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Rina8888 [55]
3 years ago
15

Planets, stars, asteroids and other celestial bodies have generally round shapes. Some are ovals, others really bumpy circles, a

nd some are a very squished oval. Why are none triangular or rectangular?
Physics
1 answer:
ss7ja [257]3 years ago
8 0
Planets revolve and rotate, they go around something and they spin, pretend they are on a road(orbit path) while they rotate and revolve. It's liking rolling a chunk of playdough in your hands to form a ball. same goes for asteroids, stars are auctally gases so that one I'm not sure how to answer.
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Use the diagram below modeling a football kicked from a horizontal surface B
djyliett [7]
B is the correct one
5 0
3 years ago
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Velocityis a vector quantity which has both magnitude and direction. Using complete sentences, describethe object's velocity. Co
mr_godi [17]

Answer:

Explanation:

A physical quantity which can be completely described by the magnitude and direction both are called vector quantities. For example, displacement, velocity, etc.

A physical quantity which can be completely explained by the magnitude only is called scalar quantity. For example, mass, time, etc.

8 0
4 years ago
Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia
USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

8 0
3 years ago
A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.
SashulF [63]

Answer:

Part a)

F = 0

Part b)

F = 0.25 N

Part c)

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

Explanation:

As we know that force on a current carrying wire is given as

\vec F = i(\vec L \times \vec B)

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

F = i(\vec L \times \vec B)

here we know that L and B is parallel to each other so

F = 0

Part b)

For 68.1 cm length wire we have

F = iLB sin\theta

here we know that

cos\theta = \frac{68.1}{166}

\theta = 65.8

so we have

F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8

F = 0.25 N

Part c)

For 151 cm length wire we have

F = iLB sin\phi

here we know that

cos\phi = \frac{151}{166}

\theta = 24.5

so we have

F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

4 0
3 years ago
Pls help will give brainlest​
Karo-lina-s [1.5K]

Respon

lqiudos ciopatmibes

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