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Svetradugi [14.3K]

3 years ago

9

A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?

what is this in rev/min?

1 answer:

andrey2020 [161]3 years ago

4 0

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19× $\frac{1}{2\pi} *60$ revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

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Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N

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We have given mass of the crate m = 49.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of static friction $\mu _s=0.584$

Coefficient of kinetic friction $\mu _k=0.399$

(a) Static friction force is given by $F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N$

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by $F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N$

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3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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ANSWER

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Parameters given:

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To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

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Therefore, the gravitational force acting between the student and the textbook is:

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That is the answer.

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