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3 years ago

8

Select the correct answer.

2 answers:

arsen [322]3 years ago

8 0

When a wave travels through a material, it loses a small amount of energy as heat. Eventually, the wave will stop when it loses all of its energy.

I think the answer is A.

andrew-mc [135]3 years ago

7 0

I think the answer would be C

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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 43.

zmey [24]

To solve this problem, we must take two important steps. First we will convert all the given units, to international system. Later we will define the torque, which is given as the product between the radius of application of the force and the Force acting on the body. Mathematically the latter is,

$\tau = r F$

Here,

r = Radius

F = Force

Now the units,

$r = 1.95cm = 1.95*10^{-2}m$

Replacing,

$\tau = (1.95*10^{-2})(43.5N)$

$\tau = 0.8483N\cdot m$

Therefore the torque that the muscle produces on the wrist is $0.85N\cdot m$

3 0

3 years ago

A ray of laser light travels through air and enters an unknown material. The laser enters the material at an angle of 36 degrees

V125BC [204]

Answer:1.27

Explanation:

Given

incident angle $i=36^{\circ}$

refracted angle $r=27.5^{\circ}$

Suppose $n_2$ is the refractive index of material then using Snell's law we can write

$n_1\sin i=n_2\sin r$

where $n_1$ =refractive index of air

$1\times \sin (36)=n_2\times \sin (27.5)$

$n_2=\dfrac{0.5877}{0.4617}$

$n_2=1.27$

3 0

2 years ago

A charge in motion is called a (n) ______.

Lubov Fominskaja [6]

<span>It's called an electric current</span>

7 0

2 years ago

Read 2 more answers

I need help with question 6 and 7!!! Also, how to do the back???

lora16 [44]

Reword the questions and you’ll get it right

7 0

1 year ago

Read 2 more answers

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

or

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago