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Nutka1998 [239]
3 years ago
12

which of the seasons is celebrated by many northern cultures with festivals, often symbolizing rebirth?

Physics
2 answers:
tensa zangetsu [6.8K]3 years ago
4 0
I’m not entirely certain, but usually, spring symbolizes rebirth.
aalyn [17]3 years ago
3 0

Spring is the season which symbolizes renewal or rebirth which is celebrated in the north by the cultural festival.

<u>Explanation:</u>

Early to 14th century, spring was referred to as the Lent. It is believed that the Goddess of fertility, Persephone returned during this time. Apart from these mythological beliefs, seeds are planted during the spring and it marks blooming of many flowers as the sun stays for long and rises early. Therefore, spring is celebrated as the season for rejuvenation and rebirth or as renewal season.

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Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted?
Iteru [2.4K]

(a) 50.4 N\cdot m

The torque exerted on the solid disk is given by

\tau=Frsin \theta

where

F is the magnitude of the force

r is the radius of the disk

\theta is the angle between F and r

Here we have

F = 180 N

r = 0.280 m

\theta=90^{\circ} (because the force is applied tangentially to the disk)

So the torque is

\tau = (180 N)(0.280 m)(sin 90^{\circ})=50.4 N\cdot m

(b) 17.2 rad/s^2

First of all, we need to calculate the moment of inertia of the disk, which is given by

I=\frac{1}{2}mr^2

where

m = 75.0 kg is the mass of the disk

r = 0.280 m is the radius

Substituting,

I=\frac{1}{2}(75.0)(0.280)^2=2.94 kg m^2

And the angular acceleration can be found by using the equivalent of Newtons' second law for rotational motions:

\tau = I \alpha

where

\tau = 50.4 N \cdot m is the torque exerted

I is the moment of inertia

\alpha is the angular acceleration

Solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{50.4 N \cdot m}{2.94 kg m^2}=17.2 rad/s^2

8 0
3 years ago
An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5 m/s2
uranmaximum [27]

Answer:

39 m/s²

Explanation:

initial Velocity (u) = 9m/s

acceleration (a) = 1.5m/s²

time(t) = 20s

V=u + at

V=9 + 1.5(20)

V=9 + 30

V=39m/s

7 0
3 years ago
Compare acids and bases. Be able to explain their properties and give examples
Ghella [55]
Acids are the EXACT OPPOSITE of bases, but they're both still deadly. Base (or basic/alkali) is something in everyday thinks like toothpaste, milk, etc. But too much intake of bases can harm you, and if water has too much alkali, the fish living in it can die. Acidic, on the other hand, is very effective. Acids are usually in oranges, sodas, even your stomach has acids in it. Too much acids can rot your teeth, can cause "heartburn", even can give you a stomachache.
7 0
3 years ago
Four point charges are on the corners of a rectangle, 8 m by 6 m in size. What is the magnitude of the resultant electric field
Minchanka [31]

Answer:

E = 1.44*10⁹* Q N/C

Explanation:

  • The Electric Field due to any of the charges, at the center of the rectangle, can be expressed as follows:

       E = \frac{K*Q}{r^{2}} (1)

       where K = 9*10⁹ N*m²/C², Q is the charge (in Coulombs) of any

       charge, and r is the distance of the charge to the center of the

       rectangle, which it is exactly half of the diagonal of the rectangle.

  • If the sides of the rectangle are 8m and 6m, the diagonal of the rectangle, just applying the Pythagorean Theorem, is as follows:

       d = \sqrt{(8m)^{2} + (6m)^{2}} = 10 m (2)

       so, r = 5 m.

  • Replacing by the givens, the electric field due to one of the charges is simply:

      E = \frac{K*Q}{r^{2}} = \frac{9e9N*m2/C2*Q}{(5m)^{2}} = 0.36*e9*Q N/C (3)

  • Due to the symmetry of the problem, the magnitude of the Electric Field due to any of the charges will be the same (assuming all charges are equal each other), so the total electric field, in magnitude, will be just four times the one due to any of the charges, as follows:

       E_{tot} = 4* E = 1.44e9*Q N/C  (4)

5 0
3 years ago
A worker pushes a wheelbarrow with a force of 40 n over a level distance of 6.0 m. if a frictional force of 24 n acts on the whe
nikklg [1K]
Net work done = 40*6 - 24 *6
=96 j
3 0
3 years ago
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