All categories

3 years ago

Spring tides have higher high tides and lower low tides when the moon, sun, and earth are aligned in a row. Why does this occur?

1 answer:

8 0

Spring tides have higher high tides and lower low tides when the moon, sun, and earth are aligned in a row. Why does this occur?

A.gravitational pull is stronger

You might be interested in

(11) The speed of radio waves is 300 000 000 m/s.

Neko [114]

Answer:

Wavelength = 9.68 meters

Explanation:

Given the following data;

Speed = 300,000,000m/s

Frequency = 31 Megahertz to Hertz = 31 * 10⁶ Hz

To find the wavelength;

Wavelength = speed/frequency

Wavelength = 300,000,000/31,000,000

Wavelength = 9.68 meters

7 0

2 years ago

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

Finally:

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

2 years ago

A circular loop of wire is in the plane of the paper. The south pole of a bar magnet is being moved from a position in front of

drek231 [11]

Answer:

Counterclockwise

explanation in attachment

5 0

3 years ago

What net force would be needed to accelerate the box at 2.3 m/s^2

Akimi4 [234]

The pertinent equation here is F=ma. You haven't shared the mass of the box, so I will use M to represent that mass.

Then F = M(2.3 m/s^2) (answer)

6 0

3 years ago

Read 2 more answers

A 12.0 cm object is 9.0 cm from a convex mirror that has a focal length of -4.5 cm. What is the distance of the image from the m

Rasek [7]

Answer:

- 3 cm

Explanation:

From the mirror formula;

1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.

1/-4.5 = 1/9 + 1/v

1/v = -1/4.5 - 1/9

= -1/3

Therefore;

v = -3 cm

Hence;

Image distance is - 3cm

5 0

2 years ago