The law of conservation of energy asserts that total energy remains constant in an isolated system. That is, energy cannot be generated or destroyed, but it may be transferred from one form to another. Frictional forces can dissipate energy, which raises the question of whether it violates the law of conservation of energy. No, it does not. When there is friction, energy is transferred from one form to another. There will be no energy loss. To illustrate this issue, consider the following scenario: two buses collide. The buses are no longer moving as a result of the collision. Where does all of this energy go? The solution is simple: the friction between buses and between buses and the road allows energy to be transferred from one form to another. You may be aware that when we rub our hands together, heat is produced; what occurs here is frictional energy converting to heat energy. This is what happens in bus accidents, where the energies can be changed to thermal energy, acoustic energy, or any other type of energy owing to friction and impact. So the energy we believed we had wasted was really converted to heat and sound. Bus collisions are not only noisy, but they also cause a lot of friction on the ground and in the bent metal. Both heat and sound are types of energy.
NOTE:
The position x of a particle moving along x axis varies with time t as x = Asin(wt) , where A and w are constants . The acceleration of the particle varies with its position as
[tex] \small \red{ \rm Nonsense = Report} \checkmark[/tex]
Let's see
[tex]\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)[/tex]
Now we know the formula of acceleration
[tex]\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)[/tex]
[tex]\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2[/tex]
From eq(1)[tex]\\ \rm\Rrightarrow \alpha=-x\omega^2[/tex]
Or
[tex]\\ \rm\Rrightarrow \alpha=-\omega^2x[/tex]
Given that the position x of a particle along X-axis varies with time t by the equation:
[tex]{:\implies \quad \sf x=A\sin (\omega t)}[/tex]
As it defines the position, so x is just displacement here, and we need to find the acceleration first for telling with what if varies, so by definition, the second differential coefficient of displacement is acceleration, so differentiating both sides w.r.t.x of the above equation in accordance with chain rule we have:
[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\cos (\omega t)\cdot \omega \cdot \dfrac{dt}{dt}}[/tex]
[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\omega \cos (\omega t)}[/tex]
Differentiating both sides w.r.t.x by chain rule again to get the 2nd order derivative
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega \sin (\omega t)\cdot \omega \dfrac{dt}{dt}}[/tex]
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega^{2}\sin (\omega t)}[/tex]
Re-write as :
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-\omega^{2}\{A\sin (\omega t)\}}[/tex]
Can be further written as
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-x\omega^{2}}[/tex]
This is the Required answer
If they ask you the differential equation for the acceleration of a wave (as the given equation was general equation of a wave), you can simply write:
[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}+x\omega^{2}=0}[/tex]
PLEASE HELP!! Lab: Electromagnetic Induction
Answer:
I think this is the answer
Explanation:
When you think about and describe the characteristics of a giraffe, what type of model are you using?
Answer:
The giraffe is the tallest of all mammals. It reaches an overall height of 18 ft (5.5 m) or more. The legs and neck are extremely long. The giraffe has a short body, a tufted tail, a short mane, and short skin-covered horns
Which is an example of radiation?
Answer:
A burning candle emits radiation in the form of heat and light. The Sun emits radiation in the form of light, heat, and particles. Uranium-238 decaying into Thorium-234 emits radiation in the form ofalpha particles.
Answer:
the sun
Explanation:
the sun emits radiation
A sound wave is produced when a medium begins to
Answer:
when a medium begins to vibrate
When ocean temperature rises,
a. excess gas is released into the atmosphere.
b. the ocean absorbs and holds carbon dioxide.
c. carbon dioxide is equalized between the ocean and atmosphere. d. no change takes place.
Answer:
see below
Explanation:
Kind of like when yo open a warm soda .....excess gas is released
HELP!! ASAP I DONT KNOW HOW TO THIS
It the mass of the foot is 10kg and it is moving at 5m/s, what is the momentum of the leg?
A sample of ore containing radioactive strontium 38Sr90 has an activity of 8.2 × 105 Bq. The atomic mass of strontium is 89.908 u, and its half-life is 28.8 yr (1 yr = 3.156 × 107 s). How many grams of strontium are in the sample?
From the activity values and the decay constant, the mass of of Strontium in the sample is:
[tex]1.62 × 10^{-7}g[/tex]
What is the decay constant of an element?The decay constant of an element is the probability of decay of a nucleus per unit time.
{λ = ln 2 / t1/2
where;
t1/2 is the half-life of the isotope.
The half-life is converted to seconds since the decay constant is asked in per seconds.
[tex]28.8 years = 28.8 × 3.156 × 10^{7} = 908928000 s \\ [/tex]
Hence;
[tex]λ = \frac{ln2}{90892800s} = 7.626 s^{-1}[/tex]
The activity of the element, A, the decay constant, λ and the number of nuclei, N are related as follows:
A = (–) λN[tex]N = \frac{8.25 ×10^{5}}{7.626×10^{-10}} = 1.082 × 10^{15} [/tex]
Molar mass of Strontium-90 is 90 g.
1 mole of Strontium-90 contains 6.02×10^23 nuclei.
The mass, m of Strontium in the sample is calculated:
[tex]m = 1.082 × 10^{15} × \frac{90 g}{6.02 × 10^{23}} = 1.62 × 10^{-7}g \\ [/tex]
Therefore, the mass of of Strontium in the sample is:
[tex]1.62 × 10^{-7} \: g[/tex]
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What is the angular acceleration experienced by a uniform solid disc of mass 5-kg and radius 20 cm when a net
torque of 50 N.m is applied? Assume the disc spins about its center.
Answer:
7.07 radians per second
A piece of fine fiber with a diameter of =6.5 m is used to prop apart the edges of two perfectly flat 3.3-cm-long pieces of glass (see diagram). When the setup is illuminated from above with light of wavelength =590 nm , an interference pattern of alternating bright and dark bands will be seen in the reflected light. If the setup is viewed from high above, how many dark bands will be seen?
We need frequency:-
[tex]\\ \rm\Rrightarrow \nu=\dfrac{c}{\lambda}[/tex]
[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8m/s}{590\times 10^{-9}m}[/tex]
[tex]\\ \rm\Rrightarrow \nu=0.0051\times 10^{17}Hz[/tex]
[tex]\\ \rm\Rrightarrow \nu=5.1\times 10^{14}Hz[/tex]
The fundamental frequency of a musical note is 330Hz. What is the frequency of the 3rd overtone?
The 3rd overtone or the third harmonic given that the fundamental frequency is 330Hz is 990Hz
What is an Harmonic?Hamonics defined as the number gotten by multiplying the nth overtone by the fundamental frequecency.
In summary, the expression for the nth Hamonic or overtone is given as
Fn = Fundamental Frequency*N
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Can anyone help me on this? It’s a water cycle.
Answer:
The water cycle shows the continuous movement of water within the Earth and atmosphere. It is a complex system that includes many different processes. Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow.
Explanation:
Please help quickly!!!
State the basic rules to draw a vector quantity
Explanation:
i dont to i want to copy and paste on my printer
You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of the crate is 245 N , and the coefficient of kinetic friction between the crate and the floor is 0.270. What must be the tension in the rope to make the crate move at a constant velocity?
Express your answer with the appropriate units.
What is the normal force that the floor exerts on the crate?
Express your answer with the appropriate units.
Hi there!
For an object on an incline with friction being pulled, the following forces are present.
Force due to Gravity Force due to FrictionForce due to tensionThe force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.
Let up the incline be positive, and down the incline be negative.
Doing a summation of forces:
[tex]\Sigma F = F_T + F_f - F_g[/tex]
For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
[tex]0 = F_T + F_f - F_g[/tex]
Now, we can express each force as an equation.
Force due to tension:
Must be solved for.Force due to gravity:
On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)This is expressed as:
[tex]F_g = Mgsin\theta[/tex]
Force due to friction:
Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:[tex]N = Mgcos\theta[/tex]
[tex]F_f = \mu Mgcos\theta[/tex]
Now, plug these expressions into the above equation.
[tex]0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T[/tex]
Mg = 245 N (weight). Plug in all values:
[tex]245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}[/tex]
The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:
[tex]N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}[/tex]
To find the tension in the rope needed to make the crate move at a constant velocity, we need to consider the forces acting on the crate.
1. Tension in the rope (T): This force is applied upwards at an angle of 20 degrees above the horizontal.
2. Weight of the crate (W): This force acts downward vertically and is equal to 245 N.
3. Normal force (N): This force is exerted by the floor on the crate and acts perpendicular to the surface of the floor.
4. Kinetic friction force (f_k): This force opposes the motion of the crate and acts parallel to the surface of the floor. The magnitude of kinetic friction is given by: f_k = μ_k * N, where μ_k is the coefficient of kinetic friction (0.270 in this case).
Since the crate is moving at a constant velocity, the net force on the crate must be zero. In other words, the forces pulling the crate forward (T and the horizontal component of the weight) must balance the forces opposing its motion (kinetic friction). The vertical forces (the vertical component of the weight and the normal force) must also balance.
Let's proceed with the calculations:
Horizontal Forces:
T * cos(20°) - f_k = 0
T * sin(20°) + N - W = 0
We can now solve for T and N:
From the horizontal forces equation:
T * cos(20°) = f_k
vertical force:
T = f_k / cos(20°)
T = (0.270 * N) / cos(20°)
From the vertical forces equation:
T * sin(20°) + N = W
T * sin(20°) + N = 245 N
Now, substitute the expression for T from the horizontal forces equation:
(0.270 * N) / cos(20°) * sin(20°) + N = 245 N
Now, solve for N:
N * (0.270 * tan(20°) + 1) = 245 N
N = 245 N / (0.270 * tan(20°) + 1)
Now, plug in the values and calculate N:
N = 245 N / (0.270 * tan(20°) + 1)
N ≈ 208.75 N
Now, we can calculate the tension in the rope (T):
T = (0.270 * N) / cos(20°)
T ≈ (0.270 * 208.75 N) / cos(20°)
T ≈ 67.24 N
So, the tension in the rope needed to make the crate move at a constant velocity is approximately 67.24 N, and the normal force exerted by the floor on the crate is approximately 208.75 N.
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If you double the mass, what happen to the acceleration?
Answer: The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.
Explanation:
a sound has a wavelength of 0.52028. If the speed is 343.00m/s, what is the musical note?
The atmosphere has mass and produces friction as objects move through it.
true or false
true
Explanation:
It has friction when objects move through it.
How long it take to change the velocity of a car from 100m/s to 250m/s if it accelerates at 5m/s^2
Answer:
30 seconds
Explanation:
The formula for calculating acceleration is
( final velocity - inital velocity) ÷ time
If we enter the values, it would be (250 - 100) ÷ t = 5m/s^2
Now we need to know 't'
So we rearrange the equation to make t the subject.
(250-100) ÷ 5 = 30s
After a boater runs out of gas a boater considers dropping the motor into the water to create waves describe the resuming motion of the boat and explains why the boaters plan will not help him reach shore ????
The boat will swing back and forth about the same location, never reaching the land. Due to which the boaters plan will not help him reach shore.
Describe boat motion?The ship rocks back and forth as the wind and waves press against it. Similar to swiveling on a chair, yaw spins the ship on an unseen central line.
Waves travelling perpendicular to the ship's motion might create this, changing the ship's heading or direction.
The motion of the motor will be identical to that of water molecules. Because there is a wave on the water surface,The motor will move in lockstep with the water particle.
The particles of the medium only vibrate perpendicular to the wave propagation as the wave travels through the water surface, but they are not moved from their initial position.
As a result, the boat will swing back and forth about the same location, never reaching the land.
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An airplane of mass 270,000 kg that is sitting on a runway has its center of gravity located at 10.0 m in front of the main landing wheels as shown. Find the magnitude of the force exerted by the runway on the nose wheel located 25.0 m from the main landing wheels.
For an airplane of mass 270,000 kg , the magnitude of the force exerted is mathematically given as
N2=1584431.13N
N1=1061568.87N
What is the magnitude of the force exerted by the runway on the nose wheel located 25.0 m from the main landing wheels.?Generally, the equation for the Force balance is mathematically given as
N1+N2=270000*g
Therefore
N2*10=N1*15
N1=10N2/15
N1=0.67N2
Hence
0.67N2+N2=270000*g
N2=1584431.13N
N1=1061568.87N
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A mass of 100kg is acted on by a 400N force acting vertically upward and a 600N force acting at a 45 degrees angle.Calculate the vertical component of the acceleration.The local acceleration of gravity is 9.8 m/s2. A.1.56 B.-1.56 C.-2.43 D.2.43 . E. 3.5
Answer:
A 1.56 see below
Explanation:
I will assume the 45 degree force is upward
vertical component = 600 sin 45 = 424.26 N
added to the other vertical force will total 824.26 N
F = ma
824.26 = 100 * a shows a = 8.24 m/s upward
Now we have to assume the mass is ALSO acted on by gravity and this value is given as 9.8 ( so downward is positive)
9.81 - 8.24 = 1.56 m/s^2
What best describes the relationship between sleep and memory? The quantity of sleep impacts memory, while the quality of sleep does not. The quantity of sleep does not impact memory, while the quality of sleep does. The quantity and quality of sleep have no impact on memory. The quantity and quality of sleep both impact memory.
Answer:
The quantity and quality of sleep both impact memory.
Explanation:
A small block sits at one end of a flat board that is 2.50 m long. The coefficients of friction between the block and the board are μs = 0.450 and μk = 0.400. The end of the board where the block sits is slowly raised until the angle the board makes with the horizontal is α0, and then the block starts to slide down the board.
Hi there!
In order for a block to begin sliding, the force due to STATIC friction must be overcome.
In this instance, the following forces are acting on the block ALONG the axis of the incline.
Force due to gravity (Fg)Force due to STATIC friction (Fs)Force due to gravity:
On an incline, the component of the force due to gravity contributing to the object's downward movement is equivalent to the horizontal (sine) component.
[tex]F_g = Mgsin\theta[/tex]
Force due to static friction:
The force due to friction is equivalent to the normal force multiplied by the coefficient of friction.
The normal force is the cosine component (perpendicular to the incline), so:
[tex]N = Mgcos\theta\\\\F_s = \mu_sMgcos\theta[/tex]
To find the minimum angle for the block to begin sliding, we can set the two forces equal to 0. They work in opposite directions (let down the incline be negative and up the incline be positive).
[tex]\Sigma F = F_s - F_g\\\\0 = F_s - F_g\\\\0 = \mu_sMgcos\theta - Mgsin\theta\\\\Mgsin\theta = \mu_sMgcos\theta[/tex]
Cancel out 'Mg' and rearrange to solve for theta.
[tex]sin\theta = \mu_scos\theta\\\\tan\theta = \mu_s\\\\\theta = tan^{-1}(\mu_s) = tan^{-1}(.45) = \boxed{24.228^o}[/tex]
A(n) __________ __________ is made with an electromagnet that is free to spin between two permanent magnets
Answer:
Like a galvanometer, an electric motor contains (a switch, an electromagnet) that is free to rotate between the poles of a permanent, fixed magnet. 9.
what is kinematics ?
explain ~
Kinematics is the branch of physics and a subdivision of classical mechanics concerned with the geometrically possible motion of a body or system of bodies without consideration of the forces involved. :)
Answer:
Kinematics is a branch of mechanics that deals with pure motion, without reference to the masses or forces involved in it.
Explanation:
Ex. X=2t+1 here x is related to t which is time(function of time)
I
same 7 kg ball.
9. A 10 N force and a 40 N force are acting in opposite direction on the
ball? What is the magnitude of the acceleration of the ball?
Answer:
Here, m=10 kg
The resultant force acting on the body is
F=(98N)2+(6N)2=10N
Let the resultant force F makes an angle θ w.r.t. 8N force.
From figure, tanθ=8N6N=43
The resultant acceleration of the body is
a=mF=10kg10N=1ms−2
The resulatnt acceleration is along the direction of the resulatnt force.
Hence, the resultant acceleration of the body is 1 ms−2 at an angle of tan−1(43) w.r.t. 8N force.

A. A cord passing over an easily turned pulley (one that is both massless and frictionless) has 7kg mass hanging from one end and a 9kg mass hanging from the other end as shown below. G 7.0 kg 9.0 ks This arrangement is known as Atwood's machine. Calculate the (i). acceleration of the masses (ii). tension in the cord. [2.5
Answer:
M g = M a = a (9 - 7) kg with the two masses going in different directions
(7 + 9) kg * g = a * 2 kg 2 kg effective in prducing acceleration
a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2
M1 a = T - M1 * g T producing upward acceleration of 1.23 m/s^2
T = 7 (a + g) = 7 * 11 = 77 N
Check:
M2 a = M2 * g - T M2 producing downward acceleration
T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N
Then, complete the riddle below by finding the matching number and writing the letter.
Newton's Second Law
44k el effe &
forces cause an object to accelerate.
3
10
Acceleration depends on I & I and M a ½ ½.
11 12
13
14 15
16
17
18
19
As the force increases, the
accelera
i on
20
21
29
30 31
increases, too.
Why did the artist paint on cement blocks instead of paper?
"
11
He wanted to create
Answer:
I have no clue I'm just trying to get points
Explanation:
:) sorry
The time period of 2 pendulums are 1.22 s and 0.54 s. Calculate the ratio of their lengths.
[tex]~~~~~\dfrac{T_1}{T_2}= \dfrac{1.22}{0.54}\\\\\\\implies \dfrac{2\pi \sqrt{\dfrac{L_1}{g}}}{2\pi \sqrt\dfrac{L_2}g}}= 2.259\\\\\\\implies \dfrac{\sqrt{L_1}}{\sqrt g} \times \dfrac{\sqrt g}{\sqrt{L_2}} = 2.259\\\\\\\implies \sqrt{\dfrac{L_1}{L_2}} = 2.259\\\\\\\implies \dfrac{L_1}{L_2} = 5.104\\\\\\\implies L_1:L_2 = 5.104[/tex]