true or false- When an atom's charge is balanced, the amount of negative charge equals the amount of neutral charge.

The distance travelled by the motorbike while it is moving at a constant speed is 150m

From the graph, we form the relationship between the speed s and the change in the time t for the journey, i.e. from t = 20sec to t = 30sec speed  s is constant = 15m/s

As the formulae states that

Speed = Distance / Time

Distance = Speed × change in time

⇒15 × ( 30 - 20 )

⇒15 × 10

⇒150m

Hence the Distance travelled by bike was 150m

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

The momentum of the baseball with a mass of 0.14 kg and velocity of 41.26 m/s is determined as 5.78 kgm/s.

The momentum of the baseball is calculated by applying the following formula as shown below;

P = mv

where;

The momentum of the baseball is calculated as follows;

mass of the baseball = 0.14 kg

velocity of the baseball = 41.26 m/s

momentum, P = mv

P = 0.14 kg  x  41.26 m/s

P = 5.78 kgm/s.

Thus, the momentum of the baseball with a mass of 0.14 kg and velocity of 41.26 m/s is determined as 5.78 kgm/s by applying the formula for linear momentum.

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The complete question is below:

Imagine a baseball pitcher and a batter. The baseball has a mass of 0.14 kg. The ball is pitched to the right with a velocity of 41.26 m/s. What is the momentum of the baseball?

Answer:

zero

Explanation:

For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.

Answer:

a

The  radial acceleration is  \(a_c = 0.9574 m/s^2\)

b

The horizontal Tension is  \(T_x = 0.3294 i \ N\)

The vertical Tension is  \(T_y =3.3712 j \ N\)

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  \(L = 10.7 \ cm = 0.107 \ m\)

     The mass of the bob is  \(m = 0.344 \ kg\)

     The angle made  by the string is  \(\theta = 5.58^o\)

The centripetal force acting on the bob is mathematically represented as

         \(F = \frac{mv^2}{r}\)

Now From the diagram we see that this force is equivalent to

     \(F = Tsin \theta\) where T is the tension on the rope  and v is the linear velocity  

     So

          \(Tsin \theta = \frac{mv^2}{r}\)

Now the downward normal force acting on the bob is  mathematically represented as

          \(Tcos \theta = mg\)

So

       \(\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}\)

=>    \(tan \theta = \frac{v^2}{rg}\)

=>   \(g tan \theta = \frac{v^2}{r}\)

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      \(a_c = \frac{v^2}{r}\)

=>  \(a_c = gtan \theta\)

substituting values

     \(a_c = 9.8 * tan (5.58)\)

     \(a_c = 0.9574 m/s^2\)

The horizontal component is mathematically represented as

     \(T_x = Tsin \theta = ma_c\)

substituting value

   \(T_x = 0.344 * 0.9574\)

    \(T_x = 0.3294 \ N\)

The vertical component of  tension is  

    \(T_y = T \ cos \theta = mg\)

substituting value

     \(T_ y = 0.344 * 9.8\)

      \(T_ y = 3.2712 \ N\)

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

       \(T = T_x i + T_y j\)

substituting value  

      \(T = [(0.3294) i + (3.3712)j ] \ N\)

The radical acceleration of the bob is 0.9575 m/s². The horizontal and vertical components of the tension force exerted by the string on the bob are 0.329 N and 3.37 N respectively.

Taking the vertical component of the tension where the weight mass is balanced, then:

T sin θ = mg

\(\mathbf{T = \dfrac{mg}{sin \theta}}\)

However, the centripetal force of the system is given by the horizontal component of the tension which can be expressed as:

T cos θ = m\(\mathbf{a_r}\)

Making \(\mathbf{a_r}\) the subject, we have:

\(\mathbf{a_r = \dfrac{Tcos \theta }{m}}\)

replacing the value of tension (T), we have:

\(\mathbf{a_r = \dfrac{ \dfrac{mg}{sin \theta}cos \theta }{m}}\)

\(\mathbf{a_r=g tan \theta}\)

where;

\(\mathbf{a_r=9.8 m/s^2 \times tan 5.58}\)

\(\mathbf{a_r=9.8 m/s^2 \times 0.0977}\)

\(\mathbf{a_r=0.9575 \ m/s^2}\)

Thus, the radical acceleration of the bob is 0.9577 m/s²

On the positive x-axis, the horizontal component of the tension force is:

\(\mathbf{T_x =Tcos \theta}\)

\(\mathbf{T_x =ma_r}\)

\(\mathbf{T_x =0.344 \ kg \times 0.9575 \ m/s^2}\)

\(\mathbf{T_x =0.329 \ N}\)

On the positive y-axis, the vertical component of the tension force is:

\(\mathbf{T_y =Tsin \theta}\)

\(\mathbf{T_y =mg}\)

\(\mathbf{T_y=0.344 \ kg \times 9.8 \ m/s^2}\)

\(\mathbf{T_x =3.37 \ N}\)

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Answer:

F = qvB sinθ

(Hope this helps! Btw, I am the first to answer. Brainliest pls! :D)

Answer:

1.21 m/s

Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

The final speed of the given ball law of conservation of energy. The final speed of the given ball is 1.21 m/s.

The law of conservation of energy,  

U₁ + K₁ + E₁ = U₂ + K₂ + E₂  

where

U₁ = initial potential energy  

K₁ = initial kinetic energy  

E₁ = initial internal energy  

U₂ = final potential energy  

K₁ = final kinetic energy  

E₁ = final internal energy  

So,    

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂  

mgh = 1/2(m + M)v² + (E₂ - E₁)  

Given that (E₂ - E₁) = change in internal energy,

ΔE = 1/2ΔK

Where

ΔK = change in kinetic energy.

So,

ΔE = 1/2ΔK = 1/2(K₂ - K₁)

ΔE = K₂/2

ΔE = 1/2(m + M)v²/2

ΔE = (m + M)v²/4  

Thus,

mgh = 1/2(m + M)v² + (E₂ - E₁)  

mgh = 1/2(m + M)v² + (m + M)v²/4  

mgh = 3(m + M)v²/4    

v² = 4mgh/3(m + M)  

Take square root of both sides,  

v = √[4mgh/3(m + M)  

put the values in the formula,  

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/3(0.514 kg + 2.63 kg)    

v = 1.21 m/s

Therefore, the final speed of the given ball is 1.21 m/s.

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The correct answer is (A) 8.3mm.

The total displacement will be the vector sum of the two displacements given in the question.

let the displacement to the south represent vector A = 5mm in magnitude directed toward south.

let the displacement to the south-west represent vector B = 4mm in magnitude directed toward south-west.

The angle between vector A and vector B is 45°, the angle between south and south-west.

SO the resultant

R = vectro A + vector B

\(R=\sqrt{A^{2}+B^{2}+2ABcos(45) } \\\\R=\sqrt{25+16+40*\frac{1}{\sqrt{2} } } \\\\R=8.3mm\)

The ant's displacement is 8.3 mm in magnitude.

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Assuming standard atmospheric conditions, the air pressure at sea level is approximately 1013.25 hectopascals (hPa) or 1 atmosphere (atm). The air pressure decreases with altitude following the barometric formula, which states that pressure decreases by about 1 hPa for every 8 meters of ascent.

Using this formula, we can estimate the air pressure at 1200 meters above sea level as follows:

1200 m / 8 m per hPa = 150 hPa

Therefore, the hiker on a mountain ridge 1200 meters above sea level would experience an air pressure of approximately 863.25 hPa (1013.25 hPa - 150 hPa) or about 0.85 atm.

The standard deviation of the sample means X, calculated earlier is the standard deviation of the population divided by the square root of the sample size.

Formula :

Mean : Mean = Sum of X values / N(Number of values)

We find mean=7068.08333

and standard deviation = 226.50003.

The standard deviation of the sample means known as the standard error of the mean is less than the population standard deviation and equal to the population standard deviation divided by the square root of the sample size. the sample mean is the average value found in the sample.

Samples are just a fraction of the total. For example, if you work for a polling company and want to know how much people spend on food each year, you don't need to survey over 300 million people. I also found that the sample mean is the arithmetic mean of all the values ​​in the sample. The sample variance measures how the data are distributed and the sample standard deviation is the square root of the variance.

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Answer:

Explanation:

d = ½at²

220 = ½a4.20²

a = 24.943310... cm/s²

v = at

v = 24.943310(4.20)

v = 104.7619047...

v = 105 cm/s

or

s = ½(0 + v)t

v = 2s/t

v = 2(220)/4.2

v = 104.7619047...

v = 105 cm/s

Answer:

greater than.

Explanation:

Since, density of the bracelet is not equal to the density of gold, then, the bracelet is not pure gold.

To know if the bracelet is pure gold, we calculate the density of the bracelet and compare it to the density of pure gold (19.3 g/cm³).

That is, for the bracelet to pure gold,

Density of bracelet ≈ 19.3 g/cm³

Density can be defined as the ratio of the mass and the volume of a substance.

The formula of Density is give as

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Hence, since the density of the bracelet is not equal to the density of gold, then, the bracelet is not pure gold.

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All of the above are risks of speeding.

Speeding is a type of aggressive driving behavior. Speeding is more than just breaking the law. The consequences are far-ranging:

Speed also affects your safety even when you are driving at the speed limit but too fast for road conditions, such as during bad weather, when a road is under repair, or in an area at night that isn’t well lit.

Speeding endangers not only the life of the speeder, but all of the people on the road around them, including law enforcement officers. It is a problem we all need to help solve.

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The change in momentum applied by a boxer to his rival when he strikes him with a force of 200N in 0.05s is 10 Ns

Momentum can be defined as "mass in motion". All objects have mass. So if an object is moving, it has momentum and its mass is moving. An object's momentum depends on two variables: How much matter is moving and how fast is it moving? It is the strength or power gained by a movement or sequence of events.

Momentum (P) equals mass (M) times velocity (v). However there are other ways to think about momentum. Force (F) is equal to change in momentum (ΔP) with respect to change in time (Δt). Also, the change in momentum (ΔP) is equal to the momentum (J). Impulse has the same units as impulse (kg*m/s or N*s).

ΔP = F × Δt

ΔP = 200 × 0.05

ΔP = 10 Ns

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A VfL (Vertical Field LED) backlighting system is commonly used in LCD (Liquid Crystal Display) televisions.

LCD TVs rely on a backlight to illuminate the liquid crystal layer, which controls the passage of light to create the visual image. The VfL technology is a specific type of LED backlighting arrangement used in certain LCD TV models. In a VfL backlighting system, the LEDs (Light-Emitting Diodes) are positioned vertically along the edges of the LCD panel.

The light emitted by these LEDs is directed across the panel using light guides or optical films, illuminating the liquid crystal layer uniformly. One advantage of VfL backlighting is its ability to provide consistent illumination across the LCD panel, reducing any potential inconsistencies in brightness or color uniformity. The vertical orientation of the LEDs allows for more precise control over light distribution, improving overall image quality.

Additionally, VfL backlighting offers potential advantages in terms of power efficiency. By selectively dimming or turning off specific zones of LEDs, local dimming techniques can be employed to enhance contrast and black levels, resulting in improved picture quality while conserving energy. It's important to note that VfL backlighting is just one of several backlighting technologies available for LCD TVs.

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Answer:

I=0.016A

Explanation:

Answer:

45 W

Explanation:

Power: This can be defined as the rate at which electrical energy is consumed in a circuit. The unit of power is Watt (W).

From the question,

The expression for power is given as,

P = VI.................. Equation 1

Where V = Voltage, I = current.

Given: V = 12 V, I = 3.75 A.

Substitute into equation 1.

P = 12(3.75)

P = 45 W.

Answer:

P = 45 Watt

Explanation:

The electrical power used by an electrical device or the electrical circuit is given by the following formulae:

P = VI

P = I²R

P = V²/R

where,

P = Electrical Power Consumed by the Device

V = The Voltage applied to the circuit or device

R = Resistance of device or circuit

I = Current passing through the device or circuit

We have the following data for our circuit:

V = 12 volts

I = 3.75 A

Therefore, it is clear from the data that we will use the first formula:

P = VI

P = (12 volts)(3.75 A)

P = 45 Watt

Supersonic flights are moving with speed greater than that of sound. This speed through air will combine the pressure waves and creates shock waves. Continuation of shock waves leads to sonic boom in air.

Sonic booms are sound created on the ground by overpressure in the air generated by fastly moving objects. Similar to someone dumping goods from a moving car, an aircraft travelling at supersonic speeds continuously produces shock waves that cause sonic booms to travel throughout its flight path.

The boom seems to be swept rearward as it moves away from the aeroplane from its point of view. The boom will strike the ground in front of the aircraft if the aircraft makes a rapid turn or pulls up.

The abrupt start and release of pressure following the building by the shock wave or "peak overpressure" is what is heard as a "sonic boom" on the ground. Only a few pounds per square foot change in pressure results from a sonic boom, which is about equivalent to the pressure shift we experience in an elevator.

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Answer:

The blue motorcycle.

Explanation:

The red one, since it has more mass, meaning it is denser, would drag on less. Less mass equals more momentum, since there is less holding it down.

Explanation

(m) is measured in kilograms (kg)

Q1) ~mass (m) is measured in kilograms (kg) 

Q2)~acceleration (a) is measured in metres per second squared (m/s²)

Q3)~force (F) is measured in newtons (N)

Hope this helped you- have a good day bro cya)

When you see yourself in a plane mirror, the image is always virtual, erect and the image formed has same size as that of the object

Besides that, the image will be at the same distance behind the mirror as the object is in the front of the mirror, so we can say that the image formed by the plane mirror is an exact copy of ourselves

Because the light does not actually pass through the image, that's why  the image formed by plane mirror is called a virtual image and the real images are formed by curved mirror and they are produced on the same side as that of the mirror

But the only difference is that the image formed by plane mirror is laterally inverted, means if raise our right hand then it would appear in the mirror as if we have raised our left hand

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

mathematical model

Explanation:

A computer model is a very complex type of mathematical model.

These complex mathematical models involves series of equations or algorithms known as set of instructions for the computer to perform difficult computations. These mathematical models include dynamical systems, statistical models, or differential equations.

Answer:

velocity is the speed of something in a particular direction while time is what is measured in hours , minutes and seconds

Explanation:

Given that,

Distance, d = -56.3 m

It strikes the ground 4.00 seconds after being thrown.

Using second equation of motion to find the speed was the rock thrown. So,

\(d=ut+\dfrac{1}{2}at^2\)

Here, a = -g

\(d=ut-\dfrac{1}{2}gt^2\\\\-56.3=u(4)-4.9(4)^2\\\\-56.3=4u-78.4\\\\u=5.52\ m/s\)

Let it will cover a distance of s meters. So,

\(s=\dfrac{v^2-u^2}{-2g}\\\\s=\dfrac{0^2-(5.52)^2}{-2\times 9.8}\\\\s=1.55\ m\)

The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The fraction of a wavelength represented by the first harmonic is 1/2.

The higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

1. The first harmonic of a standing wave on a string with a loose end occurs when there is a node near the driver end and an antinode at the loose end. To measure the frequency of the first harmonic, we need to determine the fraction of a wavelength represented by this standing wave.

The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The first harmonic of a standing wave on a string with a loose end consists of a node near the driver end and an antinode at the loose end. This configuration creates the simplest standing wave pattern.

In a standing wave, a node is a point where the amplitude of the wave is always zero, representing a point of minimum displacement. An antinode, on the other hand, is a point of maximum displacement, where the amplitude is at its highest.

Since the loose end does not invert the phase of the wave upon reflection, the reflected wave and the original wave constructively interfere at the loose end, resulting in an antinode.

In the first harmonic, there is exactly half a wavelength between the node near the driver end and the antinode at the loose end.

Therefore, the fraction of a wavelength represented by the first harmonic is 1/2.

2. To predict the frequencies of higher harmonics, we can use the relationship that the frequency of each harmonic is a multiple of the frequency of the first harmonic. The higher harmonics can be calculated as follows:

Second Harmonic: The second harmonic consists of two nodes and one additional antinode compared to the first harmonic. The fraction of a wavelength for the second harmonic is 1/2 * 2 = 1. Thus, the second harmonic has a frequency that is twice that of the first harmonic.

Third Harmonic: The third harmonic consists of three nodes and two additional antinodes compared to the first harmonic. The fraction of a wavelength for the third harmonic is 1/2 * 3 = 1.5. Thus, the third harmonic has a frequency that is three times that of the first harmonic.

Fourth Harmonic: The fourth harmonic consists of four nodes and three additional antinodes compared to the first harmonic. The fraction of a wavelength for the fourth harmonic is 1/2 * 4 = 2. Thus, the fourth harmonic has a frequency that is four times that of the first harmonic.

In general, the higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

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1. The maximum height above its launch level is 1.45 m

2. The time of flight of the ball is 1.1 s

1. How do I determine the maximum height?

From the question given above, the following data were obtained:

The maximum height can be obatianed as follow:

H = u²Sine²θ / 2g

H = [5.5² × (Sine 76)²] / (2 × 9.8)

Maximum height = 1.45 m

How do I determine the time of flight?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = [2 × 5.5 × Sine 76] / 9.8

Time of flight = 1.1 s

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Answer:

Explanation:

F = m*a = (0.5)*15 = 7.5 N