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zimovet [89]
4 years ago
12

Infer why the output force exerted by a rake must be less than input force?

Physics
1 answer:
adell [148]4 years ago
5 0
<h3><u>Answer and Explanation</u>;</h3>
  • input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
  • <em><u>The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. </u></em>
  • <em><u>The rake is not going to be able to convert all of the input force into output force, the force the rake applies to move the leaves, because of friction.</u></em>
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Given that:<br><br> = 2i + 9j ; and ⃗ = -i – 4j . Find . ⃗ ​
Marat540 [252]

Answer:

A.B = -38

Explanation:

A = 2i + 9j and B = -i - 4j.

So, A.B = (2i + 9j).(-i - 4j)

= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)

= -2i.i - 8i.j - 9j.i - 36j.j

since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have

A.B = -2(1) - 8(0) - 9(0) - 36(1)

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3 years ago
Describe what happens to pressure as the force exerted on a given area increases.
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4 0
3 years ago
) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)
natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

\tau = \vec{F} \times \vec{r}

\tau = (8i+6j)\times(-3i+4j)

\tau = (8*4)(i\times j)+(6*-3)(j\times i)

\tau = 32k +18k

\tau = 50 k

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}

cos\theta = -0.24

\theta = cos^{-1} (-0.24)

\theta = 103.88\°

Therefore the angle between the ratio and the force is 103.88°

5 0
3 years ago
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