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2 years ago

5

A motorcycle stunt driver zooms off the end of a cliff at a speed of 41.9 meters per second. If he lands after 1.62 seconds, wha

t is the height of the cliff?

1 answer:

tiny-mole [99]2 years ago

8 0

Of the cliff?

Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,

y
=
v
o
y
t
+
1
2
g
t

where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8

m
/
s
2
and
t
is time after

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The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the

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Answer: 1

Explanation:

Given

Tension is the string $T=44\ N$

mass of object $m=4\ kg$

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2 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

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Learn more: brainly.com/question/12550364

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