have a large coefficient of kinetic friction between cloth and the slide material
40(0.3+0.5)=40(0.8)=32N
F-applied=Wbook/μk
When the tire is locked, the frictional force is greater but the car cannot be steered

4 0

2 years ago

natka813 [3]

Responder:

13,01 m / s

Explicación:

Paso uno:

datos dados

masa de la persona 1 m = 80 kg

velocidad de la persona 1 v = 9 m / s

masa de la persona 2 M = 55kg

velocidad de la persona 2 v =?

Segundo paso:

la expresión del impulso se da como

P = mv

para la primera persona, el impulso es

P = 80 * 9

P = 720N

Paso tres:

queremos que la segunda persona tenga el mismo impulso que la primera, por lo que la velocidad debe ser

720 = 55v

v = 720/55

v = 13,09

v = 13,01 m / s

Por lo tanto, la magnitud de la velocidad debe ser 13.01 m / s.

4 0

2 years ago

Mamont248 [21]

Let $y_0$ be the height of the building and thus the initial height of the ball. The ball's altitude at time $t$ is given by

$y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the ground when $y=0$ after $t=2.9\,\mathrm s$ . Solve for $y_0$ :

$0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2$

$\implies y_0\approx16.258\,\mathrm m$

so the building is about 16 m tall (keeping track of significant digits).

3 0

2 years ago

mrs_skeptik [129]

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5 0

2 years ago