired horizontally, the horizontal velocity (v_horizontal) remains constant at the initial speed. Thus, the horizontal component of displacement is:
d_horizontal = v_horizontal * t
= (57 m/s) * (3.0 s)
= 171 m
The vertical component of velocity (v_vertical) will increase due to the effect of gravity. Therefore, we need to calculate the vertical component of displacement using the equation:
d_vertical = v_vertical * t + (1/2) * g * t^2
Since the initial vertical velocity is zero, the equation simplifies to:
d_vertical = (1/2) * g * t^2
= (1/2) * (9.8 m/s^2) * (3.0 s)^2
= 44.1 m
Thus, the vertical component of displacement after 3.0 seconds is 44.1 meters.
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1. A toroid filled with a magnetic substance carries a steady current of 1.76 A. The coil contains 1450 turns, has an average radius of 2.19 cm. The magnetic field through the toroid is 0.199333 T. Assume the flux density is constant. What is the magnetic field strength H within the core in the absence of the magnetic substance? Answer in units of A/m.
2. Determine the permeability of the core material. Answer in units of Wb/A m.
The magnetic field strength H within the core in the absence of the magnetic substance is 0.000001671 A/m.
The permeability of the core material is 227684.8 Wb/A m.
A toroid filled with a magnetic substance carries a steady current of 1.76 A.
The coil contains 1450 turns
average radius of coil is 2.19 cm.
The magnetic field through the toroid is 0.199333 T.
Assume the flux density is constant.
The magnetic field B through the toroid is
B = μHnI
Circumference of toroid
= 2πr
= 2 x π x 0.0219 m
= 0.1377 m
Mean length of toroid,
l = Circumference = 0.1377 m
Total number of turns,
N = 1450
n = 1450 / 0.1377
n = 10526.7 turns/m
1.
Using above values,
B = μHnI
H = B / (μnI)
H = 0.199333 / (μ x 10526.7 x 1.76)
H = 0.000001671 A/m
Hence, the magnetic field strength H within the core in the absence of the magnetic substance is 0.000001671 A/m.
2.
The permeability of the core material is,
μ = B / (HnI)
μ = 0.199333 / (0.000001671 x 10526.7 x 1.76)
μ = 227684.8 Wb/A m
Therefore, the permeability of the core material is 227684.8 Wb/A m.
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what is the emf of a battery that does 0.50 jj of work to transfer 6.0×10−2 cc of charge from the negative to the positive terminal?
The emf (electromotive force) of a battery is the potential difference between the two terminals of the battery when the circuit is open and no current is flowing. Therefore, the emf of the battery is 8.33 V.
It represents the maximum voltage that the battery can provide to a circuit when it is connected. An emf of a battery that does 0.50 J of work to transfer 6.0 × 10⁻² C of charge from the negative to the positive terminal can be calculated as follows: We know that the work done by the battery, W = 0.50 J
Charge transferred from the negative to the positive terminal, q = 6.0 × 10⁻² C, emf of the battery is given by the formula: emf = W/q
Substituting the values in the above formula we get, emf = W/q= 0.50 J/(6.0 × 10⁻² C)emf = 8.33 V. The emf of a battery can be calculated using the above formula where emf represents the potential difference between the two terminals of the battery, W represents the work done by the battery, and q represents the charge transferred from the negative to the positive terminal.
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Divers get "the bends" if they come up too fast because gas in their blood expands, forming
bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 25,000
kPa, then rises to a depth where his blood has a pressure of 5000 kPa, what will be the
volume in liters of gas in his blood?
Answer:
V= 0.25L
Explanation:
how would the strength of the force between the moon and earth change if the mass of the moon were somehow made two times greater than its actual mass? be specific, how many times larger or smaller would it be. explain your reasoning.
The force between the Moon and the Earth is calculated using the formula [tex]F = G(m_1*m_2)/r^2[/tex], where F is the force between the two objects, G is the universal gravitational constant, [tex]m_1[/tex] and [tex]m_2[/tex] are the masses of the two objects, and r is the distance between the two objects. Therefore, if the mass of the Moon is somehow made two times greater than its actual mass, the force between the Moon and the Earth would also increase.
To calculate the exact increase in force, we can use the same formula and compare the force before and after the increase in mass. Let's assume that the mass of the Moon is m before the increase and 2m after the increase. We can then use the formula to calculate the force before and after the increase, as follows:
- Before: [tex]F_1 = G\frac{mM}{r^2}[/tex]
- After: [tex]F_2 = G\frac{2mM}{r^2}[/tex]
To compare the two forces, we can divide [tex]F_2[/tex] by [tex]F_1[/tex]:
[tex]\frac{F_2}{F_1} = [G\frac{2mM}{r^2} ]/[G\frac{mM}{r^2} ][/tex]
[tex]\frac{F_2}{F_1} =2[/tex]
Therefore, the force between the Moon and the Earth would become two times greater if the mass of the Moon were somehow made two times greater than its actual mass. This is because the force of gravity is directly proportional to the masses of the objects involved.
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If an object possesses 500 J of potential energy, how much work is needed to lift this object?
a) 500 J
b) 250 J
c) 150 J
d) 1000 J
Answer:
a) 500 J
Explanation:
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where,
P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
In Science, the potential energy possessed by an object or body is the same as the work done by the object or body.
Since we know that the object possessed 500 Joules of potential energy; it would ultimately require to do a work of 500 Joules to lift the object.
Mathematically, work done = force * distance
But force = mass * acceleration due to gravity
F = mg; d = h
Substituting into the work done formula, we have;
Hence, Workdone = Fd = mgh
Compare the motions of transverse, longitudinal, and combined waves.
Explanation:
transfer to babes are always at your advice by a particular motion being a particular wave motion along didn't wave is a wave which particular is a medium move a direction parallel to the direction of the wave moves something that is similar in the surveys on the medium moves of the same direction and bathe an accident to one or two Dimensions do in London killing babe attacks in one dimension and transverse waves attacks in two Dimensions the Waze cannot be paralyzed or organized
Give Reason.......Why we use life jackets when we go in the water ???
Answer:
so we don't drown and die.
Explanation:
A solid insulating sphere has total charge Q and radius R. The sphere's charge is distributed uniformly throughout its volume. Let r be the radial distance measured from the center of the sphere.
If E = 440 N/C at r=R/4, what is E at r=4R?
The electric field (E) at r=4R is approximately 27.5 N/C. It is important to note that this calculation assumes a uniformly charged sphere and that the charge distribution is maintained throughout the volume of the sphere.
The electric field due to a uniformly charged sphere at a point outside the sphere can be calculated using the equation:
E = (k * Q * r) / (R^3)
Where:
E is the electric field at a distance r from the center of the sphere,
k is the electrostatic constant (k = 8.99 × 10^9 N m^2/C^2),
Q is the total charge of the sphere,
r is the radial distance from the center of the sphere, and
R is the radius of the sphere.
Given:
E at r=R/4 is 440 N/C.
We need to find E at r=4R.
To find E at r=4R, we can use the concept of electric field being inversely proportional to the cube of the distance.
Using the relationship:
E1 * r1^3 = E2 * r2^3
We can substitute the given values:
(440 N/C) * [(R/4)^3] = E2 * (4R)^3
Simplifying the equation:
(440 N/C) * (R^3 / 64) = E2 * (64R^3)
(R^3) cancels out, and we can solve for E2:
E2 = (440 N/C) * (64 / 64)
E2 = 440 N/C
Therefore, the electric field at r=4R is approximately 27.5 N/C.
The electric field at a radial distance of 4R from the center of a uniformly charged insulating sphere, with a known electric field of 440 N/C at r=R/4, is approximately 27.5 N/C. This relationship is derived from the formula for the electric field due to a uniformly charged sphere, which states that the electric field is inversely proportional to the cube of the distance from the center of the sphere.
By using the given electric field at r=R/4 and applying the relationship, we can find the electric field at r=4R. It is important to note that this calculation assumes a uniformly charged sphere and that the charge distribution is maintained throughout the volume of the sphere.
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When you look into your bathroom mirror, are you upside down
(inverted) or right side up (upright)?
Is this a real or virtual image?
Why?
What is the focal length of a bathroom (flat) mirror?
Answer:
When the image distance is positive, the image is on the same side of the mirror as the object, and it is real and inverted. When the image distance is negative, the image is behind the mirror, so the image is virtual and upright.
Explanation:
can you produce a real image using a 15cm and a 10cm convex lense together
Yes, it is possible to produce a real image using a 15cm and a 10cm convex lens together by properly selecting the focal lengths and positioning of the lenses .
To determine the characteristics of the image formed by two lenses, we need to consider the lens formula and the lens-maker's formula. The lens formula is given by:
1/f = 1/v - 1/u,
where f is the focal length of the lens, v is the image distance, and u is the object distance.
Let's assume the 15cm lens has a focal length of f1 and the 10cm lens has a focal length of f2. To produce a real image, the object should be placed beyond the focal point of the first lens (f1). Suppose the object distance (u1) is greater than f1. Then, using the lens formula for the first lens:
1/f1 = 1/v1 - 1/u1.
The image formed by the first lens (I1) will act as the object for the second lens. The image distance (u2) of the second lens will be equal to the image distance (v1) of the first lens. Applying the lens formula for the second lens:
1/f2 = 1/v2 - 1/u2.
To find the overall image distance (v2) and the characteristics of the image formed by the two lenses, we need to solve these equations simultaneously.
By properly selecting the focal lengths and positioning of the lenses, it is possible to produce a real image using a 15cm and a 10cm convex lens together. The specific characteristics of the image, such as its size, orientation, and location, can be determined by solving the lens formula equations for the given lens parameters.
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A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s to the west. The launching rod strikes the pinball and causes it to move in the opposite direction with a velocity of 10.0 m/s. What impulse was delivered to the pinball by the launcher?
Answer:
I = 1.8 N s, it is directed towards the right
Explanation:
For this exercise we use the relationship between momentum and moment
I = Δp
F t = p_f - p₀
in this case the initial velocity is v₀ = - 2,0 m / s and final velocity v_f = 10,0 m / s, we assume the positive right direction
I = m (v_f - v₀)
let's calculate
I = 0.150 (10.0 - (-2.0))
I = 0.150 (10 + 2)
I = 1.8 N s
as the impulse is positive it is directed towards the right