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 NThe 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:
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)
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:
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.
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 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.
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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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.
Force = mass × acceleration.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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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
How does an emergency action plan benefit your workplace
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
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
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
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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______ 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.
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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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
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
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:
Mass of the first block, = 2 kgMass of the second block, = 5 kgThe 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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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
Help me outttt pls and thanks
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 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
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.
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.Learn more about comic in: https://brainly.com/question/1418309
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
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]
Please help me.............................
Answer:
[tex]a[/tex]
Explanation:
is a corect anser
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
half-life questionnnnn:
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.
Just like our bodies, Earth's cycles tend to maintain a balance or equilibrium .
What is the amplitude of this wave ?
Hope you could get an idea from here.
Doubt clarification - use comment section.