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2 years ago

Convert 3.8 kg to Lbs

Physics

2 answers:

JulijaS [17]2 years ago

8 0

I believe it’s 8.3 to be precise

Ray Of Light [21]2 years ago

3 0

Answer:

8.37757

Explanation:

may i have brainliest

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A layer of ethyl alcohol (n= 1.361) is on top of the water ( n=1.333) . At what angle relative to the normal to the interface of

hram777 [196]

Answer:

78.35°

Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal to the interface of the two liquids is light totally reflected?

From Snell's Law,

(ni)/(nr) = Sin (θr) / Sin (θi)

Where

θi = Angle of Incidence

θr = Angle of refraction

ni = Refractive index given for ethyl alcohol

nr = Refractive index of medium from which light is refracted

ni = 1.361

nr = 1.333

, θr = 90° ( Critical Angle is reffered to as Angle of Incidence at refracted angle of 90°) (θi = θc)

(ni)/(nr) = Sin (θr) / Sin (θi)

1.361/ 1.333 = Sin (90°)/ Sin( θc)

1.021= 0.894/ Sin( θc)

Sin( θc)= (0.9794

θc = Sin⁻¹ 0.9794)

θc = 78.35°

7 0

3 years ago

A uniform disk with radius 0.650 m

VashaNatasha [74]

Answer:

$a = 13.758\,\frac{m}{s^{2}}$

Explanation:

First, the instant associated to the angular displacement is:

$(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s$

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

$\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)$

$\omega (0.240\,s) = 4.124\,\frac{rad}{s}$

The angular acceleration is obtained by deriving the previous function:

$\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}$

The resultant linear acceleration on the rim of the disk is:

$a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )$

$a_{t} = 8.190\,\frac{m}{s^{2}}$

$a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}$

$a_{n} = 11.055\,\frac{m}{s^{2}}$

$a = \sqrt{a_{t}^{2}+a_{n}^{2}}$

$a = 13.758\,\frac{m}{s^{2}}$

3 0

3 years ago

Of the eight most abundant elements in earths crust, what percentage does the six next most abundant elements of earths crust ma

Mrac [35]

Answer:

the sixth most abundant element in the earth's crust is sodium and it occupies 2.8%

5 0

3 years ago

A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te

Nitella [24]

Answer:

50 N

Explanation:

Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:

The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.

The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

6 0

3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago