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1 year ago

A 3.0 kg block is pushed 1.0 m at a constant

Physics

1 answer:

PSYCHO15rus [73]1 year ago

3 0

(a) The work done by the applied force is 32.6 J.

(b) The work done by gravity on the block is 29.4 J.

<h3>What is the force applied to the block?</h3>

The force applied to the block is determined by applying the formula for the net force on the block.

F(net) = 0

Fcosθ - mg - Fsinθμ = 0

Fcosθ - Fsinθμ = mg

F(cosθ - sinθμ) = mg

F = (mg) / (cosθ - sinθμ)

F = (3 x 9.8) / (cos26 - 0.2 x sin26)

F = 36.25 N

The work done by the applied force is calculated as;

W = Fd cosθ

W = (36.25 x 1) x cos(26)

W = 32.6 J

The work done by gravity on the block is calculated as follows;

W = mgd

W = 3 x 9.8 x 1

W = 29.4 J

The magnitude of the normal force is calculated as follows;

N = mg

N = 3 x 9.8

N = 29.4 N

Learn more about normal force here: brainly.com/question/14486416

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Answer:

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Explanation:

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Therefore, since k×k = 0

$\vec{v} \times \vec{B} = (v_x\hat{i} + v_y \hat{j} + v_z \hat{k}) \times B\hat{k}\\\vec{v} \times \vec{B} = (Bv_x \hat{i} \times\hat{k}) + (Bv_y \hat{j}\times\hat{k})$

And since i, j , k are a rigth hand system:

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j × k = i

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A 1200N load is to be lifted with 200N effort using a first class lever. At what distance

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Explanation:

Hey there!!

Let's simply work with it.

Here,

load = 1200N

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Load distance = 15cm

We have,

According to the principle of lever.

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1200×15 = 200× ED.

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$ed = \frac{18000}{200}$

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Hope it helps....

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