The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha particles directed at the foil
(1) passed through the foil
(2) remained trapped in the foil
(3) were deflected by the nuclei in gold atoms
(4) were deflected by the electrons in gold atoms
Answers
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted. While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive. And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
Answer:
Option 1
Explanation:
The correct answer is option 1
The gold foil experiment was conducted by Rutherford. This experiment was conducted to study the Atom.
In the experiment Alpha rays from the emitter are passed through gold foil and there was a receiver that was present there to intercept the alpha rays.
The outcome of the result was that most of the alpha particle pass through foil undeflected and very few rays revert back on the original path from the heavy mass present at the center.
Later this heavy mass was known Nucleus.
Hence, most alpha particles passed through the foil.
Related Questions
A particle P is projected from the origin O so that it moves along the x-axis. At time t after projection, the velocity of the particle v is given by v = 3t^2 -24t +45. ) Find an expression for the displacement of P in the first 3 seconds of its motion using calculus(integration).
Answers
Answer:
x = t³ − 12t² + 45t
Explanation:
Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 24t + 45) dt
x = t³ − 12t² + 45t + C
The particle is originally at the origin, so at t = 0, x = 0.
0 = (0)³ − 12(0)² + 45(0) + C
0 = C
x = t³ − 12t² + 45t
You have just been appointed as director of your company's corporate training division. The CEO of your company has been displeased with your company's prior training programs, so you are tasked with rehauling the entire training division. You convene a meeting of all training division managers to decide on the types of training that the division will implement. One of your managers is a firm supporter of e-training programs for employees in your company's international offices. He touts the benefits of e-programs by stressing that ________.
Answers
Answer:
He touts the benefits of e-programs by stressing that:
E-training are highly adjustable and flexible as well as employees can complete their training according to their ease.
Explanation:
The concept of E-training is becoming more and more popular now-a-days as it has the following advantages:
You can train the people all over the world with the help of internet.You can complete the training by remaining in your comfort zone.You can get the feedback of the people in a structured manner in less time.You can evaluate the performance in the real time.Due to these reasons, the manager wants the company to do the e-training programs.
How fast must a 3000 kg elephant move to have the same kinetic energy as a 65 spinter running at 10ms?
Answers
Q. : How fast must a 3000 kg elephant move to have the same kinetic energy as a 65.0 kg sprinter running at 10m/s?
Answer:
The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter
Explanation:
Kinetic Energy: This is the energy of a body in motion. It can be expressed mathematically as
Ek = 1/2mv²
Since the kinetic energy the elephant = the kinetic energy of the sprinter.
1/2m₁v₁² = 1/2m₂v₂².......................... Equation 1
Making v₁ the subject of the equation,
v₁ = √(m₂v₂²/m₁)................................. Equation 2
Where v₁ = speed of the elephant, m₁ = mass of the elephant, v₂ = speed of the sprinter, m₂ = mass of the sprinter.
Given: m₁ = 3000 kg, m₂ 65 kg, v₂ = 10 m/s
Substituting these values into equation 2,
v₁ =√(65×10²/3000)
v₁ = [tex]\sqrt{(6500)/3000[/tex]
v₁ =[tex]\sqrt{2.167}[/tex]
v = 1.47 m/s
Therefore The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter
The part of the building structure, typically below grade, upon which all other construction is built is known as:________
Answers
Answer:
horizontal structural member that supports a floor. Beams are typically wood, cold formed metal framing or steel.
Joists
Horizontal timbers, beams or bars supporting a floor.
The following images show the four terrestrial planets in our solar system (not to scale). Rank these planets from left to right based on the atmospheric pressure at the surface, from highest to lowest.
Answers
Answer:
Venus, Earth, Mars, Mercury
Explanation:
The atmospheric pressure is determined by the gravitational attraction of the planet over the atmosphere.
Since the gravitational force is defined as:
[tex]F = G\frac{m1m2}{r^{2}}[/tex] (1)
Notice from equation 1, how the gravitational force depends on the masses of the two objects that are interacting and the distance to the square between them.
For this particular case, m1 is the mass of the planet and m2 is the mass of the gasses in the atmosphere. While the r is the distance between those two.
Venus has a more dense atmosphere and than the Earth, so it has a higher atmospheric pressure. The same case can be seen for the Earth and Mars.
Mercury is the one with the lower atmospheric pressure since it has a thin atmosphere as a consequence of his small size.
The Atmospheric pressure on terrestrial planets, highest to lowest, are generally Venus, Earth, Mars, and Mercury. These rankings can change based on the specific positioning of the planets in the provided images.
The terrestrial planets in our solar system, from Mercury through Mars, have significant differences in atmospheric pressure due to several factors like their distance from the sun, their composition, size, and presence of a magnetic field. Based on the data we have:
Venus has the highest atmospheric pressure among terrestrial planets, which is about 92 times the pressure of Earth's atmosphere at sea level.
Earth comes next. Its atmospheric pressure, also known as barometric pressure, varies depending on altitude but averages about 1 bar at sea level.
Mars has a very thin atmosphere with a surface pressure less than 1% of Earth's.
Mercury, being the smallest and closest to the Sun, has virtually no atmosphere and thus has the lowest surface atmospheric pressure among the terrestrial planets.
Please note that the ranking could change if the images in question show the planets placed differently from the order mentioned.
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When entering a bicycle lane to make a right turn, within how many feet must you enter the lane before making the turn?
Answers
Answer:
200 feet
Explanation:
Most streets in US with bike lane, especially San Francisco, when entering a bicycle lane to make a right turn, you must enter 200 feet to the lane before making the turn.
Typically, when making a right turn, you should enter the bicycle lane no more than 200 feet before the turn. Always yield to cyclists before merging into the lane. These rules can vary by location.
Explanation:Regarding the query about bicycle lanes and right turns, each city or state can have different regulations, but a common rule in many regions is that a driver intending to make a right turn must enter the bicycle lane no more than 200 feet before making the turn. This allows for a safe and orderly flow of both motorized and non-motorized traffic. It's important to check if this rule applies specifically to your region though, as variations can and do exist. Always make sure to yield to any cyclists in the bike lane before merging into it.
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When the researchers connected the solution-filled glass plates of the flow chamber to the AC generator, the ITO-coated plates mostly likely functioned as:
Answers
Answer:
A capacitor
Explanation:
From experimental findings,the plates are connected to a 10Hz ac generator and a potential difference of 1.5V.The ITO is a transplant conducting material,charges accumulates on it when a voltage is applied to it.Exactly what is seen on a parallel plate capacitor
A degausser creates a magnetic field that erases data from magnetic storage media?
Answers
Explanation:
Degausser is a device used by automatically manipulating the alignment of magnetic domains on the medium to avoid data stored on computer and laptop hard disks, floppy disks, and magnetic cassette. There was a mistake. Therefore, a degausser is used to remove all audio, video and data signals from magnetic store media completely.
A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.00 × 106 m and a mass of 3.95 × 1023 kg. The cannon fires a projectile straight up at 2000 m/s. An observation satellite orbits the planet at a height of 1000 km. What is the projectile's speed as it passes the satellite?
Answers
Answer:
The projectile's speed as it passes the satellite is 1497.8 m/s.
Explanation:
Given that,
Radius of planet [tex]r=5.00\times10^{6}\ m[/tex]
Mass of planet [tex]m=3.95\times10^{23}\ kg[/tex]
Speed = 2000 m/s
Height = 1000 km
We need to calculate the projectile's speed as it passes the satellite
Using conservation of energy
[tex]E_{1}=E_{2}[/tex]
[tex]\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}[/tex]
[tex]\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}[/tex]
[tex]-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})[/tex]
[tex]v_{2}^2=v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})[/tex]
[tex]v_{2}=\sqrt{v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{2000^2+2\times6.67\times10^{-11}\times3.95\times10^{23}(\dfrac{1}{5.00\times10^{6}+1000\times10^{3}}-\dfrac{1}{5.00\times10^{6}})}[/tex]
[tex]v_{2}=1497.8\ m/s[/tex]
Hence, The projectile's speed as it passes the satellite is 1497.8 m/s.
Which telescope has a better (smaller) angular resolution: a 2-m telescope observing visible light (wavelength 5.0×10-7 m) or a 10-m radio telescope observing radio waves (wavelength 2.1×10-2 m)?
Answers
The 2-m telescope observing visible light has a better (smaller) angular resolution than the 10-m radio telescope observing radio waves.
Explanation:In order to determine which telescope has a better (smaller) angular resolution, we need to calculate the angular resolution for each telescope. The formula for angular resolution is given by θ ≈ 1.22λ/D, where λ is the wavelength of light and D is the diameter of the telescope's aperture.
For the 2-m telescope observing visible light with a wavelength of 5.0×10-7 m, the angular resolution is approximately 1.22×(5.0×10-7)/(2) = 3.05×10-7 radians.
For the 10-m radio telescope observing radio waves with a wavelength of 2.1×10-2 m, the angular resolution is approximately 1.22×(2.1×10-2)/(10) = 2.54×10-3 radians.
Therefore, the 2-m telescope observing visible light has a better (smaller) angular resolution than the 10-m radio telescope observing radio waves.
Final answer:
The 2-m optical telescope observing visible light has a better (smaller) angular resolution compared to the 10-m radio telescope observing radio waves.
Explanation:
To determine which telescope has a better (smaller) angular resolution, we can use the formula for resolving power (θ), which is given by: θ ≈ λ / D, where λ is the wavelength and D is the diameter of the telescope's aperture. For the 2-m telescope observing visible light (λ = 5.0×10⁻⁷ m), the angular resolution would be θ ≈ 5.0×10⁻⁷ m / 2 m = 2.5×10⁻⁷ radians. For the 10-m radio telescope observing radio waves (λ = 2.1×10⁻² m), the angular resolution would be θ ≈ 2.1×10⁻² m / 10 m = 2.1×10⁻³ radians. Thus, the 2-m optical telescope has a smaller (better) angular resolution than the 10-m radio telescope.
To understand how the two standard ways to write the general solution to a harmonic oscillator are related.
There are two common forms for the general solution for the position of a harmonic oscillator as a function of time t:
x(t)=Acos(ωt+ϕ) and
x(t)=Ccos(ωt)+Ssin(ωt).
Either of these equations is a general solution of a second-order differential equation (F⃗ =ma⃗ ); hence both must have at least two--arbitrary constants--parameters that can be adjusted to fit the solution to the particular motion at hand. (Some texts refer to these arbitrary constants as boundary values.)
A)
Find analytic expressions for the arbitrary constants C and S in Equation 2 (found in Part B) in terms of the constants A and ϕ in Equation 1 (found in Part A), which are now considered as given parameters.
Give your answers for the coefficients of cos(ωt) and sin(ωt), separated by a comma. Express your answers in terms of A and ϕ.
b)
Find analytic expressions for the arbitrary constants A and ϕ in Equation 1 (found in Part A) in terms of the constants C and S in Equation 2 (found in Part B), which are now considered as given parameters.
Express the amplitude A and phase ϕ (separated by a comma) in terms of C and S.
Answers
Final answer:
a. C = A cos(φ), S = -A sin(φ).
b. A = √(C²+S²), φ = atan2(S, C).
Explanation:
Both expressions given for the general solution of a harmonic oscillator can be related to each other by using trigonometric identities.
The first expression, x(t)=Acos(ωt+φ), involves an amplitude A and a phase shift φ, whereas the second one, x(t) = Ccos(ωt) + Ssin(ωt), includes two parameters C and S for the coefficients of cosine and sine respectively.
To convert between the two forms, we use the cosine of a sum identity, which states that:
cos(a + b) = cos(a)cos(b) - sin(a)sin(b).
For part (A), expressing Acos(ωt+φ) in the form of Ccos(ωt) + Ssin(ωt), we get:
C = A cos(φ),
S = -A sin(φ).
For part (B), deriving A and φ in terms of C and S from the second equation, we use the Pythagorean identity and the definitions of cosine and sine to get:
A = √(C²+S²),
φ = atan2(S, C).
The constants C and S can be expressed in terms of A and ϕ as C = A cos(ϕ) and S = -A sin(ϕ). Conversely, A and ϕ can be expressed in terms of C and S as A = √(C² + S²) and ϕ = tan⁻¹(-S/C).
Explanation:
Part A: Expressing C and S in terms of A and ϕ
The given forms are:
x(t) = A cos(ωt + ϕ)x(t) = C cos(ωt) + S sin(ωt)We can use the trigonometric identity for the sum of angles to rewrite A cos(ωt + ϕ):
x(t) = A cos(ωt + ϕ) = A[cos(ωt) cos(ϕ) - sin(ωt) sin(ϕ)]Comparing this with x(t) = C cos(ωt) + S sin(ωt), we identify:
C = A cos(ϕ)S = -A sin(ϕ)Part B: Expressing A and ϕ in terms of C and S
From the relationships above:
A = √(C² + S²) (using the Pythagorean theorem)ϕ = tan⁻¹(-S/C) (finding the angle ϕ from the ratio of S and C)2 Physics Questions WILL GIVE BRAINLIEST!!
A book is being pushed along a table at a constant speed. The coefficient of static friction is 0.75, and the coefficient of kinetic friction is 0.30. The book has a mass of 0.50 kg. What is the force of friction?
0.15 N
1.5 N
3.7 N
0.38 N
Suppose the mass of each of two spherical objects is doubled, and the distance between their centers is three times as large. How is the magnitude of the force of gravity between them affected?
It is 4/9 as large
It is 3/2 as large
It is 4/3 as large
It is 2/3 as large
It is 2/9 as large
Answers
Answer:
a) The force of friction is 1.5 N.
b) The new force is 4/9 times as large as the original force.
Explanation:
It is given that the book is moving on the table.
∴ The friction acting on the block should be kinetic.
∴ Coeffecient of kinetic friction , k = 0.3
The force of friction = kN , where 'N' is the normal force acting on the surface where friction is acting.
In this case , N = mg , where 'm' is the mass of the book and 'g' is the acceleration due to gravity.
∴ Force of friction , f = kmg = 0.3×0.5×10 = 1.5N
b) We know that ,
Gravitational force , F ∝ m , where 'm' is the mass of the body.
F ∝ [tex]x^{2}[/tex] where 'x' is the distance between the two bodies.
Since mass of both the bodies is doubled , the force will be 4 times as large compared to the original force F from the first proportionality equation.
However , as the distance is 3 times as large , the force will be 9 times as small as compared to the original force F
∴ The new force is 4/9 times as large as the old force
A stone is thrown vertically upward from the roof of a building.Does the position of the stone depend on the location chosen for the origin of the coordinate system?
Answers
Displacement is the difference between two co-ordinates so the origin doesn't matter.
Explanation:
Displacement is a physical quantity that depicts the change of position in a particle, this further is measured by the difference in position vectors limiting the duration under these two limits.Displacement = Velocity × TimeInstantaneous velocity = d/dt (position vector)
d = vt + 1/2 at² is the displacement acceleration equation.
d = u + at is the velocity displacement equation.
Displacement always has continuity with velocity and Acceleration.The average velocity is 0 if the displacement is 0.
A particle moves along the x-axis with velocity given by v(t)=3t2+6t for time tâ¥0. If the particle at position x=2 at time t=0, what is the position of the particle at time t=1?
a) 4
b) 6
c) 0
d) 11
e) 12
Answers
Answer:
b) 6
Explanation:
Given
v(t)=3t²+6t
X(0) = 2
X(1) = ?
Knowing that
v(t)=3t²+6t = dX/dt
⇒ ∫dX = ∫(3t²+6t)dt
⇒ X - X₀ = t³ + 3t²
⇒ X(t) = X₀ + t³ + 3t²
If X(0) = 2
⇒ X(0) = X₀ + (0)³ + 3(0)² = 2
⇒ X₀ = 2
then we have
X(t) = t³ + 3t² + 2
when
t = 1
X(1) = (1)³ + 3(1)² + 2
X(1) = 6
After a time, t =1, the position of the particle has been 6. Thus, the correct option is b.
The velocity of the particle moving along the x-axis has been given by:
[tex]\rm v(t)\;=\;3t^2\;+\;6t[/tex]
The differentiation of v(t) in terms of x will be:
[tex]\rm \dfrac{dx}{dt}\;=\;3t^2\;+\;6t[/tex]
[tex]\rm \int dx\;=\;\int (3t^2\;+\;6t)\;dt[/tex]
differentiation with the limits if X be x and [tex]\rm x_0[/tex]:
x - [tex]\rm x_0[/tex] = [tex]\rm t^3\;+\;3t^2[/tex]
In the given question, the value of [tex]\rm x_0[/tex] = 2:
At time t = 0
x = [tex]\rm t^3\;+\;3t^2\;+\;x_0[/tex]
x = [tex]\rm (0)^3\;+\;3(0)^2\;+\;2[/tex]
x = 2.
To find the position of the particle at time =1, given, [tex]\rm x_0[/tex] = 2.
x = [tex]\rm (1)^3\;+\;3(1)^2\;+2[/tex]
x = 1 + 3 + 2
x = 6.
Thus, after time, t =1, the position of the particle has been 6. Thus, the correct option is b.
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The term meteorology:
A) can be used interchangeably with climate because they have the same meaning.
B) is the study of the atmosphere and its related weather systems.
C) is the study of the long-term average weather conditions at a given location.
D) is the study of meteors and their effects on the atmosphere.
Answers
Final answer:
Meteorology is the study of the atmosphere and weather patterns, aiming to predict weather in the short term. It is different from climatology, which deals with long-term climate trends, and is a part of the broader field of atmospheric science.
Explanation:
The term meteorology refers specifically to the study of the atmosphere and its various phenomena, including weather patterns. This scientific field encompasses the processes and forces that contribute to the weather and aims to predict weather in the short term, which is crucial for numerous aspects of daily life and safety. Despite some common misconceptions, meteorology does not equate to the study of meteors, and it should not be confused with climatology, which is the study of climate, or long-term weather patterns, over extended periods such as decades, centuries, and millennia.
Climatology and meteorology both fall under the broader umbrella of atmospheric science, which combines these and other disciplines that focus on the atmosphere.
Therefore, the correct answer to the student's question would be option B: Meteorology is the study of the atmosphere and its related weather systems.
You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 13.2 cm. When the cylinder is rotating at 1.63 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?
Answers
Answer:
centripetal acceleration[tex](a_{c})=13.8m/s^{2}[/tex]
Explanation:
convert 1.63rev/sec to rpm by multiplying by 60
= 1.63*60=97.8rpm
Convert this to rad/sec
1rpm =π/30 rad/sec
97.8rpm = 97.8 * (π/30 rad/sec)
=10.25rad/sec
linear velocity= angular velocity *radius
radius =13.2cm=13.2/100=0.132m
v=rω
v= 0.132*10.25
v=1.35m/s
centripetal acceleration = [tex]\frac{v^{2} }{r}[/tex]
[tex]a_{c}=\frac{1.35^{2} }{0.132}[/tex]
[tex]a_{c}=13.8m/s^{2}[/tex]
Answer:
[tex]a_{c}[/tex] = 13.8 m/s².
Explanation:
The acceleration centripetal [tex]a_{c}[/tex] is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex] (1)
where v: is the tangential speed and r: is the container radius
The tangential speed is equal to:
[tex] v = \omega \cdot r [/tex] (2)
where ω: is the angular velocity
Since 1 revolution is equal to 2π rad, the velocity (equation 2) is:
[tex] v = 1.63 \frac{rev}{s} \cdot \frac{2\pi rad}{1rev} \cdot 0.132m = 1.35m/s [/tex]
Now, by entering the velocity value calculated into equation (1) we can find the acceleration centripetal:
[tex] a_{c} = \frac{(1.35m/s)^{2}}{0.132m} = 13.8m/s^{2} [/tex]
I hope it helps you!
The combined mass of the asteroids in the Solar System is comparable to the mass of__________________.
Answers
Answer:
25 times less than the Moon's mass
Explanation:
The asteroids are small, rocky objects that orbit the Sun. Although asteroids orbit the Sun like planets, they are much smaller than planets.
There are millions of boulder-size and larger rocks that orbit the Sun, most of them between the orbits of Mars and Jupiter.
The largest asteroid is Ceres with a diameter of 1000 kilometers. Pallas and Vesta have diameters of about 500 kilometers and about 15 others have diameters larger than 250 kilometers
Although in the Solar System there are millions the asteroids, the combined mass of all of the asteroids is about 25 times less than the Moon's mass.
Then the answer for your question is:
25 times less than the Moon's mass
bibliographic source: www.astronomynotes.com
Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ignored) from their low-flying planes and scoring points for having the weights land close to a target. A plane 65 m above the ground is flying directly toward a target at 46 m/s .
A.) At what distance from the target should the pilot drop the weight?B.) The pilot looks down at the weight after she drops it. Where is the plane located at the instant the weight hits the ground?not yet over the targetpast the targetdirectly over the targetnot enough information to determine
Answers
Answer:
a)165,6 m
b)Directly over the target
Explanation:
a)To calculate the time that ball hits the ground after released, we will use the free fall formula below:
[tex]h=(1/2)*g*t^2[/tex]
h=65m
g=9,8m/s^2
[tex]65=0,5*9,8*t^2\\t^2=65/(0,5*9,8)\\t=3,6 sec[/tex]
Plane has the speed of 46m/s. Due to there is no air resistance weight remains the same speed while free falling. So the distance will be:
[tex]x=46*3,6=165,6 m[/tex]
b) Due to no air resistance plane and weight will be at the same speed on the air. Therefore, when the weight hits the ground plane will be directly over the target.
The pilot should drop the weight 165.6 m prior to the target, and the plane will have passed the target by the time the weight lands.
Explanation:
In Physics, we know that the motion of the dropped weight is governed by the equations of free fall, specifically that the time t for an object to fall from rest under gravity g, ignoring air resistance, from a height h is given by t = sqrt(2h/g). In this case, g = 9.81 m/s2 (earth's acceleration due to gravity), and h= 65 m. Therefore, if we put these values into the formula, we get t = sqrt(2*65/9.81) ≈ 3.6s.
For part A, the distance from the target where the pilot should drop the weight is the same as the distance the plane will travel in the time it takes for the weight to hit the ground, which is the product of the plane's speed (46m/s) and the time of flight t of the weight. Therefore, the distance is 46*3.6 = 165.6 m.
The answer for part B is that the plane will have passed the target when the weight hits the ground as it continues to move forward at 46 m/s.
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The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass 40.0 kg resting on the output plunger. The piston and plunger are nearly at the same height, and each has a negligible mass. By how much is the spring compressed from its unstrained position?
Answers
The spring in the hydraulic chamber is compressed by 24.5 cm from its unstrained position when a 40.0 kg rock is resting on the output plunger. This is calculated using Hooke's Law.
Explanation:The key to solving this problem is understanding that the force exerted by a spring follows Hooke's Law, which states that F = -kx, where F is the force, k is the spring constant, and x is the displacement of the spring from its equilibrium position. Given that the spring constant k = 1600 N/m and the force exerted by the rock of mass 40.0 kg is F = mg = 40.0 kg * 9.81 m/s² = 392.4 N, we can find the displacement as follows:
x = F/k = 392.4 N / 1600 N/m = 0.245 m or 24.5 cm
Therefore, the spring is compressed by 24.5 cm from its unstrained position when the 40.0 kg rock is resting on the output plunger.
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The spring is compressed by 24.5 cm from its unstrained position.
Explanation:The amount by which the spring is compressed from its unstrained position can be calculated using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In this case, the force exerted by the spring is equal to the weight of the rock.
Using the equation F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring, we can rearrange the equation to solve for x:
x = F / k = (m * g) / k
Plugging in the values, we have x = (40.0 kg * 9.8 m/s²) / 1600 N/m.
Simplifying the equation, we get x = 0.245 m = 24.5 cm.
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If we found an Earth-sized planet orbiting very close to another star (causing it to be very hot), which planet would its surface resemble the most?
Answers
Answer:
Venus is definitely the answer
Explanation:
If we found an Earth-sized planet orbiting very close to another star (causing it to be very hot), which planet would its surface resemble the most?
Venus is known as the earth twin because it is similar in shape to the earth. Although, it is not the closest to the sun, it traps the suns heat in its atmosphere making it the hottest planet in the solar system. the temperature is as high as 465 degrees Celsius. Venus is known to be the second brightest body in the solar system, the moon being the first. A day in venus could be longer than a year on the earth. It revolves in the opposite direction around the sun compared to other planets. It has an atmospheric pressure which ninety times grater than what we have here on earth.
A steel scale measures the length of a copper rod as 80cm when both are at 20°c , the caliberation temperature of the scale . what would the scale read for the length of the rod when both are at 40°c,
given coefficient of linear expansion of steel =1.1×10^-5°c^-1
coefficient of linear expansion of copper =1.7×10^-5°c^-1
a. 80.0096 cm
b. 80.0272 cm
c. 1 cm
d. 25.2 cm
Answers
Answer:
a) 80.0096 cm
Explanation:
This is an exercise of thermal expansion, we must calculate how much each material is dilated and the difference is the expansion of the measurement
The expression for linear dilation
ΔL = α L₀ ΔT
Let's start with the steel
ΔL / L₀ = 1.1 10⁻⁵ (40 -20)
ΔL / L₀ = 22 10⁻⁵
Copper
ΔL / L₀ = 1.7 10⁻⁵ (40-20)
ΔL / L₀ = 34.0 10⁻⁵
What made the copper bar high is the differentiated being read
ΔD = ΔL / L₀_copper - ΔL / L₀_Steel
The initial length of the steel balance is made equal to the length of the copper rod
ΔL = (34 - 22) 10⁻⁵ / 80
ΔL = 0.0096
For which the final length of the copper bar
[tex]L_{f}[/tex] = L₀ +ΔL
[tex]L_{f}[/tex] = 80 +0.0096
[tex]L_{f}[/tex] = 80.0096 cm
The correct answer is a
A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the door’s center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.
Answers
Each hin-ge exerts a horizontal force of 165 N on the door.
Explanation:To find the horizontal components of the force exerted on the door by each hin-ge, we need to divide the weight of the door evenly between the two hing-es. The weight of the door is 330 N, so each hin-ge will support 330 N / 2 = 165 N.
Next, we need to determine the horizontal component of the force exerted by each hin-ge. Since the door's center of gravity is at its center and the hin-ge is 0.50 m from the top and 0.50 m from the bottom, the horizontal distance between the hin-ge and the center of gravity is the same for both hin-ges.
Using the formula for torque, τ = force × distance, we can calculate the torque exerted by the weight of the door about each hin-ge. Since the torque is equal to the force times the perpendicular distance from the axis of rotation to the line of action of the force, and the horizontal component of force is perpendicular to the distance, we can set up the following equation: torque = force × distance
Solving for the horizontal component of force, we have: force = torque / distance
By plugging in the torque value for each hin-ge and the distance value, we get:
force = 165 N × 0.50 m / 0.50 m = 165 N
Therefore, the horizontal components of force exerted on the door by each hin-ge are both 165 N.
At an age of 380,000 years, the temperature of the universe had fallen to 3000 K, and electrons could then combine with protons to produce hydrogen gas instead of roaming freely through space.
What major transition occurred as a consequence of this change in the universe at this time?
a) The present laws of physics were applicable to the properties of the universe time.
b) The universe became transparent to light for the first time.
c) Nuclear fusion no longer occurs below this temperature, and so, general fusion throughout the universe would have ceased.
d) The universe would have lost its electrical charge suddenly to become elect.
Answers
Answer:
The major transition occurred as a consequence of this change in the universe at this time is that b)The universe became transparent to light for the first time.
Explanation:
For the first 380,000 years or so, the universe was essentially too hot for light to shine. The heat of creation smashed atoms together with enough force to break them up into a dense plasma, an opaque soup of protons, neutrons and electrons that scattered light like fog. Then 380,000 years after the Big Bang, matter cooled enough for atoms to form during the era of recombination, resulting in a transparent, electrically neutral gas.
This set loose the initial flash of light created during the Big Bang, which is detectable today as cosmic microwave background radiation. However, after this point, the universe was plunged into darkness, since no stars or any other bright objects had formed yet.
You must give a signal either by hand and arm or by a signal device:______
A) Only if other traffic is affected by your movement
B) Only if you are driving a truck or car
C) Only at night
D) Anytime you change lanes
Answers
Answer:
D) Anytime you change lanes
Explanation:
Correct way to execute a change of direction maneuver :
1 - The driver must warn by means of optical signals any maneuver that implies a lateral or backward displacement of his vehicle, as well as his intention to immobilize it or to slow down his gear in a considerable way. Such optical warnings will be made well in advance of the start of the maneuver, and, if they are bright, they will remain in operation until the end of the maneuver.
2 - Unless the road is conditioned or signposted to perform it in another way, it will be as close as possible to the right edge of the road, if the change of direction is to the right, and to the left edge, if it is to the left and the road is One way. If it is on the left, but the road through which it circulates is two-way traffic, it will adhere to the longitudinal mark of separation between the senses or, if it does not exist, to the axis of the road, without invading the area destined to the Wrong Way; when the road is two-way traffic and 3 lanes, separated by dashed longitudinal lines, should be placed in the center. In any case, the placement of the vehicle in the appropriate place will be carried out with the necessary advance and the maneuver in the least possible space and time.
3 - If the change of direction is on the left, it will leave the center of the intersection on the left, unless it is conditioned or marked to leave it on the right .
You must give a signal either by hand and arm or by a signal device anytime changing lanes. The correct option is (D).
Traffic lights, turn signals, horn blowing, and hand gestures are examples of signals used in the context of driving and transportation to express intentions and maintain order on the road.
Providing signals while driving is important for the safety of oneself and others. Signals are a safety measure to avoid accidents.
It is crucial to give other drivers advance notice of your intended lane change in order to safeguard their safety. Other drivers can anticipate your actions and modify their driving accordingly if you signal. It is a fundamental component of prudent and secure lane switching.
Hence, You must give a signal either by hand and arm or by a signal device anytime changing lanes. The correct option is (D).
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An amplifier has an input power of 5 mW. The output power is 100 mW. What is the decibel gain to the nearest decibel?
Answers
Answer:
The decibel gain to the nearest decibel = 13 db
Explanation:
Gain of an amplifier: This is defined as the measure of the ability of an amplifier to increase the power of a signal from the input to the output port by adding energy converted from some power supply to the signal.
It is represented mathematically as.
db = 10×log (p₂/p₁)............................. Equation 1
Where dp = decibel gain of the amplifier, p₁ = input power of the amplifier, p₂ = output power of the amplifier.
Given: p₁ = 5 mW, p₂ = 100 mW.
Substituting these values into equation 1
db = 10×log(100/5)
db = 10×log(20)
dp = 10 ×1.301
dp = 13 db gain
Therefore the decibel gain to the nearest decibel = 13 db
What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.
Answers
Incomplete question.The complete question is
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.
Answer:
Total gauge pressure= 3724Pa or 3.724kPa
Explanation:
Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.
Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.
Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa
When you shoulder your shotgun, what part of your body should fit snugly against the stock?
Answers
Answer:
Cheek
Explanation:
Typically the scope is a bead on the gun's edge. Our eye must be in alignment with the muzzle, so that the proper placement of our head on the stock is crucial. The stock will fit snugly to our cheek as we put the pistol to our face with our head on other side just above gun's center line.
The correct answer is cheek.
Tarzan swings on a 31.0-m-long vine initially inclined at an angle of 42.0° with the vertical. What is his speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s? (Pick the answer closest to the true value.)A. 13.9 m/sB. 12.5 m/sC. 8.4 m/sD. 6.9 m/sE. 11.0 m/s
Answers
Answer:
Speed of Tarzan at the bottom of the swing is 12.5 m/s.
Explanation:
Given that,
Length of the vine, L = 31 m
The swing is inclined at an angle of 42 degrees with the vertical. We need to find the speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s. using the conservation of mechanical energy as :
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]gh=\dfrac{1}{2}v^2[/tex]
h is the height of the height.
[tex]h=L-L\ cos\theta[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2gL(1-\ cos\theta)}[/tex]
[tex]v=\sqrt{2\times 9.8\times 31(1-\ cos(42))}[/tex]
v = 12.492 m/s
or
v = 12.5 m/s
So, his speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s is 12.5 m/s. Hence, this is the required solution.
Option b.
By making use of the conservation of mechanical energy principle and using trigonometric relations, we find that Tarzan's final speed as he reaches the bottom of his swing is approximately 12.55 m/s, hence answer choice (B) 12.5 m/s is the closest.
Explanation:To solve this problem, we need to recognize that this is a problem about the conservation of mechanical energy. Initially, Tarzan has both kinetic energy (because he pushes off with a speed of 6.00 m/s) and potential energy (because he is at a height above the lowest point of the swing). At the bottom of the swing, all of this energy will have been transformed into kinetic energy, as he is now at the lowest point of the swing, and hence has no potential energy.
The kinetic energy at the top of the swing is (1/2)*m*(6.00 m/s)^2 and the potential energy at the top of the swing is m*g*h. We can write the height 'h' in terms of the length of the vine (31.0 m) and the angle it makes with the vertical, theta (42.0°), as h = L*(1 - cos(theta)) = 31.0 m*(1 - cos(42.0°)).
As mechanical energy is conserved, the sum of kinetic and potential energy at the top of the swing will equal the kinetic energy at the bottom of the swing, which we can write as (1/2)*m*v_f^2. Equating and simplifying gives us a value of v_f = sqrt(2*g*L*(1 - cos(theta))(2*9.8 m/s^2*31.0 m*(1 - cos(42.0°)))).
Popping these numbers into a calculator gives an answer of approximately 12.55 m/s. Hence, the closest answer, and the one we should choose, is (B) 12.5 m/s.
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Federal Law mandates that the steering or helm area of a power boat less than 20 feet in length must have which of the following?a. Latex gloves.b. Capacity Plate.c. Ladder stand.d. Rear
Answers
Answer:Federal Law mandates that the steering or helm area of a power boat less than 20 feet in length must have Capacity Plate. The correct option is B.
Explanation: The capacity Plate indicates the maximum weight capacity and/or the maximum number of people that the boat can carry safely in good weather. Information on the capacity Plate includes the maximum number of adult persons, the maximum gross load, and the maximum size of engine, in horsepower,
Answer:
b. Capacity Plate.
Explanation:
According to the Federal Law, boats of less than 20 feet in length need to have a capacity plate that has to be in the steering area all the time. Also, this capacity plate has to show the maximum people capacity, weight and horsepower recommended. Because of this, the answer capacity plate.
A train has a constant speed of 10 m/s around a track with a diameter of 45 m what is the centripal acceleration?
Answers
Answer:
[tex]a_{c}= 4.44\frac{m}{s}[/tex]
Explanation:
When an object goes on a circular movement, it has a centripetal acceleration that always points toward the center of the circle, it is the responsible of the change of direction in the movement of the object. and that centripetal acceleration is related with the speed in the next way:
[tex]a_{c}=\frac{v^{2}}{r} [/tex], with v the speed, r the radius of the track that is half of the diameter (22.5 m)
[tex]a_{c}=\frac{10^{2}}{22.5} [/tex]
[tex]a_{c}= 4.44\frac{m}{s}[/tex]