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Svetach [21]
2 years ago
10

The path of the sun on the celestial sphere is called.

Physics
1 answer:
Ede4ka [16]2 years ago
7 0

Answer:

Ecliptic

Explanation:

The path the sun, moon, and planets take across the sky as seen from Earth. It defines the plane of the Earth's orbit around the sun. The name "Ecliptic" comes from the fact that eclipses take place along this line.

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How would you describe the behavior of particles in a solid?
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3 years ago
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What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0
3 years ago
Naoki's bicycle has a mass of 9 kg. If Naoki sits on her bicycle and starts pedaling with a force of 107.2 N, causing an acceler
Murrr4er [49]

Answer:

Explanation:

F = ma

m = F/a

m = 107.2/1.6 = 67 kg

so Naoki's mass must be the total mass - mass of the bike

so her mass is 67 - 9 = 58 kg...B

6 0
3 years ago
An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with const
andreev551 [17]

Answer:

0 N

Explanation:

This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N

5 0
2 years ago
A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the s
tamaranim1 [39]

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.  

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j

(a) The velocity of boat with respect to ground is

\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s

(b) The direction is given  by

tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o

7 0
2 years ago
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