In a certain part of the North America Nebula, the amount of interstellar extinction in the visual wavelength band is 1.1 magnitudes. The thickness of the nebula is estimated to be 20 pc and it is located 700 pc from Earth. Suppose that a B spectral class main-sequence star is observed in the direction of the nebula and that the absolute visual magnitude of the star is known to be M(V) = -1.1 from spectroscopic data. Neglect any other sources of extinction between the observer and the nebula. Show all your work, assumptions, equations, and units.1. Find the apparent visual magnitude of the star if it is lying just in front of the nebula.2. Find the apparent visual magnitude of the star if it is lying just behind the nebula.

Answer:

Focal length (f) = 1/2 (Object distance + Image distance)

f = 1/2 (12.2 cm + 25.0 cm)

f = 18.6 cm

The frequency of oscillation of the swing is 0.25 Hz.

The time period of oscillation of the swing is 4 s.

The length of the vine of the swing is 3.97 m.

The restoring force acting on Mohammed is 692.9 N.

Mass of Mohammed, m = 50 kg

Angle made by the vine with the vertical, θ = 45°

Number of complete swings made by Mohammed, n = 30

Time taken for this swing, t = 2 minutes = 120 seconds

a) The frequency of the swing is defined as the number of complete oscillations in one second.

So, the frequency of oscillation of the swing is,

f = n/t

f = 30/120

f = 0.25 Hz

b) The time period of oscillation of the swing is,

T = 1/f

T = 1/0.25

T = 4 s

c) The expression for the time period is given by,

T = 2π√(l/g)

T² = 4π² x (l/g)

l/g = T²/4π²

Therefore, the length of the vine of the swing is,

l = T²g/4π²

l = 4² x 9.8/4 x (3.14)²

l = 3.97 m

d) The restoring force acting on Mohammed,

F = mg sinθ

F = 50 x 9.8 x sin 45°

F = 490 x 1/√2 = 490/1.414

F = 692.9 N

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Answer: Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another. ... During the 1930s Bohr helped refugees from Nazism.

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:

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.

\(F_g = Mgsin\theta\)

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:
\(N = Mgcos\theta\\\\F_s = \mu_sMgcos\theta\)

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).

\(\Sigma F = F_s - F_g\\\\0 = F_s - F_g\\\\0 = \mu_sMgcos\theta - Mgsin\theta\\\\Mgsin\theta = \mu_sMgcos\theta\)

Cancel out 'Mg' and rearrange to solve for theta.

\(sin\theta = \mu_scos\theta\\\\tan\theta = \mu_s\\\\\theta = tan^{-1}(\mu_s) = tan^{-1}(.45) = \boxed{24.228^o}\)

The mass of the wire of lowest A on a piano is 0.00165 kg.

The frequency of a vibrating string is given by the equation:

f = (1/2L) * sqrt(T/μ)

where f is the frequency of the string, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string (mass per unit length).

We know the frequency of the lowest A on a piano is 27.5 Hz. We also know that one half wavelength occupies the string, so the length of the string is half the wavelength:

L = (1/2) * λ

The wavelength of a sound wave is given by:

λ = 2L/n

where n is the number of nodes (points of zero displacement) in the wave. For the lowest A on a piano, n = 1, so we can write:

λ = 2L

Substituting this into the equation above for L, we obtain:

L = λ/2

Now we can substitute these values into the first equation:

27.5 = (1/2)(λ/2) * sqrt(308/μ)

Simplifying, we get:

λ = 4L

308/μ = 4(27.5)^2 (1/4)

μ = 0.000824 kg/m.

Since μ = m/L, where m is the mass of the wire and L is its length, we can find the mass of the wire by multiplying the linear mass density by the length of the string:

m = μL

The length of the string is given as 2.00 m, so we can write:

m = 0.000824 kg/m * 2.00 m = 0.00165 kg

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Answer:

T = 3.14 10⁻²  s

Explanation:

The general equation for a traveling transverse wave is

           y = A sin (k x - wt +Ф)

where w is the angular velocity that is related to the period and Ф is the phase of the oscillation

           w = 2π / T          (1)

In this case they indicate that the expression of the wave

         y = 0.003 cos (20 x + 200t)

if we use the relationship of the double angles

          a = kx - wt

          b = Ф

          sin (a + b) = sin a cos b + sin b cos a

for the case Ф = 90 we have

          sin (a + b) = cos b

we substitute

          y = A cos (kx -wt)

with an initial phase of fi = 90º

if we compare the terms of the two expressions

         A = 0.003 m  

         k = 20 m⁻¹

         w = -200  rad/s

the negative sign indicates that the wave goes to the left

if we clear from equation 1

          T = 2π / w

          T = 2π / 200

          T = 3.14 10⁻²  s

Answer:

The water cycle shows the continuous movement of water within the Earth and atmosphere. ... Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow. Water in different phases moves through the atmosphere (transportation).

Explanation:

It's the water cycle.

If the dog walked at a constant speed the whole way, the graph of the dog's position versus time would be a straight line. This is because the dog's velocity (which is the derivative of position with respect to time) would be constant, and the acceleration (which is the derivative of velocity with respect to time) would be zero.

A straight line on a position-time graph indicates that the object is moving at a constant velocity. The slope of the line would be equal to the velocity of the dog.

If the graph is a horizontal line, it would indicate that the dog is at rest. If the line slopes upward, the dog is moving in the positive direction (for example, to the right in a position-time graph), and if the line slopes downward, the dog is moving in the negative direction.

In all, A constant speed means a constant velocity and the line is a straight line with a particular slope.

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The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.

Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.

The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:

P = ρgh + Po

where:

P is the absolute pressure at the given depth

ρ is the density of the water

g is the acceleration due to gravity (assumed to be 9.81 m/s²)

h is the depth of the ocean

Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)

Substituting the given values, we get:

P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa

P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa

P = 1.002 x 10^8 Pa.

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At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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Answer:

The magnitude of the normal force is equal to the weight of the box multiplied by the cosine of the angle θ.

Explanation:

When a box is pulled along a frictionless surface at an angle θ, the normal force acting on the box is perpendicular to the surface and opposes the force of gravity.

In this case, the box is accelerating to the right, which means the force F is greater than the force of gravity pulling the box downward.

The magnitude of the normal force (N) can be determined by breaking down the forces in the vertical direction:

Sum of vertical forces = 0

N - mg * cos(θ) = 0

Here, mg * cos(θ) represents the vertical component of the weight of the box, which is balanced by the normal force.

Rearranging the equation, we have:

N = mg * cos(θ)

Answer:

Explanation:

When the gun is fired, the bullet moves forward with a great momentum while the gun moves backwards with an equal momentum. The formula for calculating the recoil of the gun is expressed as

- mv = MV

where

m and M are the masses of the bullet and gun respectively

v and V are the velocities of the bullet and gun respectively

Thus,

V = - mv/M

From the information given,

m = 30g

we would convert 30g to kg. Recall, 1000g = 1kg. Thus,

30g = 30/1000 = 0.03 kg

M = 4

v = 90

Thus,

V = - 0.03 x 90/4

V = - 0.675 m/s

The recoil of the gun is - 0.675 m/s

Answer:

(i) 20 m/s

(ii) 3.5 m/s

(iii) 6.5 m/s

(iv) 85 m/s

Explanation:

1(kilometer/hour) = 1000(meters) / 3600(seconds) can also be expressed as 1(kilometer/hour) = 5/18 (meters/second), which is its simplified form. To convert km/h to m/s, directly multiply the given value of speed by the fraction 5/18.

This information is from: https://www.cuemath.com/measurement/km-h-to-m-s/

The maximum range of the water jet is\(Rmax = h(sqrt(2)),\) and it is achieved when the height of the hole is \(yo = h/2.\)

We can use the equations of motion for an object under constant acceleration to derive the formula for the horizontal distance R that a jet of water will travel before striking the ground. The acceleration of the water jet is due to gravity, and it is constant and equal to the acceleration due to gravity, g.

Let t be the time it takes for the water jet to hit the ground after leaving the hole. We can use the equation of motion for the vertical direction:

\(y = yo + voy t - (1/2)gt^2\)

where y is the vertical displacement of the water jet, yo is the initial vertical displacement (the height of the hole from the bottom of the tank), voy is the initial vertical velocity (which is zero), and g is the acceleration due to gravity.

Solving for t, we get:

\(t = sqrt((2(yo - y))/g)\)

Now we can use the equation of motion for the horizontal direction:

\(x = v_{0} x t\)

where x is the horizontal displacement (which is R), and vox is the initial horizontal velocity (which is constant and equal to the velocity of the water jet as it emerges from the hole).

We can express \(v_{0} x\) in terms of the vertical displacement y and the time t:

\(v_{0} x = R/t = R(sqrt(g/(2(yo - y))))\)

Substituting for t and simplifying, we get:

\(v_{0} x = R(sqrt(2g/h))\)

Now we can express the range R in terms of the tank height h and the height of the hole yo:

\(R = (v_{0} x^2/h) = 2h(sqrt(yo/h))\)

To derive the formula for an identical tank on the moon where the acceleration due to gravity is Jm = 1/4 of the acceleration due to gravity on Earth (g), we can substitute g/4 for g in the equation for the horizontal velocity vox. This gives:

\(vox = R(sqrt(g/(8h)))\)

Substituting into the equation for R, we get:

\(R = (vox^2/h) = 8h(sqrt(yo/h))\)

To show that the maximum range of the water jet is achieved when yo = h/2, we can differentiate R with respect to yo and set the result equal to zero:

\(dR/dyo = (4/h)(sqrt(yo/h))(h/2 - yo)\)

Setting this equal to zero and solving for yo, we get:

\(yo = h/2\)

To find the maximum range Rmax, we substitute yo = h/2 into the equation for R:

\(Rmax = 2h(sqrt(h/2h)) = h(sqrt(2))\)

Therefore, the maximum range of the water jet is \(Rmax = h(sqrt(2))\), and it is achieved when the height of the hole is yo = h/2.

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Answer:

100 m

Explanation:

The frequency ratio between two frequencies f1 and f2 is given by:

frequency ratio = f2 / f1

Substituting the given values, we get:

frequency ratio = 576 Hz / 320 Hz = 1.8

Therefore, the frequency ratio between the two frequencies f1 = 320 Hz and f2 = 576 Hz is 1.8.

(b) If we go down from f1 by an interval of a fourth, we get a frequency that is 5 semitones lower. The frequency ratio for a single semitone is given by:

2^(1/12) ≈ 1.05946

Going down by 5 semitones, we get a frequency ratio of:

1 / (2^(5/12)) ≈ 0.84090

Therefore, the frequency ratio between f1 and the frequency that is a fourth below it is approximately 0.84090.

(c) To find the frequency of f3, we need to know the interval between f1 and f3. Let's assume that f3 is a fifth above f2. The frequency ratio for a fifth is given by:

2^(7/12) ≈ 1.49831

Therefore, the frequency of f3 is:

f3 = f1 * (2^(7/12)) * (2^(7/12)) = 320 Hz * 1.49831 * 1.49831 ≈ 716 Hz

Therefore, the frequency of f3 is approximately 716 Hz.

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A projectile is an object thrown into the air and subject only to gravity and, if applicable, air resistance. The time taken by disc C to reach its maximum height is approximately 1.53 seconds, and its maximum height above the ground is around 11.48 meters. The time from when disc C was thrown upwards until ball B hits the disc is roughly 1.29 seconds.

3.1 Explanation of the term projectile:

A projectile refers to an object that is launched or thrown into the air and is subject only to the forces of gravity and air resistance (if applicable). The motion of a projectile can be analyzed independently of its mass, shape, or any other physical property. The key characteristic of a projectile is that it follows a curved path known as a trajectory.

When a projectile is launched, it moves along a parabolic trajectory due to the combination of its initial velocity and the force of gravity acting vertically downward. The horizontal motion of a projectile remains constant and unaffected by gravity, while the vertical motion is influenced by the acceleration due to gravity.

The path of a projectile can be described mathematically by considering its initial velocity, angle of projection, and the acceleration due to gravity. Projectile motion finds applications in various fields, such as sports, engineering, and physics, where objects are launched or thrown.

3.2.1 Time taken by disc C to reach its maximum height:

To determine the time taken by disc C to reach its maximum height, we can use the kinematic equation for vertical motion. The equation is:

vf = vi + at

Where:

vf = final velocity (which is zero at the maximum height)

vi = initial velocity

a = acceleration (in this case, acceleration due to gravity, -9.8 m/s²)

t = time

Since the disc is thrown vertically upwards, its initial velocity is 15 m/s. We want to find the time it takes for the disc to reach its maximum height, so we'll use the equation and solve for time (t):

0 = 15 + (-9.8)t

Rearranging the equation, we get:

9.8t = 15

t = 15 / 9.8

Calculating this, we find:

t ≈ 1.53 seconds

Therefore, it takes approximately 1.53 seconds for disc C to reach its maximum height.

3.2.2 Maximum height above the ground reached by disc C:

To determine the maximum height reached by disc C, we can use another kinematic equation for vertical motion:

vf² = vi² + 2ad

Where:

vf = final velocity (which is zero at the maximum height)

vi = initial velocity

a = acceleration (in this case, acceleration due to gravity, -9.8 m/s²)

d = displacement (maximum height)

Since we know the initial velocity (vi) and acceleration (a), we can solve for the displacement (d), which represents the maximum height:

0² = 15² + 2(-9.8)d

Rearranging the equation, we get:

0 = 225 - 19.6d

19.6d = 225

d = 225 / 19.6

Calculating this, we find:

d ≈ 11.48 meters

Therefore, the disc C reaches a maximum height of approximately 11.48 meters above the ground.

Calculating the time from the moment that disc C was thrown upwards until the time ball B hits the disc:

To find the time it takes for ball B to hit disc C, we need to calculate the time it takes for both objects to reach the same height.

Since disc C was thrown upwards from the edge of the roof and ball B was shot vertically upwards from the foot of the building, we need to consider the additional height of the building (30 meters).

The time it takes for disc C to reach the ground is the same as the time it takes for ball B to reach a height of 30 meters above the ground.

Using the kinematic equation for vertical motion, we can calculate the time for ball B:

d = vit + 0.5at²

Where:

d = displacement (30 meters)

vi = initial velocity (40 m/s)

a = acceleration (acceleration due to gravity, -9.8 m/s²)

t = time

30 = 40t + 0.5(-9.8)t²

Rearranging the equation, we get:

4.9t² + 40t - 30 = 0

Solving this quadratic equation, we find:

t ≈ 1.29 seconds or t ≈ -5.82 seconds

Since time cannot be negative in this context, we discard the negative solution.

Therefore, it takes approximately 1.29 seconds from the moment that disc C was thrown upwards until ball B hits the disc.

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(a) the shell's horizontal range = 285839 m.

(b) the amount of time the shell is in motion is: 273.11 sec.

The length of the horizontal axis is a projectile's horizontal range. In addition, it would move before returning to the original vertical position.

Horizontal range is equal to Initial Speed in the horizontal direction, multiplied by the time the shell is in motion.  Therefore, both questions can be solved at the same time.

Step 1.  Find Velocity in the Horizontal direction by:  

Vx = (1.70 x 10³) × cos(52°)

Vx = 1046.6 m/sec

Step 2.  Find Velocity in the Vertical direction by:

Vy = (1.70 x 10³) × sin (52°)

Vy = 1339.6 m/sec

Step 3. Find the amount of time the shell is in motion by:  

t= (2 × Vy)/gravity

t = (2 × 1339.6) / 9.81       [ Gravity = 9.81m/s²]

t = 273.11 sec

Step 4. Find the shell's horizontal range:  

Distance = Vx × t

or, distance = (1046.6 × 273.11)

or, distance = 285839 m

Thus, (a) the shell's horizontal range = 285839 m.

(b) the amount of time the shell is in motion is: 273.11 sec.

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12. The time taken for the journey is 400 s

13. The time taken for the train is 200 s

14. The time taken is 7 h

15. The time taken for the beetle is 12 s

16. The time taken for the journey is 0.0068 h

The time taken in each case as given by the question can be obtain as follow:

12. The time taken for the journey

Time taken = Distance / Speed

Time taken = 26000 / 65

Time taken = 400 s

13. The time taken for the train

Time taken = Distance / Speed

Time taken = 3200 / 16

Time taken = 200 s

14. The time taken to travel

Time taken = Distance / Speed

Time taken = 672 / 96

Time taken = 7 h

15. The time taken for the beetle

Time taken = Distance / Speed

Time taken = 1.08 / 0.09

Time taken = 12 s

16. The time taken for the journey

Time taken = Distance / Speed

Time taken = 35 / 5147

Time taken = 0.0068 h

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Answer:

a) El valor de la densidad es 0.79 \(\frac{g}{cm^{3} }\) o 790 \(\frac{g}{cm^{3} }\)

b) El peso especifico es 7749.9\(\frac{N}{m^{3} }\)

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

\(d=\frac{m}{V}\)

En este caso:

Reemplazando:

\(d=\frac{237 g}{300cm^{3} }\) →  d=0.79 \(\frac{g}{cm^{3} }\)

\(d=\frac{0,237 kg}{0,0003m^{3} }\)→  d=790 \(\frac{g}{cm^{3} }\)

El valor de la densidad es 0.79 \(\frac{g}{cm^{3} }\) o 790 \(\frac{g}{cm^{3} }\)

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 \(\frac{m}{s^{2} }\)= 2,32497 N

el peso especifico es calculado como:

\(Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }\)

Pe= 7749.9\(\frac{N}{m^{3} }\)

El peso especifico es 7749.9\(\frac{N}{m^{3} }\)

The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)

Pressure (P) = Density (ρ) × g × Depth (h)

Given: Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Depth (h) = 6 m

Using the given values, one can calculate the pressure:

Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure

= 58800 kg·m⁻¹·s⁻²

Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x \(10^5\) Pa:

Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x\(10^5\) Pa)

Pressure = 580.124 atm

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Electricity

Energy transfer is the movement of energy from one point to another. This can be heat energy, kinetic energy, etc.

There are three forms of energy transfer, namely:

Conduction

Convection

Radiation

Conduction is the transfer of heat energy in which there is no actual movement of the molecules involved. This usually happens in solid materials

Convection is the transfer of energy in fluids. This involves the movement of the molecules through which the energy is transferred.

Radiation is the transfer of energy in which no material is involved.

Note that electricity only involves the movement of charges, it is not a mode of energy transfer.

Answer:

Metals are good conductors of heat and electricity. ... As metals are good in conducting heat, it can transfer heat very quickly and evenly. Generally, copper, aluminum metals are used for cooking vessels as they are very good conductors of heat. Due to this reason's metals are used in cooking vessels.

Answer:

h= 63.5 m

Explanation:

       \(\Delta h = \frac{1}2} * g* t^{2} = \frac{1}{2} *(-9.8 m/s2)*(3.6s)^{2} = -63.5 m\)

        y₀ = h = 63.5 m

Answer: When a cube is divided into smaller cubes of equal size, the length, height, and width of each new cube are one-third of the original cube.

So each new cube will have a length, height, and width of 4/3 cm.

To calculate the surface area of each new cube, we need to find the area of each face and then add them up. Each cube has six faces, so we can calculate the surface area of one cube by multiplying the area of one face by six.

The area of one face of the new cube is (4/3)*(4/3) = 16/9 cm².

So the surface area of one new cube is 6*(16/9) = 96/9 cm² = 10.67 cm² (rounded to two decimal places).

Since we have 8 new cubes, the total surface area is 8 times the surface area of one new cube:

Total surface area = 8 * 10.67 cm² = 85.33 cm² (rounded to two decimal places).

Therefore, the surface area of the 8 new cubes is 85.33 cm².

Answer:

B is the answer bc it's no way c an a and d can maybe c

Answer:

5.95m/s to 2 decimal places

Explanation:

In physics speed is measured in metres per second so convert 8mins to seconds

8x60=420 seconds

The formula needed:

Speed (m/s)= Distance (m)/Time (s)

2500/420=5.95m/s

The coin's acceleration is 0.37 m/s²

Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.

Calculation:-

V² = U -2aS

a = U/2S

  = 2/2×2.7

   =  0.37 m/s²

Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.

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