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BigorU [14]
3 years ago
12

What are some of the names given to the Moon by the Algonquin?

Physics
1 answer:
frutty [35]3 years ago
6 0
The names given to the Moon by the Algonquin are as follows:

January- Full Wolf Moon
February- Full Snow Moon
March- Full Worm Moon
April- Full Pink Moon
May- Full Flower Moon
June- Full Strawberry Moon
July- Full Buck Moon
August- Full Sturgeon Moon
September- Full Corn Moon; Full Harvest Moon
October- Full Hunter's Moon
November- Full Beaver Moon
December- Full Cold Moon; Full Long Nights Moon
You might be interested in
What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
3 years ago
Read 2 more answers
The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0
3 years ago
008 (part 3 of 4) 3.0 points
motikmotik

Car A take a time of 2.55hr and car B take a time of 2.14 hr

We know that distance divide by time is speed

here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr​

so speed=distance/time

s=d/t

t=d/s

=189/74

=2.55hr

In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr  

s=d/t

t=d/s

=199.8/93

=2.14hr  

so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h

learn more about Speed here brainly.com/question/13943409

#SPJ9

5 0
1 year ago
30 points plz help ill do anything... literally anything.
ruslelena [56]

Answer:

1. 2.5s

Explanation:

1. For time, divide Distance / speed

25m / 10

=2.5s

3 0
3 years ago
Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh
ahrayia [7]

Answer:

i) FALSE,  ii) TRUE,  iii) FALSE, iv)  FALSE

Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

           n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

Let's apply these equations to our case where

          nₓ <n_y

i) The expression of the refractive index

          nₓ < n_y

         \frac{c}{v_x} = \frac{c}{n_y}

         v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

 the statement of FALSE

8 0
3 years ago
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