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3 years ago

Could you help me with this question?

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

Magnitude of static friction force is 70 sin40° = 44.99 N.

No, it is not necessary that it is maximum static friction.

Normal force is equal to 70 cos40° = 53.62 N.

Explanation:

We apply newton law of moton equation along the plane and perpendicular to plane;

Along the plane,

70 sin 40° = $f_{r}$ ---------------(1)

70 cos 40° = N --------------(2)

$f_{r}_{max}$ = μN -----------------(3)

So, it depends on the value of μ that the friction is maximum or not .

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A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

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A ball is thrown forward at 5 m/s. How would the path of the ball differ on Earth than on the moon? The ball would follow a curv

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Answer:

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