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3 years ago

The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?

1 answer:

4 0

Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

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Latitude and longitude are the two coordinates that determine a specific point on the Earth’s surface. How does knowing the loca

DedPeter [7]

Answer:

hey sandra! i hope this will help you!

Explanation:

The geographic coordinate system is a reference system that uses the two angular latitude (north or south) and longitude (east or west) coordinates to determine the positions of the land surface points. These two angular coordinates measured from the center of the Earth are from a spherical coordinate system that is aligned with its axis of rotation. These coordinates are usually expressed in sexagesimal degrees:

Latitude measures the angle between any point and the equator. Latitude lines are called parallel and are circles parallel to the equator on the Earth's surface. Latitude is the distance between any point and Ecuador, measured on the meridian that passes through that point.

• All points located on the same parallel have the same latitude.

• Those who are in the north of Ecuador receive the denomination North (N).

• Those who are in the south of Ecuador receive the denomination South (S).

• It is measured from 0º to 90º.

• The latitude of 0º corresponds to Ecuador.

• The North and South poles have latitude 90º N and 90º S respectively.

The length measures the angle along the equator from any point on Earth. It is accepted that Greenwich in London is length 0 in most modern societies. The lines of length are maximum circles that pass through the poles and are called meridians.

• All points located on the same meridian have the same length.

• Those who are east of Meridian Zero receive the designation East (E).

• Those who are west of Meridian Zero receive the denomination West (O).

• It is measured from 0º to 180º.

• The Greenwich meridian corresponds to the length 0º.

By combining these two angles, the position of any point on the Earth's surface can be expressed.

6 0

3 years ago

In an oscillating lc circuit, when 75.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of th

olga2289 [7]

<span>(a) E = Â˝ QÂ˛/C, so .. (b) E(max) = Â˝LiÂ˛ (i=current), so .</span>

4 0

3 years ago

The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque

Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago

The pulse site located at the point where the upper leg bends is called the

VLD [36.1K]

The pulse site located at the point where the upper leg bends is called the femoral. It is an artery found in the thigh. It is large and is deemed as the main arterial supply for the lower part of the body. It is known as the second artery that is the largest. It is being used as the catheter access artery. From it, diagnostics for the heart, brain, arms, kidney and other parts can be directed to the other arterial system. It can also be used as a source to draw blood that is from the arteries when there is low blood pressure.

3 0

3 years ago