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3 years ago

Please answer this. Science 7th grade.

Physics

1 answer:

nikdorinn [45]3 years ago

6 0

Answer:

the answer would be 2

Explanation:

it would be 2 because if u look at the diagram the darkest arrow is pointsin towards earth and the moon and when the moon is infront of the sun it cause's an eclispe

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The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an ang

V125BC [204]

Answer:

2954.6 N/C, 46.36 degree from positive axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E = 1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

5 0

3 years ago

Global Precipitation Measurement (GPM) is a tool scientists use to forecast weather. Which statements describe GPM? Select three

lyudmila [28]

Answer:

B.It is a satellite that collects data about rain and snow

C.Its orbit covers 90 percent of Earth’s surface

F.The sensors measure microwaves

5 0

3 years ago

If the time for F18 to take off doubles, the acceleration will ____.

kherson [118]

Answer:

decreases

Explanation:

a=[v-u]/t

as time increases acceleration decreases and vice versa

8 0

3 years ago

an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul

kolbaska11 [484]

Answer:

$r=\frac {mv}{qb}$

Explanation:

In this case, since the charged particle moves in circular motion, the centripetal force is equivalent to the magnetic force.

$\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}$

5 0

3 years ago

A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s

3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

$F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s$

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0

3 years ago