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kodGreya [7K]
3 years ago
5

I will mark as the brainliest answerplz 8,9,10​

Physics
1 answer:
blagie [28]3 years ago
5 0

Answer:

8.  acceleration = \dfrac{d(velocity)}{d(time)}  = 1 unit .

9. acceleration = \dfrac{d(velocity)}{d(time)}  = -1 unit.

10. acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units.

Explanation:

8. i) acceleration = velocity / time

  ii) In this figure velocity = time

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)} = 1 unit .

9. i) acceleration = velocity / time

  ii) In this figure 4 = m + 5, therefore m = -1

     therefore velocity = (-0.5 \times time) + 5

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)}  = -1 units.

10.) velocity is constant at 2

     therefore acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units

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Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N
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First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say F_{1}, and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
F_{1} = 7i, where i = Unit vector in the x-direction.

2) Take the second force, say F_{2}, and assume that it is making an angle \alpha with the first force F_{1}.

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

F_{2} = (15*cos \alpha)i + (15*sin \alpha )j

Now from simple vector addition, we know that,
F_{R} = F_{1} + F_{2} --- (A)

Where F_{R} = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find F_{1}+F_{2} first!
F_{1}+F_{2} =  7i+(15*cos \alpha)i + (15*sin \alpha )j

=> F_{1}+F_{2} =  (7+15*cos \alpha)i + (15*sin \alpha )j

Now the magnitude of F_{1}+F_{2} is,
| F_{1}+F_{2}| = \sqrt{ (7+ 15*cos \alpha)^{2} +  (15*sin \alpha )^{2}}

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}

Since (sin \alpha)^{2} + (cos \alpha)^{2} = 1, therefore,

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Since  | F_{1}+F_{2}| = |F_{R}|, and the magnitude of the resultant force is 20N, therefore,

 |F_{R}| = | F_{1}+F_{2}|
20 = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Take square on both sides,
400 = 49 + 225 + 210*(cos \alpha)
(cos \alpha) =  \frac{3}{5}

\alpha = 53.13^{o}

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

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The distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.


<h3>What is distance?</h3>

Distance can be defined as the horizontal length between two points.

To calculate the distance at which the string must be attached to balance the rod, we use the formula below.

Formula:

  • m₁gx = m₂g(1.2-x)................. Equation 1

From the question,

Given

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  • m₂ = 2.5 kg
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Substitute these values into equation 1

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Solve for x

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Collect like terms

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Hence, the distance from the 2.0 kg mass at which a string should be attached to balance the rod is 0.67 m.

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