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I am Lyosha [343]
1 year ago
7

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.

(c) The magnitude of the normal force is 29.4 N.

<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

#SPJ1

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Part b)

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Part a)

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so here frequency of light will remain same in both water and glass and it will be same as that in air

f = \frac{v}{\lambda}

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Part b)

As we know that the refractive index of water is given as

\mu_w = 4/3

so the wavelength in the water medium is given as

\lambda_w = \frac{\lambda}{\mu_w}

\lambda_w = \frac{656 nm}{4/3}

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Similarly the refractive index of glass is given as

\mu_w = 3/2

so the wavelength in the glass medium is given as

\lambda_g = \frac{\lambda}{\mu_g}

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Part c)

Speed of the wave in water is given as

v_w = \frac{c}{\mu_w}

v_w = \frac{3 \times 10^8}{4/3}

v_w = 2.25 \times 10^8 m/s

Speed of the wave in glass is given as

v_g = \frac{c}{\mu_g}

v_g = \frac{3 \times 10^8}{3/2}

v_g = 2 \times 10^8 m/s

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