While certain parts of your question cannot be answered without specific data such as the spring constant, we can deduce from the half frequency that the mass of the initial object must also be 300g. More mass on the spring would pull the equilibrium further down but we can't quantify without more details.
Explanation:This question involves simple harmonic motion and Hooke's Law. Hooke's Law is F = -kx where F is the force, k is the spring constant, and x is the displacement. Simple Harmonic Motion deals with oscillations like that of a spring. However, without values for the spring constant (k) or initial speed of the object, we cannot calculate the frequency of the oscillation or speed of the object at any given point.
As for part (c), the frequency of the oscillation halves when a second mass is introduced. This implies that the system's mass has doubled as frequency is inversely proportional to the square root of the mass. Therefore, the initial object's mass would be 300g as well.
For part (d), without any specific figures such as the spring constant, the new equilibrium position below yi cannot be determined. Commonly, the force produced by the weight of the objects will be counteracted by the spring force when the system is in equilibrium. More mass would pull the spring further down until this equilibrium is achieved.
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The frequency of oscillation can be determined using the formula f = 1 / T. The speed of the object when it is 8 cm below the initial position can be calculated using the formula v = A * ω. The mass of the first object can be found by setting up two equations and solving for M. The new equilibrium position is slightly lower than the initial position yi.
Explanation:(a) The frequency of an oscillation can be determined using the formula:
frequency (f) = 1 / period (T)
Since the object oscillates up and down, it completes one full cycle (2π radians) every time it goes from the highest point to the lowest point and back. The period (T) is therefore the time it takes for one complete cycle. The object reaches its lowest position when it is 10 cm below yi, so the total distance traveled is 10 cm. Since the object moves up and down symmetrically, we can calculate the period by dividing the total distance traveled by 2 times the amplitude (A):
T = 2π * sqrt(m / k)
where m is the mass of the object and k is the spring constant. Given that the spring is massless, the mass of the object does not affect the period. Therefore, we can calculate the period as:
T = 2π * sqrt(k)
From the period, we can compute the frequency using the formula provided above.
(b) The speed of the object when it is 8.0 cm below the initial position can be determined using the formula for the speed of an object in simple harmonic motion:
speed (v) = amplitude (A) * angular frequency (ω)
The angular frequency is related to the period by the formula:
angular frequency(ω) = 2π / period (T)
Using the given values, we can calculate the angular frequency and then substitute it into the equation for speed.
(c) When an object of mass 300 g is attached to the first object, the system oscillates with half the original frequency. Let's denote the mass of the first object as M and the spring constant as K. We can set up two equations to solve for the mass of the first object:
Equation 1: f = (1 / (2π)) * sqrt(K / M) (original frequency)
Equation 2: (f/2) = (1 / (2π)) * sqrt(K / (M + 0.3)) (new frequency)
By solving these equations simultaneously, we can find the mass of the first object.
(d) The new equilibrium position, with both objects attached to the spring, is determined by the combined mass of the two objects. Since the first object is now heavier due to the addition of the second object, the equilibrium position will be slightly lower than the initial position yi. The exact displacement can be calculated using the formulas for equilibrium position in simple harmonic motion.
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The total lifetime of the Sun is about 10 billion years, from when it was born to when it dies, and the half-life of 238U is 4.5 billion years. Suppose a particular meteoroid contained 24 micrograms of uranium-238 (238U) when the Sun was born. If the meteoroid is still in orbit when the Sun dies, it will contain _____
Answer:
a little less than 6 micrograms
Explanation:
The half-life of an element is defined as the time in which half of the isotope emits its radiation and becomes a different element. Therefore after 4.5 billion years the meteoroid will contain 12 micrograms of uranium, after 9 billion years will contain 6 micrograms. After 10 billion years will contain a little less than 6 micrograms.
Label 1a lies on the rim of a large crater, and 1b lies on the rim of a smaller one. Which crater must have formed first?
Answer:
Label 1a
Explanation:
The craters are usually referred to as those large depressions that are created due to the impact by meteorites, asteroids or other astronomical bodies. These are of variable size and shape, which are controlled by factors such as the mass of the impact body, velocity at which the impact body strikes.
According to the rule of thumb, the geologically older surfaces are exposed to various types of impacting bodies such the meteoroids, asteroids, and comets, for a longer duration of time in comparison to the surfaces that are younger in age.
The bigger craters are formed first, and the smaller craters are formed later. Thus, the age of the craters is directly proportional to the size of the craters.
Hence, the larger crater (label 1a) was formed first.
The law of superposition implies that the larger crater, labeled as 1a, must've formed first, as new craters form on top of existing surfaces. Therefore, the rim of the larger crater existed prior to the formation of the smaller crater, 1b.
Explanation:In the study of craters, the principle that is usually applied is the Law of Superposition. This law states that in an undisturbed sequence, the oldest layers lie at the bottom and the youngest layers at the top. Applying this to your question: if a crater (labelled 1a) locates on the rim of a large crater, and another crater (labelled 1b) lies on the rim of a smaller one, the larger crater would have formed first. This is because any new crater that forms would be on top of, or superimposed on, the existing surface. Therefore, the rim of the larger crater must have been in existence for the smaller crater (1b) to have formed on it.
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earth's troposphere hydrosphere and lithosphere contain large amounts of which element?
Answer: the element is Oxygen.
Explanation: Earth's troposphere hydrosphere and lithosphere contain large amounts of Oxygen. it is the most abundant element in the Earth's crust. Oxygen makes up 467,100 ppm (parts per million) of the Earth's crust, or 46.6%.
Oxygen is the primary element found in Earth's troposphere, hydrosphere, and lithosphere, as part of the air, water (H2O), and silicate materials in many minerals and rocks.
Explanation:The Earth's troposphere, hydrosphere, and lithosphere all contain large amounts of the element oxygen. The troposphere, part of the atmosphere, contains oxygen in the air we breathe, while the hydrosphere, which encompasses all water on Earth, would have oxygen as a component of H2O. In the lithosphere, which covers the crust and the upper part of the mantle, oxygen is found in many minerals and rocks as part of silicate materials.
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Which layer of the atmosphere absorbs the most ultraviolet rays
Answer:
The stratosphere
Explanation:
In the lower portion of the stratosphere there is a high concentration of ozone in relation to other parts of the atmosphere. This region is known as the ozone layer. This layer absorbs most of the Sun's ultraviolet radiation, especially ultraviolet B rays, allowing ultraviolet A rays to pass.
Answer:
TROPOSPHERE
Explanation:
A particular star cluster contains stars all with the same apparent magnitude of +4. Near the cluster (and at the same distance from the Earth) is a single star with an apparent magnitude of +1. The star's brightness appears to be the same as the collective brightness of the entire cluster.
How many stars are in this cluster?
Answer:
3,000 stars.
Explanation:
Subtract the magnitude of the single star from the magnitude of the star cluster.
(+4) - (+1) = +3
Then multiply your answer by 1000
= 3,000 is a stars.
Why are summer days longer than winter days on earth?
Answer:
Tilt of the Earth
Explanation:
The Earth's rotational axis is titled at an angle 23.4°. When then northern hemisphere is pointed towards the Sun it is summer in the northern hemisphere and winter in the southern hemisphere.
In the northern hemisphere a day would be longer and gets more sunlight than the southern hemisphere.
The same happens when it is summer in the southern hemisphere.
Answer:
The sun is father away during the winter.
Explanation:
In the northern hemisphere, during the winter it is summer in the southern hemisphere. The sun is farther away for us so the days get shorter in the winter since the sun inches away faster.
The resistance (R) of a copper wire varies directly as its length (L). Write this relation as a formula using k as the constant of variation.
Answer:
Explanation:
According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.
i.e. [tex]V\propto R[/tex]
[tex]V=R I[/tex]
where R=resistance
[tex]R\propto L[/tex]
[tex]R\propto \frac{1}{d^2}[/tex]
whee L and d are length and Diameter
thus [tex]R=k \frac{L}{d^2}[/tex]
where k=constant of Variation
What types of compounds are involved in solutions that are used to grow crystals? In your own words please
Answer:
This is a table of common chemicals that produce nice crystals. The color and shape of the crystals are included. Many of these chemicals are available in your home. Other chemicals in this list are readily available online and are safe enough for growing crystals at home or in a school. Recipes and specific instructions are available for hyperlinked chemicals.
Ionic compounds, held together by electrostatic forces, form crystalline structures that exhibit rigid and brittle properties. They tend to have high melting and boiling points, reflecting the strength of the ionic bonds. Though ionic solids are poor conductors in their solid state, they readily dissolve in water, becoming good conductors when melted or dissolved.
Explanation:Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.
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Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object of mass M > m, which is initially at rest on the surface?
The collision is elastic.
A. All of the initial kinetic energy of the less-massive object is lost.
B. The momentum of the objects that are stuck together has a smaller magnitude than the initial momentum of the less massive object.
C. The speed of the objects that are stuck together will be less than the initial speed of the less-massive object.
D. The direction of motion of the objects that are stuck together depends on whether the hit is a head on collision.
Answer:
C. The initial momentum should be equal to the final momentum due to the conservation of momentum.
[tex]P_{initial} = mv_0\\P_{final} = (M+m)v_1\\v_1 = \frac{m}{M+m}v_0[/tex]
Since m/(M+m) < 1, v_1 > v_0.
Explanation:
Wrong -> A. Since the smaller particle still moves after the collision, it has a kinetic energy.
Wrong -> B. The total initial momentum is equal to the momentum of the smaller particle. Therefore, the momentum of the objects that stuck together is equal to that of the smaller object.
Wrong -> D. Since the bigger object is initially at rest and the surface is frictionless, the direction of motion will be the same as the direction of the smaller particle.
Please help, it’s much needed
Answer:
Displacement = 72m
Acceleration = 4 [tex]m/s^{2}[/tex]
Explanation:
The jet ski increases its speed from zero to 24 m/s in 6 seconds.
We are supposed to find the displacement and its acceleration rate.
Displacement is the distance it covers in this time interval.
Acceleration is the rate of increase of speed.
Acceleration = [tex]\frac{change in velocity}{time}[/tex]
Acceleration = [tex]\frac{24 - 0}{6}[/tex]
Acceleration = 4 [tex]m/s^{2}[/tex]
According to equation of motion ,
[tex]v^{2}-u^{2} = 2\times a\times s[/tex]
[tex]24^{2} - 0 = 2\times 4\times s[/tex]
s = [tex]\frac{576}{8}[/tex]
Displacement , s = 72 m
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the amplitude is increased to 40 cm. What is the coefficient of static friction between the two blocks?
Answer:
0.72
Explanation:
[tex]T[/tex] = Time period of oscillation = 1.5 s
Angular frequency is given as
[tex]w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s[/tex]
[tex]A[/tex] = Amplitude of oscillation = 40 cm = 0.40 m
[tex]\mu[/tex] = Coefficient of static friction = ?
[tex]a[/tex] = acceleration of the block
[tex]m[/tex] = mass of the block
Maximum acceleration of the block is given as
[tex]a = Aw^{2}[/tex]
frictional force is given as
[tex]f = \mu mg[/tex]
As per newton's second law
[tex]f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72[/tex]
The coefficient of static friction between the two blocks is 0.72.
How to calculate the coefficient?From the information given, the angular frequency will be:
= 2π/T
= 2(3.14)/1.5
= 4.2rad/s
The maximum acceleration of the block is Aw² and the frictional force is umg. In this case,
umg = ma
u(9.8) = 0.40 × 4.2².
u = (0.40 × 4.2²) / 9.8
u = 0.72
Therefore, the coefficient of static friction between the two blocks is 0.72.
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Consider a telescope with a small circular aperture of diameter 2.0 centimeters.A) If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resolved if their angular separation is 3.0×10−5 radians?
Answer:
[tex]4.91803\times 10^{-7}\ m[/tex]
Explanation:
[tex]\theta[/tex] = Angular seperation = [tex]3\times 10^{-5}\ rad[/tex]
[tex]\lambda[/tex] = Wavelength
We have relation
[tex]sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow \lambda=\dfrac{dsin\theta}{1.22}\\\Rightarrow \lambda=\dfrac{2\times 10^{-2}sin(3\times 10^{-5})}{1.22}\\\Rightarrow \lambda=4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength of the telescope is [tex]4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
Given data:
The diameter of the telescope is, d = 2 cm = 0.02 m.
The angular separation of the two point source of light is, [tex]\theta = 3.0 \times 10^{-5} \;\rm radians = 3.0 \times 10^{-5} \;\rm radians \times \dfrac{\pi}{180}\\\theta = 5.23 \times10^{-7}^{\circ}[/tex].
The angular separation of two point source is also known as apparent distance between the sources, and it is defined as the angle between the two objects as viewed directly by the observer.
The standard relation for the angular separation of telescope is given as,
[tex]sin \theta = 1.22 \times \dfrac{\lambda}{d}[/tex]
here, [tex]\lambda[/tex] is the maximum wavelength of light.
Solving as,
[tex]sin (5.23 \times 10^{-7}) = 1.22 \times \dfrac{\lambda}{0.02}\\\\\lambda = \dfrac{sin (5.23 \times 10^{-7}) \times 0.02}{1.22} \\\\\lambda = 1.49 \times 10^{-10} \;\rm m[/tex]
Thus, we can conclude that the maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
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Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth?
5m
Explanation:
Given parameters:
Weight of object = 50N
Work done in lifting object = 250J
Unknown:
Vertical height = ?
Solution:
The work done on an object is the force applied to lift a body in a specific direction.
Work done = force x distance
Weight is a force in the presence of gravity;
Work done = weight x height of lifting
Height of lifting = [tex]\frac{work done }{weight}[/tex]
Height of lifting = [tex]\frac{250}{50}[/tex] = 5m
The vertical height through which the object was lifted is 5m
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Final answer:
To calculate the vertical height that a 50 N object is moved with 250 J of work done against Earth's gravity, divide the work by the force. This gives a height of 5 meters.
Explanation:
To find the vertical height an object is moved, given the work done against the Earth's gravitational field, we use the formula for work: Work = Force × Distance, where the force in this context is the weight of the object due to gravity. Given the work done (250 J) and the weight of the object (50 N), we can rearrange the formula to solve for distance, which is the vertical height moved: Distance = Work / Force.
Distance = 250 J / 50 N = 5 meters.
Therefore, the object was moved through a vertical height of 5 meters. This calculation assumes that the work done is solely against the gravitational pull of the Earth, and it highlights the direct relationship between work, force, and distance in physical interactions.
When following a motorcycle, drivers must remember that motorcycles can ____________ than other vehicles.
Answer:
Stooping more Quickly
Explanation: Stooping more quickly means that Motor cycles can easily bend and stop,this is due to it's design, Motor cycle have only two wheels,they contain just two tyres one in front and one at the back,they are not as firm on the ground when compared to other vehicles such Motor car. They have a lower weight which makes them easily bent or stopped once there's any barrier.
Motorcycles can stop quicker than other vehicles.
Explanation:When following a motorcycle, drivers must remember that motorcycles can stop quicker than other vehicles. Due to their smaller size and weight, motorcycles have a shorter stopping distance. This means that drivers should maintain a safe following distance to avoid rear-ending the motorcycle.
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The theory of force generation proposed in the passage is best supported by which of the following observations about Amoeba locomotion?
Question:
The theory of force generation proposed is best supported by which of the following observations about Amoeba locomotion?
a. amoeboid movement stops upon expose to cytochalasins
b. amoeboid movement cannot occur if mitosis is blocked
c. moving amoeba cells produce more troponin than do stationary ones
d. the rate of movement is inversely proportional to the viscosity of the medium in the Amoeba moves
Answer:
The amoeba location can be supported by :The amoeboid movement stops upon expose to cytochalasins
Explanation:
The amoeboid movement common in eukaryotic cells. cytochalasins are the type of drug that inhibits the growth of the micro filaments, micro filaments capability to elongate are both implemented as important to make the force for movement in amoeba. further Cytochalasins are known as the fungal metabolites which can bind the actin filaments and also restrict polymerization and elongation of actin. It can also permeate cell membranes, preventing the cellular translocation and cause cells to enucleate.
Earl is using his hands to hold a metal pan 10 centimeters above a hot burner. How can this scenario be changed to demonstrate conduction between the pan and the burner?
Answer:
Earl needs to touch the pan to the burner.
Explanation:
Earl is demonstrating heat transfer by conduction.
Conduction is the transition of heat energy by direct contact. Hotter molecules have more momentum and they move faster.These hot molecules collide with the cooler ones, transfer their energy to cooler particles , this process is continued to pass it on to other molecule near to them.
Earl needs to touch the pan to the burner.
Answer:
D the answer is D
Explanation:
Tarzan steps out of his house and grabs a hanging vine to swing to the ground. If his house is 11.3 m above the ground, what is Tarzan's speed when he reaches the ground?
Answer:
v = 14.88 m / s
Explanation:
For this exercise let's use energy conservation. We place the reference system on the ground
Start. Tarzan before jumping
Em₀ = U = m g h
Final. Upon reaching the ground
[tex]Em_{f}[/tex]= K = ½ m v²
Energy is conserved
Em₀ = [tex]Em_{f}[/tex]
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 11.3)
v = 14.88 m / s
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
A. on the half of the road with traffic going in the opposite direction
B. on the half of the road that the vehicle is traveling
C. anywhere on a two-way street
Answer: the correct option is B ( on the half of the road that the vehicle is traveling).
Explanation: according to Georgia Code About Pedestrians; The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is
on the half of the road that the vehicle is traveling.
Answer:
B
Explanation:
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
A. on the half of the road with traffic going in the opposite direction
B. on the half of the road that the vehicle is traveling
C. anywhere on a two-way street
According to the law regrading pedestrian, the pedestrian has the right of way immediately she is crossing on the crosswalk and is half of the road that the vehicle is travelling.
These rules are there for the safety of the driver, pedestrians and other road users
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011 pascals. How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters.
Answer:
[tex]\Delta l=0.015m[/tex]
Explanation:
We have given initial length of the steel guitar l = 1 m
Cross sectional area [tex]A=0.5mm^2=0.5\times 10^{-6}m^2[/tex]
Young's modulus [tex]\gamma=2\times 10^{11}Pa[/tex]
Force F = 1500 N
So stress [tex]=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa[/tex]
We know that young's modulus [tex]=\frac{stress}{strain}[/tex]
So [tex]2\times 10^{11}=\frac{3\times 10^{9}}{strain}[/tex]
[tex]strain=1.5\times 10^{-2}=0.015m[/tex]
Now strain [tex]=\frac{\Delta l}{l}[/tex]
[tex]0.015=\frac{\Delta l}{1}[/tex]
[tex]\Delta l=0.015m[/tex]
Answer:
Explanation:
L = 1 m
A = 0.5 mm² = 0.5 x 10^-6 m²
Y = 2 x 10^11 Pa
F = 1500 N
ΔL = ?
Use the formula for the young's modulus
[tex]Y = \frac{FL}{A\Delta L}[/tex]
[tex]\Delta L = \frac{FL}{AY}[/tex]
[tex]\Delta L = \frac{1500\times 1}{0.5\times10^{-6}\times 2\times 10^{11}}[/tex]
ΔL = 0.015 m
ΔL = 0.02 m
Which element has properties of electrical conductivity and luster and exists as a liquid at STP?
Answer:
Mercury - hydrargyrum (Hg)
Explanation:Mercury is a chemical element with the symbol Hg and atomic number 80. It is commonly known as quicksilver and was formerly named hydrargyrum.
Symbol: Hg
Atomic number: 80
Atomic mass: 200.59 u
Melting point: -38.83 °C
Boiling point: 356.7 °C
Electron configuration: [Xe] 4d^14 5d^10 6s^2
Group :12
Ionization energy 1007.1J
Electronegativity :2
Crust Abundance : 0.085ppm
Electron Affinity : Unknown eV
Name: Mercury , the first planet in the solar system (Hg from former name hydrargyrum , from greek hydr - water and argyros silver .
I hope it helps. :).
A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizontal. To make his job easier, he oiled the bottom of the box, reducing the coefficient of friction to 0.250. What is the magnitude of the acceleration of the box?
Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.
Question:
Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.
A typical asteroid in the asteroid belt, a trojan asteroid, a typical kuiper belt object, a typical oort cloud object
Answer:
Ranking of the objects from left to right based on their orbital period around the sun from shortest to longest
a typical asteroid in the asteroid belta Trojan asteroida typical Kuiper belta typical Oort cloud objectExplanation:
Half the mass of the belt is present in the largest four asteroids, they are ceres, vesta, pallas and hygiea. The asteroid belt total mass is 4% than that of moon, 22% Pluto. The three dwarf planets are present in the Kuiper belt Pluto, Haumea and Makemake. The spherical layer of icy objects around the sun are Oort cloud. Objects in the Oort cloud are made up of water ice, ammonia, and methane. It is spherical from outer and and torous shaped from inside.
A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. All of the following are true EXCEPT:_________
I. the angle of reflection is 30 degrees
II. the angle of refraction is 30 degrees
III. total internal reflection will result, depending on the critical angle
Answer:
All of the following are true EXCEPT, the angle of refraction is 30 degrees
Explanation:
It is given that,
The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. We know that when a ray of white light moves through the air and strikes the surface of water in a beaker, refraction occurs. Let r is the angle of refraction. It can be calculated using Snell's law as :
[tex]n=\dfrac{sin\ i}{sin\ r}[/tex]
[tex]sin\ r=\dfrac{sin\ i}{n}[/tex]
[tex]sin\ r=\dfrac{sin(30)}{1.33}[/tex]
[tex]r=22.08^{\circ}[/tex]
So, the angle of refraction is 22.08 degrees.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. Since, it is incident at an angle of 30 degrees, so the angle of reflection will be 30 degrees. Also, total internal reflection will result, depending on the critical angle.
So, the wrong statement is the angle of refraction is 30 degrees. It is equal to 22.08 degrees.
Let v⃗ A be the velocity of the car at point A. What can you say about the acceleration of the car at that point?
Answer and Explanation:
Let the velocity of the car at a point is A is v
We have to tell about the acceleration at that point
Acceleration at that point will be perpendicular to the velocity and directed inside of the track
We can tell anything about the magnitude of the acceleration as magnitude of acceleration is the rate of change of velocity and here we have no information about time
While a car's velocity at a given point only describes its speed and direction at that moment, its acceleration at that point accounts for changes in speed and direction over time. To ascertain acceleration at a point, the velocities before and after that point need to be compared.
Explanation:The velocity of a car at a given point, such as point A, tells us it's direction and speed at that instant. However, the acceleration of the car at that point is a different matter. Acceleration is defined as the change in velocity per unit time, and it doesn't only involve speed, but also direction. A car could be moving at a constant speed, but if it's changing direction, it's considered to be accelerating. Therefore, to ascertain acceleration at point A, we would need to know not just the velocity at point A, but also at points immediately before and after.
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A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds and in what direction?
Answer:
v = 2 m/s (West)
Explanation:
Given
m = 2 Kg
v initial = 4 m/s (East)
F = 6 N (West)
t = 2 s
We can use the formula
v = v initial + a*t
if F = m*a ⇒ a = F/m = 6 N / 2 Kg
a = 3 m/s² (West)
then
v = v initial + a*t = (4 m/s) + (-3 m/s²)(2 s)
v = - 2 m/s = 2 m/s (West)
The velocity is a vector quantity. The velocity of the given cart is 2 m/s to the west.
To solve the given problem, use the kinematic equation,
[tex]v = s + at[/tex]........................1
Where,
[tex]s[/tex] - initial speed = 4 m/s
[tex]a[/tex] - acceleration
[tex]t[/tex] - time - 2 s
Acceleration can be calculated by the formula,
[tex]F = ma[/tex] or
[tex]a = \dfrac Fm[/tex]
Where,
m = mass = 2 Kg
F = force = 6 N
So,
[tex]a = \dfrac {6}2\\\\a = 3\rm \ m/s^2[/tex]
Put the values in equation 1,
[tex]v = 4 +3\times 2\\\\v = \bold {2 m/s}[/tex]
Therefore, the velocity of the given cart is 2 m/s.
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A solar eclipse will occur Group of answer choices
1. at every full Moon. during every new Moon.
2. at full Moon only when the Moon is on the ecliptic.
3. at new Moon only when the Moon is on the ecliptic.
Answer:
3. at new Moon only when the Moon is on the ecliptic.
Explanation:
Solar eclipse is the condition when the moon comes in between the sun and the earth. In this condition the moon casts its shadow on the earth.Whether the eclipse is a total solar eclipse, a partial solar eclipse or an annular solar eclipse depends on various factors, but the position of the moon must be on the same orbital plane as that of the earth's orbit around the sun.The sun is about 400 times larger than the moon in size and the sun is almost 400 times farther from the earth than the moon is, this makes it possible for the moon to cover the sun completely leading to a complete solar eclipse.As we know that the orbit of the earth around the sun and the orbit of the moon around the earth is elliptical which leads to a variation in the distance from their rotating centers, so not of every eclipse the moon covers the sun completely developing an annular eclipse.When the moon is close enough to the earth on the ecliptic but not completely aligned in between the sun and the earth leads to a partial solar eclipse.A solar eclipse occurs during a new Moon, exclusively when the Moon is on the ecliptic path. This is when the Moon moves between the Earth and the Sun and blocks the sunlight. A solar eclipse can't occur during a full Moon.
Explanation:A solar eclipse happens during a new Moon when the Moon is on the ecliptic path. This event occurs when the Moon moves between the Earth and the Sun and blocks the Sun's light. The Moon must be on the ecliptic, which is the apparent path of the Sun's motion on the celestial sphere, for the alignment to be precise. If the Moon is not on this path, then it will not block the Sun, and a solar eclipse will not occur. It's also important to note that during a full Moon, the Moon is opposite the Sun in our sky, so a solar eclipse cannot occur.
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Please help me I need the answer before tomorrow and Please explain how you got your answer!!! While playing basketball in PE class, 67 kg Johan lost his balance after making a lay-up at 5.4 m/s and collided with the padded wall behind the basket and came to rest in 0.24 seconds. What force acted on his body?
A: -1300
B: -1500
C: -1700
D: -1900
Answer:
Option B
-1500 N
Explanation:
We know that force, F=ma where m is the mass of Johan and a is the acceleration
Since acceleration, [tex]a=\frac {v-u}{t}[/tex] where v and u are final and initial velocities, t is time taken. Considering that he comes to rest, v=0 and u is given as 5.4 m/s, t=0.24 s.
Therefore
[tex]F=ma=m\frac {v-u}{t}=67\times \frac {0-5.4}{0.24}=-1507.5\approx -1500 N[/tex]
The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train average speed over the course of the trip was 100km/h how many kilometers long was the third stage
Answer:
the third stage was 480 km long
Explanation:
Stage 1:
Time = 1 hours
Speed = 80km
Stage 2:
Time = 2 hours
Speed = 200km
Stage 3:
Time = 4 hours
Let the Distance at the stage 3 be x
Average speed of the train route = 100 km/h
So
[tex]\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0[/tex]
[tex]\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100[/tex]
Lets find the speed at stage 1
Speed = [tex]\frac{Distance }{Time}[/tex]
Speed = [tex]\frac{80}{1}[/tex]
Speed 1= 80 km/hr
The speed at stage 2
Speed = [tex]\frac{Distance }{Time}[/tex]
Speed = [tex]\frac{200}{2}[/tex]
Speed 2 = 100 km/hr
The speed at stage 3
Speed = [tex]\frac{Distance }{Time}[/tex]
Speed = [tex]\frac{x}{4}[/tex]
Speed 3 = [tex]\frac{x}{4}[/tex]
we kow that average is ,
[tex]\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100[/tex]
[tex]\frac{ 80 + 100+ \frac{x}{4} }{3} = 100[/tex]
[tex]\frac{ 180 + \frac{x}{4} }{3} = 100[/tex]
[tex]\frac{ \frac{720 +x}{4} }{3} = 100[/tex]
[tex]\frac{720 +x}{4} \times \frac{1}{3} = 100[/tex]
[tex]\frac{720 +x}{12} = 100[/tex]
[tex]720 +x = 100 \times 12[/tex]
[tex]720 +x = 1200[/tex]
[tex]x = 1200- 720[/tex]
x = 480
To calculate the length of the third stage of the train's journey, we use the average speed over the total trip. Summing the known distances of the first two stages and subtracting this from the total distance traveled, which is 700 km, gives us the length of the third stage as 420 km.
Explanation:To find the length of the third stage in a train's journey when the average speed over the entire trip is known, we can use the formula for average speed, which is the total distance traveled divided by the total time taken. According to the given problem, the first two stages took 1 and 2 hours and were 80 km and 200 km long, respectively. Adding the times gives us 3 hours for the first two stages.
The third stage took 4 hours, making the total time for the full journey 7 hours (3 hours + 4 hours). Since the average speed of the train over the course of the trip was 100 km/h, we multiply this by the total time to find the total distance. Thus, the total distance is 700 km (100 km/h * 7 h).
To find the distance covered in the third stage, we subtract the sum of the first two stages (80 km + 200 km) from the total distance. The length of the third stage is 700 km - 280 km = 420 km.
Technician A says that the connecting rods should be marked before disassembly. Technician B says that pistons are cam ground so that when operating temperature is reached, the piston will have expanded enough across the piston pin area to become oval shaped. Who is right?
A) Technician A only B) Technician B only
C) Both technicians A and B D) Neither technician A nor B
Answer:
C.
Explanation:
Both technicians A & B are right since the connecting rods must be marked to indicate which surface faces front. And a cam-ground piston machine must also be configured oval shape so that it will be able to fit the cylinder better throughout its operating temperature range.
Technician A is correct that the connecting rods should be marked before disassembly. Technician B is incorrect about the cam ground pistons. The correct answer is A) Technician A only.
Explanation:Technician A is correct that the connecting rods should be marked before disassembly. This is important in order to ensure that the rods are reinstalled in the same location and orientation as before, which helps maintain balance and proper engine function.
Technician B is incorrect. Pistons are not cam ground to become oval shaped when operating temperature is reached. Instead, pistons are typically designed with expansion clearances to account for thermal expansion, ensuring that they maintain proper shape and function at operating temperatures.
Therefore, the correct answer is A) Technician A only.
why is 21-cm radiation so important to the study of interstellar matter and the galaxy
Answer:
Because its the most suitable radiation for that use
Explanation:
Astronomers use the 21 cm radiation for interstellar and cosmological research as this type of radiation is best suitable for the purpose its intended for. It can easily go through gas and dust molecules and other interstellar bodies that causes a blur vision of the content of space.