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3 years ago

Starting from rest, a car travels 18 meters as it accelerates uniformly for 3.0 seconds. What is the magnitude of

Physics

1 answer:

omeli [17]3 years ago

8 0

Answer:

$a=4\frac{m}{s^2}$

Explanation:

Hello.

In this case, for this uniformly accelerated motion in which the car starts from rest at 0 m/s and travels 18 m in 3.0 s, we can compute the acceleration by using the following equation:

$x_f=x_0+v_0t+\frac{1}{2}at^2$

Whereas the final distance is 18 m, the initial distance is 0 m, the initial velocity is 0 m/s and the time is 3.0 s, that is why the acceleration turns out:

$a=\frac{2(x_f-v_ot)}{t^2} =\frac{2(18m-0m/s*3.0s)}{(3.0s)^2}\\ \\a=4\frac{m}{s^2}$

Best regards.

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

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The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

<em>Length of the string, L = 100 cm</em>

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The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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Let masses (M) and (m) of two objects are distance (d) apart. Let F be the attractional force between two masses.

Importance of The Universal Law of Gravitation

It binds us to the earth.

It is responsible for the motion of the moon around the earth.

It is responsible for the motion of planets around the Sun.

Gravitational force of moon causes tides in seas on earth.

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When an object falls from any height under the influence of gravitational force only, it is known as free fall.

Acceleration Due to Gravity

When an object falls towards the earth there is a change in its acceleration due to the gravitational force of the earth. So this acceleration is called acceleration due to gravity.

The acceleration due to gravity is denoted by g.

The unit of g is same as the unit of acceleration, i.e., ms−2

Mathematical Expression for g

From the second law of motion, force is the product of mass and acceleration.

F = ma

For free fall, acceleration is replaced by acceleration due to gravity.

Therefore, force becomes:

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Factors Affecting the Value of g

As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

As we go at large heights, value of g decreases.

To Calculate the Value of g

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Mass of the earth, M = 6 × 1024 kg, and

Radius of the earth, R = 6.4 × 106 m

Putting all these values in equation (iii), we get:

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Difference between Gravitation Constant (G) and Gravitational Acceleration (g)

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Its value is 6.67×10-11Nm2/kg2.

Its value is 9.8 m/s2.

It is a scalar quantity.

It is a vactor quantity.

Its value remains constant always and everywhere.

Its value varies at various places.

Its unit is Nm2/kg2.

Its unit is m/s2.

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Let an object is falling towards earth with initial velocity u. Let its velocity, under the effect of gravitational acceleration g, changes to v after covering the height h in time t.

Then the three equations of motion can be represented as:

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Height covered in t seconds, h = ut + ½gt2

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The weight of a body changes from place to place, depending on mass of object.

The SI unit of weight is Newton.

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Weight of an object is given as,

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Try the following questions:

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