A 12-kg block on a horizontal frictionless surface is attached to a light
spring (force constant = 0.80 kN/m). The block is initially at rest at its
equilibrium position when a force (magnitude P = 80 N) acting parallel to
the surface is applied to the block, as shown. What is the speed of the
block when it is 13 cm from its equilibrium position?"


Answers

Answer 1

The speed of the block at the displacement from the equilibrium position is 1.062 m/s.

The given parameters:

Mass of the block, m = 12 kgSpring constant, k = 0.8 kN/mExtension of the spring, x = 13 cm = 0.13 mApplied parallel force, F = 80 N

The speed of the block is calculated by applying the principle of conservation of mechanical energy as shown below;

[tex]\frac{1}{2} mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{800 \times 0.13^2}{12} } \\\\v = 1.062 \ m/s[/tex]

Thus, the speed of the block at the displacement from the equilibrium position is 1.062 m/s.

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Answer 2

Answer:

The speed of the block at the displacement from the equilibrium position is 1.1266 m/s.

Step-by-step explanation:

Solution :

Using principle of conservation of mechanical energy formula to find the speed of the block :

[tex]\begin{gathered} \longrightarrow{\pmb{\sf{\frac{1}{2} mv^2 = \frac{1}{2}kx^2}}}\end{gathered}[/tex]

»» m = Mass of the block, »» k = Spring constant,»» x = Extension of the spring»» F = Applied parallel force

As per given data information in the question we have :

✧ Mass of the block = 12 kg✧ Spring constant = 0.8 kN/m✧ Extension of the spring = 0.13 m✧ Applied parallel force = 80 N

Substituting all the given values in the formula to find the speed of the block

[tex]\longrightarrow{\sf{ \: \:\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: \cancel{\dfrac{1}{2}}mv^2 = \cancel{\dfrac{1}{2}}kx^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: mv^2 = kx^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v^2 = \dfrac{kx^2}{m}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: \sqrt{{v}^{2} } = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13}^{2}}{12}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13} \times 0.13}{12}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times 0.0169}{12}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{13.52}{12}}}}[/tex]

[tex]{\star{\underline{\boxed{\rm{\red{ v \approx 1.1266 \: m/s}}}}}}[/tex]

Hence, the speed of block is 1.1266 m/s.

[tex] \rule{300}{1.5}[/tex]


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Just like our bodies, Earth's cycles tend to maintain a balance or equilibrium .

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The Correct answer is Equilibrium

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a plane crashes with a deceleration of 185 m/s. How many g’s is this?

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Answer:

26 g's

Explanation:

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The density of water is 1000 kg/m the pressure pf water at 10 m depth is about

Answers

Answer:

pressure in liquids is given as:

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Explanation:

From the formula above

p = 10 X 1000 X 10

p = 100000N/m

If you have to measure the temperature > above 80°c, do you use an alcohol or mercury thermometer? why?

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Answer:

I use mercury thermometer to measure the temperature above 80°C because the boiling point of alcohol is 78°C . So, it can't measure the temperature above 78°C.

A fan blade Spins at 3,000 revolutions per minute.
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Imagine that there is a small rocky body caught by Earth’s gravity. Draw a comic-strip cartoon

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To create this comic strip you can use a narration describing each step and illustrate each one with one image or drawing.

Creating a comic strip involves using images and short texts to explain a specific idea or phenomenon. In the case of the process for a meteor to enter Earth you can use the following ideas.

A meteoroid approaches the Earth at high speed and draw a meteor traveling near to different planets and approaching Earth.What is that? and draw the Earth wondering who or what is approaching.The meteoroid enters the atmosphere of the Earth and becomes a meteor and draw the rocky body burningThe rocky body crashes with the surface becoming a meteorite and draw the zone where the meteorite crashed.

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Which of the following is most likely to be a secondary source

Answers

Answer:

analyze, assess or interpret an historical event, era, or phenomenon,.

Explanation:

Secondary sources are works that analyze, assess or interpret an historical event, era, or phenomenon, generally utilizing primary sources to do so. Secondary sources often offer a review or a critique. Secondary sources can include books, journal articles, speeches, reviews, research reports, and more.

What is the amplitude of this wave ?

Answers

Amplitude of a wave is the maximum displacement of the particles of a medium from their mean positions during the propagation of a wave.Here, the maximum displacement of the particles during wave propagation is equal to the height of the crests, i.e., EN or the depth of the troughs, i.e., MC.So, EN and MC are the amplitudes of the wave.The measure of EN is 1 mm.Hence, the amplitude of the wave is 1 mm.Note : I haven't considered -1 mm because amplitude cannot be negative.

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Which is true of gamma radiation? O A. It increases the number of protons. O B. It is the heaviest of the three types. O C. It does not cause transmutation. O D. It has a positive charge.

Answers

Answer:  Your answer Is A)

Explanation:    

Its direction of deflection shows it possitively charged

It brings one element into another by bombardment(transmutation)

______ uses radioactive materials to create contrast in the body and help form images of the structure and function of organs.

Answers

Answer:

Nuclear medicine or PET scanning (Positron Emission Tomography)

Explanation:

Not super sure what your question is looking for but I think it can be either. PET is a technology used in the nuclear medicine field, and nuclear medicine is a broad field of using the technology the question described.

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Answer:

Weathering and erosion

Explanation:

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Anything can happen at any moment. And if you have a plan for anything it will make things easier for you.


A wheel with radius 41.5 cm rotates 5.13 times every second.
Find the period of this motion.
What is the tangential speed of a wad of chewing gum stick to the rim of the wheel?

Answers

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:

[tex]v = 2\pi\cdot r\cdot f[/tex] (1)

Where:

[tex]r[/tex] - Radius of the wheel, in centimeters[tex]f[/tex] - Frequency, in hertz

If we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:

[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]

[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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Which is an electromagnetic wave A. The waves that heat a cup of water in a microwave oven B. A flag waving in the wind C. Turning on a flashlight D.The changes in the air that result from blowing a horn

Answers

Answer:

The answer would be A. The waves that heat a cup of water in a microwave oven

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms will emit electromagnetic radiation and visible light can be observed. If this light passes through a diffraction grating, the resulting spectrum appears as a pattern of four isolated, sharp parallel lines, called spectral lines. Each spectral line corresponds to one specific wavelength that is present in the light emitted by the source. Such a discrete spectrum is referred to as a line spectrum.
What is the wavelength of the line corresponding to n =4 in the Balmer series? Express your answer in nanometers to three significant figures. EVO AV Om ? X (n) = 4.86.10? By the early 19th century, it was found that discrete spectra were produced by every chemical element in its gaseous slale. Even though these spectra were found to share the common feature of appearing as a set of isolated lines, it was observed that each element produces its own unique pattern of lines. This indicated that the light emitted by each element contains a specific set of wavelengths that is characteristic of that element. Submit Previous Answers Request Answer X Incorrect; Try Again; 19 attempts remaining

Answers

Answer:a)   λ = 4.862 10⁻⁷ m,  b)  λ = 4.341 10⁻⁷ m

Explanation:

The spectrum of hydrogen can be described by the expression

         

in the case of the initial state n = 2 this series is the Balmer series

a) Find the wavelength for n = 4

       

let's calculate

          = 1,097 10⁷ ()

         \frac{1}{ \lambda} = 1.097 10⁷ 0.1875 = 0.2056 10⁷

          λ = 4.862 10⁻⁷ m

b) n = 5

           

    \frac{1}{ \lambda} = 1,097 10⁷ ()

    \frac{1}{ \lambda} = 1.097 10⁷ 0.21 = 0.23037 10⁷

     λ = 4.341 10⁻⁷ m

An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. Find its density.

Answers

Answer:

Weight of object = 11.2 N

Apparent weight = 3.83 N     when immersed

Weight of water displaced = 11.2 - 3.83 = 7.37 N      

d (density) W / V       weight / volume      the weight density

Wo = Vo do    weight of object

Ww = Vo dw    where Ww is weight of equivalent volume of water = 7.37

Wo / Ww = do / dw    dividing previous equations

do = 11.2 / 7.37 dw = 1.52 dw

The density of the object is 1.52 that of water

The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3

So the weight density is 14900 N/m^3

An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.

To find the density, the given values are,

Weight in air = 11.20 N

Weight in water = 3.83 N

density of water = 1000 kg/m³

What is meant by Density?

According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.

Loss in weight of the object = Weight of object in air - weight of object in water

Loss in weight = 11.20 - 3.83 = 7.37 N

Volume of body x density of water x g = 7.37

Let V be the volume of body

V x 1000 x 9.8 =7.37

V = 7.5× 10⁻⁴ m³

Weight in air = Volume of body x density of body x g

11.20 = 7.5× 10⁻⁴ x d x 9.8

d = 1523 kg/m³.

Thus, the density of body is 1523 kg/m³.

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Answers

Answer:

Explanation:

15 km(1000 m / km) / (20 min(60 s/min)) = 12.5 m/s

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Answers

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical height of 19 m. If the sphere has a radius of 0.60 m, what is the angular speed of the sphere at the bottom of the incline plane

Answers

Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

I = 2/5 m R^2     inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

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ω = 16.3 / .6 = 27.2 / sec

A 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2 , what is the coefficient of kinetic friction between the box and the inclined plane?

Answers

Answer:

Explanation:

                            F = ma

mgsinθ - μmgcosθ = ma

      gsinθ - μgcosθ = a

                   μgcosθ = gsinθ - a

                              μ = (gsinθ - a) / gcosθ

                              μ = (9.81sin24 - 0.245) / 9.81cos24

                              μ = 0.4178906...

                              μ = 0.418

The coefficient of kinetic friction will be equal to 0.418.

What is friction?

Friction is the force that prevents solid surfaces, fluid layers, and material elements from sliding against each other. There are various kinds of friction: Dry friction is the force that opposes the relative lateral motion of two in-touch solid surfaces.

Given that a 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2

The coefficient of kinetic friction will be calculated as,

F = ma

mgsinθ - μmgcosθ = ma

gsinθ - μgcosθ = a

μgcosθ = gsinθ - a

Solve for the value of the coefficient of friction,

μ = (gsinθ - a) / gcosθ

μ = (9.81sin24 - 0.245) / 9.81cos24

μ = 0.4178906...

μ = 0.418

Therefore, the coefficient of kinetic friction will be equal to 0.418.

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A mass of 0.01 kg of steam at a quality of 0.9 is contained in the cylinder as shown in Figure 1 below. The spring just touches the top of the piston. Heat is then added until the air expanded, and the spring is compressed by 15.7 cm. Calculate the final temperature of the steam in this cylinder. Sketch the process on a P-y and T-v diagrams with respect to saturation lines.​

Answers

A mass of 0.01 kg of steam at a quality of 0.9 is contained in the cylinder as shown in Figure 1 below. The spring just touches the top of the piston. Heat is then added until the air expanded, and the spring is compressed by 15.7 cm. Calculate the final temperature of the steam in this cylinder. Sketch the process on a P-v and T-v diagrams with respect to saturation lines K = 50 kN/m 160 kg 20 cm Figure 1 Piston cylinder with spring

A body is moving along a circular path with variable speed, it has both radial and tangential acceleration.

Select one:
True
False

Answers

Answer:

True;   ar = v^2 / R      Radial acceleration because it moves in a circular path

    at = α R = tangential acceleration because its speed changes

a = at + ar    total acceleration equals sum of radial and tangential

Help me outttt pls and thanks

Answers

ANSWER: C. Weight
EXPLANATION: Weight is a non-contact force because gravity exerts its force through a field. An object does not need to be touching the Earth to have a weight.

2. The system shown is accelerated by applying a tension Ti to the right-most cable. Assume the system is
frictionless. Solve for the tension in the cable between the blocks, T2, in terms of T. (not a).

Answers

The tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex]

The given parameters:

Mass of the first block, = 2 kgMass of the second block, = 5 kg

The net force on the block system is calculated by applying Newton's second law of motion as follows;

Total mass of the block-system = 2 kg + 5 kg = 7 kg

The acceleration of the block-system;

[tex]a = \frac{T_i}{7}[/tex]

The tension in the cable between the blocks, T₂, is calculated as;

[tex]T_2 = m_2 a\\\\T_2 = 2a\\\\T_2 = 2 \times \frac{T_i}{7} \\\\T_2 = \frac{2T_i}{7}[/tex]

Thus, the tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex].

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AP Physical problem the wording is really throwing me off and im totally lost on how to do this. I would love some help please and thank you! ​

Answers

Explanation:

a) Here is the free-body diagram. Note that I included the components of the weight mg (shown in dotted arrows) for use in the other parts of the problem.

b) The component of the weight parallel to the plane (shown in the diagram as a dotted arrow along the x-axis) is [tex]mg\sin15[/tex] and it is equal to

[tex]mg\sin15 = (25\:\text{kg})(9.8\:\text{m/s}^2)\sin15 = 63.4\:\text{N}[/tex]

c) Applying Newton's 2nd law to the y-axis, we can write

[tex]y:\;\;\;N - mg\cos15 = 0 \Rightarrow N = mg\cos15[/tex]

[tex]N = (25\:\text{kg})(9.8\:\text{m/s}^2)\cos15 = 236.7\:\text{N}[/tex]

d) The component of the weight mg into the plane is the same as the normal force, hence it's also 236.7 N.

e) To solve for the coefficient of friction, we apply Newton's 2nd law to the x-axis:

[tex]x:\;\;\;mg\sin15 - F_f = 0[/tex]

[tex]\Rightarrow F_f = mg\sin15\;\;(2)[/tex]

where [tex]F_f[/tex] is the frictional force defined as [tex]F_f = \mu N[/tex] so we can use Eqn(1) on Eqn (2) to write

[tex]\mu (mg\cos15) = mg\sin15[/tex]

Solving for [tex]\mu,[/tex] we get

[tex]\mu = \dfrac{\sin15}{\cos15} = \tan15 = 0.27[/tex]

Your friend's Frisbee has become stuck 19 m above the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 4.1 m/s when it hits the Frisbee.

If you release the rock 1.8 m above the ground, with what minimum speed must you throw it?

Answers

Answer:

18.36 m/s

Explanation:

We can solve this using conservation of energy. The energy in the system will be conserved since there are no outside forces acting upon it so the potential energy and kinetic energy will be equal. Giving us this formula to start:

1/2mv^2=mgh

m=mass

g=gravity

h=height

v=velocity

We can start by figuring out the total height the rock travels which we can do by subtracting the height of the frisbee by the height the rock started at.

19m-1.8m=17.2m

Now we can plug in our variables to solve for velocity.

First we negate mass since its on both sides and cancels out leaving us with.

1/2v^2=gh

Plug in.

1/2v^2=(9.8)(17.2)

1/2v^2=168.56

v^2=337.12

v=18.36m/s

When a hot metal cylinder is dropped into a sample of water, the water molecules

Answers

Answer:

I believe the answer is speed up.

Explanation:

this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll

If a person climbed Mt. Everest has a mass of 105 kg and a weight of 625 N what would be the acceleration due to gravity?

Answers

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

Force = mass × acceleration.

From the parameters given:

Mass = 105 kgForce = 625 N

By replacing the given values into the above equation, we can determine the acceleration.

625 N = 105 kg × acceleration.

[tex]\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}[/tex]

acceleration = 5.95 N/kg

Since 1 N/kg = 1 m/²

acceleration = 5.95 m/s²

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A gazelle is grazing while standing in a fixed location. When it's startled by a predator, the gazelle accelerates uniformly for 16.8 s until it reaches a speed of 21.9 m/s. The gazelle then runs in a straight line at constant speed for an additional 16.4 s. Finally, it uniformly slows to a stop at a rate of 1.8 m/s/s. What is the total distance traveled by the gazelle in meters

Answers

Answer:

Explanation:

acceleration phase

average speed was 21.9/2 = 10.95 m/s

distance covered is 10.95 m/s(16.8 s) = 183.96 m

distance at top speed 21.9 m/s(16.4 s) = 359.16 m

distance while decelerating (0² - 21.9²)/(2(-1.8)) = 133.225 m

total = 183.96 + 359.16 + 133.225 = 676.345 = 676 m

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