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]
Help me outtttt jejjejejeje
Answer:
do it got a picture
Explanation:
AnswAnswer This!!!!!!
I'll give brainliest to whoever gets it right.
A stereo speaker produces a pure "A" tone, with a frequency of 220.0 Hz.
What is the period of the sound wave produced by the speaker?
T=
What is the wavelength water of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?
λwater=
What is the wavelength air of this sound wave as it travels through air with a speed of about 341 m/s?
λair=
(a) The period of the sound wave is 0.005 s.
(b) The wavelength of the wave when the speed of the wave is 1480 m/s is 6.73 m.
(c) The wavelength of the sound wave as it travels through air is 1.55 m.
The given parameters;
Frequency of the wave, F = 220 HzThe period of the sound wave is calculated as follows;
[tex]T = \frac{1}{f} \\\\T = \frac{1}{220} \\\\T = 0.005 \ s[/tex]
The wavelength of the wave when the speed of the wave is 1480 m/s is calculated as follows;
[tex]v = f\lambda \\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{1480}{220} \\\\\lambda = 6.73 \ m[/tex]
The wavelength of the sound wave as it travels through air with a speed of about 341 m/s;
[tex]\lambda = \frac{v}{f} \\\\\lambda = \frac{341}{220} \\\\\lambda = 1.55 \ m[/tex]
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Which of the following describes the motion of a block while it is in equilibrium? The block:
A. moves at a constant speed
B. slows down gradually to stop
C. speeds up for a bit, then moves at a constant speed
D. Accelerates constantly
The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
EQUILIBRIUM:A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.
In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.
Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
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At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, how much time does it take the puck to come to rest?
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is 4.51 sec.
What is speed?The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.
Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,
force = ma = μmg, putting the value,
a = - 1.175 m/sec²
now using equation of motion
v = u + at
t = 4.51 sec
So, time taken by the puck to come to rest is 4.51 sec.
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help with this please
Answer:
I think that those are all true but I'm not completely sure.
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.
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
what type of stretching is beneficial for sports performance and involves momentum
Answer:
Dynamic stretching
Explanation:
Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.
A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. The momentum of the chair
a. was zero while stationary and increased when the person stood.
b. was greatest while the person sat in the chair.
c. remained the same.
d. was zero when the person got out of the chair and increased while the person sat.
Answer:
a. was zero while stationary and increased when the person stood.
Explanation:
momentum is mass times velocity.
initial velocity was zero
final velocity was NOT zero.
A 10kg object is 15 meters up a hill. Find its potential energy
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J
Hey there!
having a question which is from ( JEE Advanced 2014 )
[tex]\huge\color{black}\boxed{\colorbox{pink}{Question♡}}[/tex]
A rocket is moving in a gravity free space with a constant acceleration of 2 −2 along +x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 −1 relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 −1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is?
[tex] \\ \\ [/tex]
proper explanation needed~
thankyou~
The time taken for the two balls to hit each other is 8 s.
The given parameters:
Acceleration of the rocket, a = 2 m/s²Length of the chamber, s = 4 mSpeed of the first ball, = V1 = 0.3 m/sSpeed of the second ball, V2 = 0.2 m/sThe time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;
[tex](V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the two balls to hit each other is 8 s.
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I need help. please look at the image below and let me know I need this by 7:20 am pst.
Answer:
3(1.5) = 4.5 V
Explanation:
which one would cool down faster, water or a block of concrete? why?
Answer:
If you mean by temperature
Water, cause say the water just sits there, it'll take a few minutes to cool down. But for concrete having all the hot sun rays on it during summer, can make it very hot, if you put 'cold water' there, it'll only last there for a moments (and that's by effect not waiting), if you wait for it to cooldown by itself, it'll take quite awhile.
If you mean solidify:
The stuff required to make concretes takes HOURS and HOURS to finish solidifying, unlike water being turn into ice normally only takes about half an hour or an hour (maybe 2) in the freezer (I don't know how long for ice, but not long at all in comparison to concrete).
Explanation:
Explanation was mostly part of the answer
3. A car travelling at 12 m/s into a stationary truck of about 10 times the cars mass. a. If the collision was completely inelastic, what velocity would the two travel at if the stuck together? b. If the collision was completely elastic, what would be the velocities of the car and truck after the collision? c. In order to exert a force of only 3500N on the truck during the collision, how much time would the collision have to take?
(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.
(b) For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c) The time taken to exert the given force is 0.00625 m (s).
The given parameters;
Initial velocity of the car, u₁ = 12 m/sInitial velocity of the truck, u₂ = 0Mass of the car, = mMass of the truck, = 10m(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]
(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]
Apply one-directional velocity equation:
[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]
Substitute the value of [tex]v_2[/tex] into equation (1);
[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]
Solve for [tex]v_2[/tex];
[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]
Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c)
The change in the momentum of the truck is calculated as;
[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]
The time taken to exert the given force is calculated as follows;
[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]
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For northern hemisphere observers, which celestial object would be above the horizon for the greatest
amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is
70° above the celestial equator, or one that is 40" below the celestial equator? Which one would be above
the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.
Answer:
Explanation:
For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.
For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
What is the equator?The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.
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Explain how the linear rate spring operates?
Which of these do not affect fluid friction?
The surface of an object
The viscosity (thickness) of fluid an object is in
The shape of an object
The weight of an object
A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 25 J. The maximum speed of the block is:
Answer:
Explanation:
easy way
when system is all kinetic energy, velocity is at a maximum
E = ½mv²
v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s
harder way
ω = √(k/m) = √(80/0.5) = √160 rad/s
When the system is entirely spring potential, the amplitude A is
E = ½kA²
A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m
maximum velocity is ωΑ = 0.79√160 = 10 m/s
If you apply a net force of 100 N to the hoverboard, and it accelerates
2m/s/s, how much mass does it have?
Answer:
50 kgExplanation:
The mass of an object given only the force acting on it and it's acceleration can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force in N
a is the acceleration in m/s²
From the question
f = 100 N
a = 2 m/s²
We have
[tex]m = \frac{100}{2} = 50 \\ [/tex]
We have the final answer as
50 kgHope this helps you
A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?
[tex]a_c = 3.14\:\text{m/s}^2[/tex]
Explanation:
First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:
[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]
The centripetal acceleration [tex]a_c[/tex] is defined as
[tex]a_c = \dfrac{v^2}{r}[/tex]
Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as
[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]
[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]
A student catches a can of soda dropped from some unknown height by Mr. Fineman. If
the can was dropped from rest, and it is traveling at 8.9 m/s before it arrives in the
student's hand...
Answer:
v² = u² + 2gh
8.9² = 0² + 2(9.8)h
h = 4.0 m
v = at
t = 8.9/9.8
t = 0.91 s
a heat engine is a device that uses to produce useful work
A heat engine is a device that uses to produce useful work.
Definition - a device for producing motive power from heat, such as a gasoline engine or steam engine.
So..
If this is a true or false question.. Your answer is:
TRUE
Answer:
HEAT
Explanation:
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 wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with just enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon
Answer:
Explanation:
An impulse results in a change of momentum
If the wagon and dog both stop, they must have had equal and opposite momentums
FΔt = mΔv
F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)
v₁ = t₀ = 0
F = m(v₀)/t₁
F = 55(2.1)/0.1 = 1155 N
We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.
WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!
A 225 g hockey puck is sliding on ice in an arena towards the end boards that are 15.7 m away. The puck is travelling 12.0 m/s when it slides into some rough ice (coefficient of kinetic friction= 0.550).
Determine:
a) the acceleration of the puck on the rough ice.
b) the distance from the end boards the puck is when it comes to a stop.
Please show work.
Answer:
Explanation:
a) the acceleration of the puck on the rough ice.
a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²
(comes from μ = F/N = ma/mg = a/g)
b) the distance from the end boards the puck is when it comes to a stop.
v² = u² + 2as
0² = 12.0² + 2(-5.40)s
s = 13.3 ft
so distance from the boards is
15.7 - 13.3 = 2.4 m
by the way...that's some VERY rough ice...more like sand.
Which car is experiencing negative acceleration?
A. a car sitting at a red light
B. a car slowing down as it approaches a red light
C. a car driving in a circle at a constant speed
D. a car changing its speed from 0 km/hr to 10 km/hr
Answer:B
Explanation:
Answer: b
Explanation: b as it slows down and decelerates
The force shown in the figure(Figure 1) moves an object from x = 0 to x = 0.75 m.
1/How much work is done by the force?
2/How much work is done by the force if the object moves from x = 0.20 m to x = 0.55 m ?
Answer:
(a) The force changes its magnitude with respect to displacement, hence the total work will be sum of increment of work in three steps:-
step 1 . from 0 to 0.25m .
force = 0.6 N
displacement= 0.25m
work done =( force × displacement) = (0.25 × 0.6 ) = 0.15 joule.
step 2:- .
work done in moving from 0.25 to 0.50 m.
work done = ( force × displacement) = ( 0.4 × 0.25) = 0.10 Joule. .
step 3 :-
work done in moving from 0.50 to 0.75 m
work done = 0.8 × 0.25 = 0.200 joule.
hence total work done = ( 0.20+0.10+0.15) = 0.45 joule. ans
(b) similar concept you have to use here also.
step 1:
from 0.20 to 0.50, force of magnitude 0.4 N acts on the object.
Work done = ( 0.50-0.20)× 0.4 = 0.30 × 0.4 = 0.12 joule.
step :- 2
from 0.50 to 0.55 , force of magnitude 0.8 N acts on the block.
work done = 0.8× ( 0.55- 0.50) = 0.04 joule
total work done = 0.04 + 0.12 = 0.16 joule. ans
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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is it true that playing badmenton help you to become a better person?
Answer:
There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.
Answer:
It's true because playing any sport makes a person happy. So a happy person is a better person.
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