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posledela
3 years ago
10

3 examples when friction is helpful?

Physics
1 answer:
Svetllana [295]3 years ago
4 0
Some examples of when friction is helpful are: to help the movement of tires. When you walk, and also, when you erase. :)
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What will the resistance be for a lamp that draws 4.6 amps of current from a 120-volt outlet? A. 550 ohms B. 115 ohms C. 26 ohms
Paraphin [41]
R = V/I = 120 / 4.6 ≈ 26 ohms.

Option C.
3 0
2 years ago
Read 2 more answers
Alicia está a punto de perder su bus. En un desesperado intento, corre a una velocidad constante de 5 m/s. Cuando está a 15 m de
pochemuha

Answer:

Si logra alcanzar el bus.

Explanation:

Para poder solucionar este problema debemos de tener en cuenta que Alicia corre a velocidad constante para poder alcanzar el bus. La formula de la cinematica que tiene en cuenta la velocidad constante es la siguiente:

x_{f} = x_{o}+(v*t)

donde:

Xf = Ubicacion del punto donde se encuentra el bus [m]

Xo = Ubicacion desde donde esta Alicia [m]

v = velocidad constante = 5 [m/s]

t = tiempo [s]

Xf - Xo = 15 [m]

15 = 5*t

t = 3 [s]

Ahora con el tiempo podemos encontrar la velocidad del bus por medio de la siguiente ecuacion de cinematica para la aceleracion constante:

v_{f} = v_{i}+(a*t)

donde:

Vf = velocidad del bus despues de los 3 [s]

Vi = velocidad inicial = 0

a = aceleracion = 0.5 [m/s^2]

Vf = 0 + (0.5*3)

Vf = 1.5 [m/s]

La velocidad del bus es menor que la velocidad de Alicia, por ende Alicia alcanzara el bus.

7 0
3 years ago
In the case of the vacuum tube and movable stopper, what force moves the stopper?
sergejj [24]
I'm pretty sure it's the<span> impact of the air molecules on the outside of the stopper. They exert a net inward force, which is not resisted by anything on the other side.</span>
6 0
2 years ago
Read 2 more answers
A spacecraft travels at 1.5 X 108 m/s relative to Earth. A process onboard the
kvasek [131]

Answer:

73.6 minutes

Explanation:

relative time = time interval / √(1 - observer velocity² / speed of light²)

we have relative time. we want time interval.

rearrange

time interval = relative time x √(1 - observer velocity² / speed of light²)

convert 85 mins into seconds

85 x 60 = 5100

1.5 x 10⁸ as a number is 150000000

for c = 299 792 458

time interval = 5100 x √(1 - 150 000 000² / 299 792 458²)

for c = 3 x 10⁸

time interval = 5100 x √(1 - 150 000 000² / 300 000 000²)

time interval = 5100 x 0.866

time interval = 4415.71

divide by 60 for back into minutes

time = 73.6 minutes

4 0
2 years ago
2. An object goes from a speed of 9 m/s to a total stop (Om/s) in 3 s. What
xxMikexx [17]
Acceleration = (0-9) / 3 = -3m/s^2
5 0
3 years ago
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