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VladimirAG [237]
3 years ago
8

Which of the following is the smallest conceivable amount of time that could pass between a lunar eclipse and a solar eclipse .

( 1point )
a . bout 14 days

b. bout 28 days

c . about 183 days

d. about 365 days
Physics
1 answer:
pickupchik [31]3 years ago
5 0
The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
because earth is making shade on moon

in order for solar eclipse to occur the line is now:
Sun-Moon-Earth
because moon is making a shade on earth (blocking sun = solar eclipse)

Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14
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Electricity.
When electricity flows through wire (such as a battery circuit) it creates a magnetic field around the wire.
6 0
3 years ago
A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and
victus00 [196]

Answer:

a)  t=195.948N.m

b)  \phi=13.6 \textdegree

Explanation:

From the question we are told that:

Density \rho=1.225kg/m^2

Velocity of wind v=14m/s

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient \mu=2.05

a)

Generally the equation for Force is mathematically given by

F=\frac{1}{2}\muA\rhov^2

F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2

F=163.29

Therefore Torque

t=F*r*sin\theta

t=163.29*1.2*sin90

t=195.948N.m

b)

Generally the equation for torque due to weight is mathematically given by

t=d*Mg*sin90

Where

d=sin \phi

Therefore

t=sin \phi*Mg*sin90

195.948=833sin \phi

\phi=sin^{-1}\frac{195.948}{833}

\phi=13.6 \textdegree

5 0
3 years ago
A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo
denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

\mu _smg=610

\mu _s\times 93\times 9.8=610

\mu _s=0.669

5 0
3 years ago
What is a lol diagram.
Novosadov [1.4K]

Explanation:

where is your diagram? lol

7 0
2 years ago
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hichkok12 [17]
0.76kg !!!!! I hope it right
3 0
2 years ago
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