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?"

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

The given parameters:

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.**

Learn more about **conservation **of **mechanical energy** here: https://brainly.com/question/6852965

**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 forceAs 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 NSubstituting 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]

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.

**Answer: Your answer Is A)**

**Explanation: **

**Its direction of deflection shows it possitively charged**

**It brings one element into another by bombardment(transmutation)**

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

The Correct answer is Equilibriumthe earth's cycle cycle tend to maintain the balance or equilibrium of the earth by maintaining the proper amount of water , nitrogen , carbon , phosphorous on earth . Each of the comonent of earth's cycle is equally important because they support life on earth. They are basis of life as carbon is found in abundant amount in all the life forms. Phosphorous, nitrogen and Water is equally important in the human beings as well as in plants and animals Hence , they maintain the equilibrium of earth

Please help me.............................

**Answer:**

[tex]a[/tex]

**Explanation:**

is a corect anser

Qué velocidad –en m/s– tiene un móvil, que recorre 15 km en 20 minutos

(es para hoy por faaaa)

**Answer:**

**Explanation:**

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

half-life questionnnnn:

120 and 150 and 180 and 210

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?

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

**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.

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

**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.

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

**Answer:**

pressure in liquids is given as:

P= hpg

where h is the depth

where p is the density

where g is 10

**Explanation:**

From the formula above

p = 10 X 1000 X 10

p = 100000N/m

What is the amplitude of this wave ?

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.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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!

**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]

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

**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

v = R ω

ω = 16.3 / .6 = 27.2 / sec

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).

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

The given parameters:

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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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.

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

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

**Answer:**

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

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?

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.

From the parameters given:

Mass = 105 kgForce = 625 NBy 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/²acceleration = **5.95 m/s² **

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Help me outttt pls and thanks

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.

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.

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.

**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

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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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

**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**

Imagine that there is a small rocky body caught by Earth’s gravity. Draw a comic-strip cartoon

to illustrate its journey as it travels through space toward Earth, enters Earth’s atmosphere, and

lands on Earth. Describe your illustration with narration or speech/thought bubbles. Include the

use of these key terms: atmosphere, meteor, meteoroid, meteorite.

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.

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

Select one:

True

False

**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

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?

**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**

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?

**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.

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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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

**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

How does an emergency action plan benefit your workplace

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

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

**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.

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

**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

A fan blade Spins at 3,000 revolutions per minute.

How

many degrees does it rotate in one second?

18,000 degrees in one second i believe

SOMEONE PLEASE HELP MEEEEEEE

**Answer:**

Weathering and erosion

**Explanation:**

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

**Answer:**

26 g's

**Explanation:**

Hope this helps~

Have a great day

Zero

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