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WINSTONCH [101]
3 years ago
6

When Coach Kwan notices that a player is getting tired, she takes out the tired player and substitutes a fresh player.

Physics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

is replacement

Explanation:

Wittaler [7]3 years ago
5 0
The answer is a




If you can’t get the correct answer, this is on edg


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Metallic bonds are responsible for many properties of metals, such as conductivity. Why is this possible? (1 point)
qaws [65]

Answer:

The bonds can shift because valence electrons are held loosely and more freely

Explanation:

Please give brainliest if you can,have a good day<3 :)

4 0
3 years ago
I don’t know what to do help me
kirill115 [55]

Answer:

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Explanation:

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4 0
3 years ago
A power plant uses Uranium<br> to produce energy
antoniya [11.8K]

Answer:

how is that a question?

Explanation:

yeah i dunno the answer cause thats not a question

7 0
3 years ago
Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

6 0
3 years ago
A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i
Lubov Fominskaja [6]

Answer:

11.8 m/s

Explanation:

At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).

Sum of forces in the centripetal direction:

∑F = ma

mg − N = m v²/r

At the maximum speed, the normal force is 0.

mg = m v²/r

g = v²/r

v = √(gr)

v = √(9.8 m/s² × 14.2 m)

v = 11.8 m/s

3 0
3 years ago
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