What energy store is in the human
BEFORE he/she lifts the hammer?​

\(\\ \bull\sf Frequency=f=6\times 10^{14}Hz\)

\(\\ \bull\sf Wavelength=\lambda=3.75\times 10^{-7}m\)

Now

\(\\ \sf\longmapsto v=f\lambda\)

\(\\ \sf\longmapsto v=6\times 10^{14}s^{-1}\times 3.75\times 10^{-7}m\)

\(\\ \sf\longmapsto v=22.5\times 10^7ms^{-1}\)

Answer:

velocity=v = 22.5 * 10⁷ m s ^ - 1

Explanation:

\(\sf\longmapsto \: Frequency = 6 * 10 ^ {14 } H z\)

\(\sf\longmapsto \: Wavelength = 3.75 \times 10 {}^{-7} m\)

Now let's solve out!!

\(\sf\longmapsto \: velocity = frequency \lambda\)

\(\sf\longmapsto \: velocity=6 \times 10 {}^{14} s {}^{ - 1} \times \ 3.75 \times 10 {}^{ - 7} m\)

\(\sf\longmapsto \: velocity= 22.5 \times 10 ^ 7 m s ^ { - 1}\)

Note:

\(\lambda\) is the Greek letter known as lambda.

A block of 0.5 kg is placed on top of another wooden block which weighs 1.0 kg. The coefficient of static friction between the two blocks is 0.35, whereas the coefficient of kinetic friction between the lower block and the level table is 0.20.

To calculate the maximum horizontal force that can be applied to the lower block, we need to determine the limiting frictional force between the two blocks.

Since the upper block is not moving, the force of static friction is acting on it. We can calculate this force as follows:

`F_static = friction coefficient * normal force`

where, normal force = weight of upper block = 0.5 kg * 9.81 m/s^2 = 4.905 N

`F_static = 0.35 * 4.905 = 1.718 N`

Therefore, the static frictional force acting on the upper block is 1.718 N.

Now, we need to find the maximum force that can be applied to the lower block before it starts moving. This force is equal to the force of static friction acting on the lower block.

Since the upper block is not moving, the force of static friction acting on the lower block is equal to the force of static friction acting on the upper block.

`F_static(lower block) = F_static(upper block) = 1.718 N`

This means that the maximum horizontal force that can be applied to the lower block is 1.718 N.

However, if the applied force exceeds this value, the lower block will start moving and the force of kinetic friction will be acting on it, which is equal to:

`F_kinetic = friction coefficient * normal force`

`F_kinetic = 0.20 * 4.905 = 0.981 N`

Hence, if the applied force exceeds 1.718 N, the lower block will start moving and the force of kinetic friction will act on it, which is 0.981 N.

Therefore, the maximum horizontal force that can be applied to the lower block is 1.718 N.

Answer:

Explanation:

To determine the maximum horizontal force that can be applied to the lower block without causing the blocks to move, we need to calculate the maximum static friction force between the two blocks. This force is given by:

F_friction = coefficient of static friction * normal force

where the normal force is the force perpendicular to the surface of contact between the blocks. Since the blocks are resting on a level table, the normal force acting on the lower block is equal to the weight of both blocks, which is:

N = (m1 + m2) * g

where m1 is the mass of the lower block, m2 is the mass of the upper block, and g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the given values, we have:

N = (1.0 kg + 0.5 kg) * 9.81 m/s^2 = 14.715 N

The maximum static friction force is then:

F_friction = 0.35 * 14.715 N = 5.15025 N

Therefore, the maximum horizontal force that can be applied to the lower block without causing the blocks to move is 5.15025 N. If a greater force is applied, the blocks will start to move and the kinetic friction force will take effect, which is given by:

F_kinetic = coefficient of kinetic friction * normal force

where the coefficient of kinetic friction is 0.20 in this case.

Answer:

nothing moves if anything other doesn't move it.

natural position of things is the resting

Explanation:

yeah that's it

Answer:

The book does not move because all the forces acting on the book are in equilibrium, creating a net force of zero magnitude on the book. Since the net force on the book is zero, this means that the book will remain in its present state of motion (which is that the book is motionless).

Explanation:

Newton's first law states that all body will remain in  their present state of motion unless acted upon by an external force. In order word, a moving body should continue moving forever, and a body at rest should remain like that forever, unless they are acted upon by a force to either cause them to stop stop or to start moving respectively.

The process of refraction of light​ occurs when light rays bends when travelling between media of different densities.

What is refraction of light?

Refraction of light is the bending of light rays or the change in the direction of light rays as it travels between media of different densities.

Light waves travel faster in media of less density than media of more density.

The change in density of the media makes light waves to be bend towards or away from the normal.

For example, when light travels from less dense air to more dense water, the light rays are bent towards the normal. However, when light rays travel from water to air, the light rays are refracted away from the normal.

In conclusion, refraction of light waves occur when light crosses the boundary of media of different densities.

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Answer:

F=ma

Explanation:

Force is equal to mass multiplied by area

Force= mass × area

In turn the formula can be twisted so

Force/mass= area

Force/area= mass

Answer:

It states that the rate of change of velocity of an object is directly proportional to the force applied and takes place in the direction of the force. It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s²). Thus, an object of constant mass accelerates in proportion to the force applied.

Answer:

The gravitational force Asteroid A experiences is greater than the gravitational force Asteroid C experiences

Answer:

b

Explanation:

the more heat the hotel it get which cause more heat

Answer: C. Entropy

Explanation: a pex :)

The bullet fired at an angle of 80° with the horizontal and an initial velocity of 420 m/s can travel up to a height of 825.4 meters and a horizontal distance of 80.1 meters after 2 seconds.

To determine the height and horizontal distance traveled by a bullet fired at an angle of 80° with the horizontal and an initial velocity of 420 m/s after 2 seconds, we can use the equations of motion.

Firstly, we can break down the initial velocity of the bullet into its horizontal and vertical components. The horizontal component remains constant throughout the motion and is given by:

Vx = Vcosθ

where V is the initial velocity and θ is the angle of projection. Substituting the given values, we get:

Vx = 420cos80° = 40.05 m/s (approx.)

The vertical component of the initial velocity can be calculated as:

Vy = Vsinθ

Substituting the given values, we get:

Vy = 420sin80° = 416.95 m/s (approx.)

Now, we can use the following equations of motion to determine the height and horizontal distance traveled by the bullet after 2 seconds:

Vertical motion:

y = yo + Voyt + (1/2)gt^2

where y is the vertical displacement, yo is the initial height (assumed to be zero), Voy is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Substituting the given values, we get:

y = 0 + 416.95(2) - (1/2)(9.8)(2)^2

y = 825.4 m (approx.) Therefore, the bullet can travel up to a height of 825.4 meters after 2 seconds. Horizontal motion:

x = xo + Voxt

where x is the horizontal displacement, xo is the initial horizontal position (assumed to be zero), Vox is the initial horizontal velocity, and t is the time.

Substituting the given values, we get:

x = 0 + 40.05(2)

x = 80.1 m (approx.)

Therefore, the bullet can travel a horizontal distance of 80.1 meters after 2 seconds.

In summary, the bullet fired at an angle of 80° with the horizontal and an initial velocity of 420 m/s can travel up to a height of 825.4 meters and a horizontal distance of 80.1 meters after 2 seconds.

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Answer:

Approximately \(9.39 \; {\rm m\cdot s^{-1}}\) after the sphere has travelled a distance of \(5\; {\rm m}\).

Approximately \(16.3\; {\rm m\cdot s^{-1}}\) right before touching the ground (a distance of \(15\; {\rm m}\).)

Assumption: \(g = 9.81\; {\rm N\cdot kg^{-1}}\).

Explanation:

Weight of the sphere: \(m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}\), downwards.

Drag on the sphere: \(10.0\; {\rm N}\) upwards.

Net force on the sphere: \(98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}\) downwards.

Acceleration of the sphere: \(a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}\).

Apply the SUVAT equation \(v^{2} - u^{2} = 2\, a\, x\), where \(v\) is the final velocity, \(u\) is the initial velocity (\(0\) in this case, as the sphere was released from rest,) and \(x\) is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for \(v\):

\(\displaystyle v = \sqrt{2\, a\, x + u^{2}}\).

For example, after the ball travelled a distance of \(5.00\; {\rm m}\), \(x = 5.00 \; {\rm m}\):

\(\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}\).

Similarly, \(x = 15.0\; {\rm m}\) right before landing, such that:

\(\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}\).

The net force on the cart is 7 N to the right. The unit of force is Newtons (N).

Unlike the frictional force , which operates in the opposite direction to the cart's velocity, gravitational force on the cart always points in the same direction, regardless of the cart's motion.

A pull is a force that causes an object to move closer to you. On the other side, it is a push if it travels away. An object may experience a push or pull as a result of interacting with another object, which is a common definition of force. Hence, any force is essentially a push or a pull.

Net force = 10 N to the right - 3 N to the left

Simplifying, we get:

Net force = 10 N - 3 N

Net force = 7 N to the right

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Answer:

ωf = 2.95 rad/sec

Explanation:

       \(L = I * \omega (1)\)

       \(L_{o} = L_{f} (2)\)

        \(L_{o} = I_{o} * \omega_{o} = \frac{1}{2}* M* R^{2}* \omega_{o} =\\ \frac{1}{2} * 77.0 kg* (2.70m)^{2}* 7.40 rad/sec = 2076.92 kg*m2*rad/sec (3)\)

       \(I_{f} = I_{o} +( m_{p} * R^{2}) = (\frac{1}{2} * M* R^{2}) + ( m_{p} * R^{2}) =\\ (\frac{1}{2} * 77.0 kg* (2.70m)^{2}) +( 58.0 kg * (2.70m)^{2}) = \\ 280.67 kg*m2 + 422.82 kg*m2 = \\ 703.49 kg*m2 (4)\)

        \(L_{f} = I_{f} * \omega_{f} (5)\)

       \(\omega_{f} = \frac{L_{o} }{I_{f}} = \frac{2076.92kg*m2*rad/sec}{703.49kg*m2} = 2.95 rad/sec (6)\)

The statement shows a case of rotational motion, in which the disc decelerates at constant rate.

i) The angular acceleration of the disc (\(\alpha\)), in revolutions per square second, is found by the following kinematic formula:

\(\alpha = \frac{\omega_{f}-\omega_{o}}{t}\) (1)

Where:

If we know that \(\omega_{o} = \frac{5}{12}\,\frac{rev}{s}\), \(\omega_{f} = 0\,\frac{rev}{s}\) y \(t = 15\,s\), then the angular acceleration of the disc is:

\(\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}\)

\(\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}\)

The angular acceleration of the disc is \(\frac{1}{36}\) radians per square second.

ii) The number of rotations that the disk makes before it stops (\(\Delta \theta\)), in revolutions, is determined by the following formula:

\(\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}\) (2)

If we know that \(\omega_{o} = \frac{5}{12}\,\frac{rev}{s}\), \(\omega_{f} = 0\,\frac{rev}{s}\) y \(\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}\), then the number of rotations done by the disc is:

\(\Delta \theta = 3.125\,rev\)

The disc makes 3.125 revolutions before it stops.

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The mass of the child is 33.26 kg.

Mass can be defined the quantity of matter a body contained.

The S.I unit of mass is kilogram (kg)

To calculate the mass of the  child, we use the formula below.

Formula:

Where:

Make m the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, the mass of the child is 33.26 kg.

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Answer:

Earth's magnetic field (and the surface magnetic field) is approximately a magnetic dipole, with the magnetic field S pole near the Earth's geographic north pole (see Magnetic North Pole) and the other magnetic field N pole near the Earth's geographic south pole (see Magnetic South Pole). This makes the compass usable for navigation. The cause of the field can be explained by dynamo theory. A magnetic field extends infinitely, though it weakens with distance from its source. The Earth's magnetic field, also called the geomagnetic field, which effectively extends several tens of thousands of kilometres into space, forms the Earth's magnetosphere. A paleomagnetic study of Australian red dacite and pillow basalt has estimated the magnetic field to be at least 3.5 billion years old

The difference is that Hall effect sensor detects the strength of a magnetic field perpendicular to it, while a permanent magnetic sensor detects the angle of a parallel magnetic field.

The permanent magnetic sensor tends to be more accurate as it has a bigger detectable area that has detects layout error.

What is a magnet?

A magnet is any material that produces a magnetic field. A magnet has north and south poles at opposite ends.

The difference between a hall effect magnet and a permanent magnet sensor is that hall effect detects  magnetics fields perpendicular to it while a permanent magnet detects magnetic fields parallel to it.

In conclusion, the permanent magnet is more accurate as it has a wider detectable area.

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The elasticity and Poisson's ratio for this metal is 0.232.

What is ratio?

The realation between two numbers which shows how much bigger one quantity is than another.

Sol-

As per the given question

P=190KN

d=16 mm

Lo=50mm

X=50.1349-50=0.1349mm

Y=15.99-16=-0.01mm

The formula-

E=ó/€

Ó=P/A

A=r/4 d^2 =π/4(16)^2=201.062 mm

ó={190(1000)}201.062=944.982 Mpa

E=944.982/0.002698=350.253 GPa

€y=-0.000625

v=0.232(answer)

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Answer:

Velocity of the two balls after collision: \(60\; \rm m \cdot s^{-1}\).

\(100\; \rm J\) of kinetic energy would be lost.

Explanation:

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass \(m\) and velocity \(v\) is: \(p = m \cdot v\).

Momentum of the two balls before collision:

Based on the assumptions, the sum of the momentum of the two balls after collision should also be \(30\; \rm kg \cdot m \cdot s^{-1}\). The mass of the two balls, combined, is \(0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg\). Let the velocity of the two balls after the collision \(v\; \rm m \cdot s^{-1}\). (There's only one velocity because the collision had sticked the two balls to each other.)

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about \(v\):

\(0.5\, v = 30\).

\(v = 60\).

In other words, the velocity of the two balls right after the collision should be \(60\; \rm m \cdot s^{-1}\).

The kinetic energy of an object of mass \(m\) and velocity \(v\) is \(\displaystyle \frac{1}{2}\, m \cdot v^{2}\).

Kinetic energy before the collision:

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

Sum of the kinetic energies of the two balls right after the collision:

\(\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J\).

Therefore, \(1000\; \rm J - 900\; \rm J = 100\; \rm J\) of kinetic energy would be lost during this collision.

C) a bicyclist riding up a steep hill

The metaphor for a system with rising gravitational potential energy is "a bicyclist riding up a steep hill." Let's get into greater detail:

A cyclist faces resistance from gravity as they ride up a steep slope. The cyclist's elevation, or height above the ground, rises as they cycle and climb uphill. Gravity is pulling the cyclist down the hill by exerting downward force. The cyclist must apply force to the pedals in order to move forward and overcome the pull of gravity. In order to do this, the bicyclist must transform chemical energy from their body into mechanical energy. The distance of the cyclist from the centre of the Earth grows as they ride up the hill. The height and mass of an object affect its gravitational potential energy. In this scenario, as the bicyclist's height rises, their gravitational potential energy also rises.

     Due to the higher elevation, the energy input from the biker is stored as increased potential energy. When the bicycle descends the hill or does work, this potential energy can be transformed back into kinetic energy or other types of energy.

The coefficient of kinetic friction between the block and the incline is 0.168.

The coefficient of kinetic friction is determined from the principle of conservation of energy as follows;

distance traveled on the incline = L

L = h/sin35

L = 1.743 h

Since it takes twice the time travel down,

velocity to slide = 1.743 h/2t = 0.871 v

K.E of the block when it slides = ¹/₂mv₁²  = ¹/₂m(0.871v)² = 0.871²(K.E to fall) = 0.758K.E to fall straight down

1 - 0.758 = 0.24

μmgcos(35) = 0.24(mgsin(35)

μcos(35) = 0.24(sin(35)

μ = 0.168

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a) 12.1 Average.

B) It is important to record measurements to the correct number of significant figures because otherwise, the integrity of the number is compromised.

A) Determine the average length and express the answer using the correct number of significant figures:

12.2+12.1+12=36.3/3 = 12.1 Average.

B)  Based on your average length calculation in part a, discuss the importance of recording measurements to the appropriate number of significant figures:

It is important to record measurements to the correct number of significant figures because otherwise, the integrity of the number is compromised.

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The force 'F' on the block is equal to  120 N. Therefore, option D is correct.

The frictional force is described as the force formed by two surfaces that slide against each other. The frictional force is affected by the texture of the surface, the angle, as well as the position of the object.

Friction can be defined as the force that resists motion when one surface comes in contact with the other surface. The mechanical advantage will reduce by the force of friction therefore the ratio of output to input will be reduced because of friction.

The formula of the frictional force can be represented as:

F = μmg

Given, the weight of the block, W =  500 N

W = mg

500 = m × 9.8

m = 51 Kg

The acceleration of the block, a = 0.4 m/s²

The frictional force of the block, F = 100 N

F(net) = F (applied) - F (frictional)

ma = F - 100

51 ×0.4 = F - 100

F = 100 + 20.4

F = 120.4 N

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Explanation:

Given:

Diameter = 200 m

Radius, r = 200/2 = 100 m

Time taken, t = 30 seconds

Formula to be used:

Distance traveled, = circumference of circle = 2πr

Answer:

Putting all the values, we get

Distance traveled = 2πr

So, the distance traveled by Raju is 628.57 m.

Now, magnitude of the displacement,

At the end of 30 seconds, Raju will come to starting position or initial position, so displacement is zero.

The acceleration at the moment when the stone is left from the hand would be equal to:

a). g

Thus, option a is the correct answer.

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None of the statements are true of all proteins. Here is a correct statement about proteins:

Proteins are large, complex molecules that perform a wide variety of functions in the body, such as catalyzing metabolic reactions, replicating DNA, responding to stimuli, and transporting molecules from one location to another. They are made up of chains of amino acids, which are linked together in a specific sequence determined by the genetic code. Protein structure and function is highly dependent on the specific sequence of amino acids in the protein, and proteins can take on a wide variety of shapes and sizes, ranging from small, globular proteins to large, fibrous proteins.

Answer:

I=Q/T

2.8=Q/3

8.4C=Q

Explanation:

a. There are two main types of pluralism: cultural and political

b. Yes, the United States is becoming more pluralistic

c. Ethnic enclaves can be both a positive and negative aspect of pluralism in the US.

a. Pluralism refers to the coexistence of multiple cultures, religions, and ethnicities within a society. There are two main types of pluralism: cultural and political. Cultural pluralism emphasizes the preservation of distinct cultural traditions, while political pluralism emphasizes the equal representation of different groups within the political system.

Cultural pluralism is often associated with the idea of multiculturalism, which recognizes and celebrates the diversity of different cultures. This type of pluralism acknowledges the importance of cultural differences and encourages individuals to maintain and express their cultural identities. In contrast, political pluralism emphasizes the importance of equal representation of different groups within the political system. This can take the form of proportional representation in government or equal access to political power for different groups.

b. The United States is becoming more pluralistic, as the population becomes more diverse and the importance of multiculturalism and political pluralism is increasingly recognized. The country is home to many different ethnic and cultural groups, and these groups are increasingly visible in politics and society. This is evident in the increasing number of ethnic and cultural enclaves, which are areas where different ethnic groups live in close proximity to one another.

c. Ethnic enclaves can be both a positive and negative aspect of pluralism in the US. On the one hand, they allow individuals to maintain their cultural traditions and create supportive communities. On the other hand, they can also lead to segregation and exclusion, as different groups may be reluctant to interact with one another. In order to promote a positive form of pluralism, it is important to encourage interaction and understanding between different groups, while also respecting and celebrating their distinct cultural identities.

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Let's see

\(\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)\)

Now we know the formula of acceleration

\(\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)\)

\(\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2\)

\(\\ \rm\Rrightarrow \alpha=-x\omega^2\)

Or

\(\\ \rm\Rrightarrow \alpha=-\omega^2x\)

Given that the position x of a particle along X-axis varies with time t by the equation:

\({:\implies \quad \sf x=A\sin (\omega t)}\)

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:

\({:\implies \quad \sf \dfrac{dx}{dt}=A\cos (\omega t)\cdot \omega \cdot \dfrac{dt}{dt}}\)

\({:\implies \quad \sf \dfrac{dx}{dt}=A\omega \cos (\omega t)}\)

Differentiating both sides w.r.t.x by chain rule again to get the 2nd order derivative

\({:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega \sin (\omega t)\cdot \omega \dfrac{dt}{dt}}\)

\({:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega^{2}\sin (\omega t)}\)

Re-write as :

\({:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-\omega^{2}\{A\sin (\omega t)\}}\)

Can be further written as

\({:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-x\omega^{2}}\)

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:

\({:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}+x\omega^{2}=0}\)