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Dmitriy789 [7]
3 years ago
8

An object's acceleration is given by a(t)=a(t)=60t m/s260t m/s2 . if it begins at rest, how far has it gone after 10 seconds?

Physics
1 answer:
Alex Ar [27]3 years ago
6 0
The object's acceleration is given by the expression: a(t) = 60t m/s^2. The distance it traveled after 10 seconds after starting from rest can be obtained by integrating the expression, a(t), twice in order to get the expression for the distance as a function of time. 

The first integral of a(t) is the velocity, v(t), of the object which is then equal to: v(t) = 30t^2. The integral of v(t) is then distance, d(t), of the object which is then equal to: d(t) = 10t^3. With the distance equation available, we simply plug in t = 10 seconds into the equation. It is then determined that the object has traveled 10,000 m or 10 km after 10 seconds starting from rest.    
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The charged particles are often deflated in a magnetic field.

<h3>What is a magnetic field?</h3>

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What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011
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Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

<h3>What is electric force?</h3>

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4 0
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8
Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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