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1 year ago

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What is the angle of refraction when a ray of light passes frm ethanol to air

1 answer:

77julia77 [94]1 year ago

5 0

Answer: Your answer is<u> 1.36.</u>

Hope this helps!

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A car pulled by a tow truck has an acceleration of 2.0 m/s^2. What is the mass of the car if the net force on the car is 3,500 N

Nana76 [90]

Answer:

ans: 1750 kg

Explanation:

F = ma

so ,

m = F/a

= 3500 / 2

= 1750 kg

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2 years ago

a log splitter puts out 1 100W of power for every 1500 W put into it. the efficiency of the machine is what

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Answer:400 is the answer 1500-1100 is 400

Explanation:

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1 year ago

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2 years ago

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

3 0

2 years ago

A length of copper wire has a resistance 44 Ω. The wire is cut into three pieces of equal length, which are then connected as pa

WINSTONCH [101]

Answer:

$\frac{R}{1} = \frac{44}{9}\ohm$

Explanation:

Let us imagine that there are three wire of length equal length having equal resistances each of 44/3 Ω

Now connect these wires in parallel to so that their equivalent resistance is R.

then

$\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R} = \frac{3}{44}+\frac{3}{44}+\frac{3}{44}$

$\frac{1}{R} = \frac{9}{44}$

⇒ $\frac{R}{1} = \frac{44}{9}\ohm$

5 0

2 years ago

Read 2 more answers