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katovenus [111]
3 years ago
15

Hello !! I am new here. Are you all from USA ?​

Physics
2 answers:
Alex787 [66]3 years ago
7 0

Answer:

No

Explanation:

Anni [7]3 years ago
5 0

Answer:

nope

Explanation:

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An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds
Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s

So, the average speed of the object is 6 m/s.

3 0
3 years ago
A rock is placed into a graduated cylinder containing 80 mL of water. What is the volume of the rock if the water level rises to
erastovalidia [21]
Ok so the expression that you will be doing is water-water+object. The actual expression is 120-80. The answer would be 40mL. Remember, don't forget your units! :)
6 0
3 years ago
Historia de baloncesto internacional resumen
xz_007 [3.2K]

Answer:

El baloncesto tuvo sus inicios en el año de 1891, en los Estados Unidos de Norte América justamente en la ciudad de Springfield, el mismo surgió como alternativa al Hockey sobre hielo, el cual solo se realizaba durante la época del invierno. La YMCA fue la primera asociación en practicar baloncesto cuando su pionero y fundador era entrenador en dicha asociación.

Además de los jugadores, el equipo esta conformado por masajistas, medicos, Entrenador, director y asistente técnico, los cuales tienen funciones específicas para prestar asistencia y dirección a los jugadores.

Explanation:

5 0
3 years ago
A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
3 years ago
two bumper cars at an amusement park collide together abd get stuck together. assuming that the system of the two bumper cars is
Fiesta28 [93]

The answer to this problem is M1v1-m2v2=(m1+m2)v

7 0
3 years ago
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