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Oduvanchick [21]

3 years ago

10

Please Answer: Which lists the waves in order of frequency, from highest to lowest? A.X-rays, gamma rays, ultraviolet B.radio wa

ves, visible light, X-rays C.ultraviolet, microwaves, gamma rays D.visible light, infrared, microwaves

1 answer:

sergeinik [125]3 years ago

4 0

Answer:

D.visible light, infrared, microwaves

Explanation:

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A snail moves 40 cm in<br> 20 seconds. How fast<br> did it move?

nasty-shy [4]

Answer:

too fast to be a snail

......

2 cm per second

5 0

3 years ago

The ideal gas law includes all of the following properties, except the

Verizon [17]

The correct answer among the choices is option D. Density is not one of the properties included in the ideal gas law. The law is expressed as: PV=nRT. As we can see, the pressure, the volume and the temperature of the gas are included in the law.

3 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

2 years ago

Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos

soldi70 [24.7K]

<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

F₁ = 100 cos 20 i + 100 sin 20 j

F₁ = 93.97 i + 34.20 j

That is force 2 is 150 pound with x axis at 60°

F₂ = 150 cos 60 i + 150 sin 60 j

F₂ = 75 i + 129.90 j

F₁ + F₂ = 93.97 i + 34.20 j + 75 i + 129.90 j

F₁ + F₂ = 168.97 i + 164.10 j

$\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0$

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0

3 years ago

An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0

3 years ago