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3 years ago

12

A truck driver is

2 answers:

Gekata [30.6K]3 years ago

7 0

Answer:

Because he was not driving his truck! He’s walking on the sidewalk.

Explanation:

Mazyrski [523]3 years ago

6 0

Because the truck driver is walking

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Solve for (b) how many revolutions it takes for the cd to reach its maximum angular velocity in 1.36s?

Digiron [165]

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

$\omega = 30 rad/s$

The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

$\theta = (\frac{\omega_0 + \omega}{2})t$

where:

$\theta$ is the angular displacement of the cd during the time interval t

$\omega_0$ is the initial angular velocity of the cd

$\omega$ is the final angular velocity

Here we have:

t = 1.36 s

$\omega_0 = 0$ (assuming the cd starts from rest)

Therefore, the angular displacement of the cd during this time is:

$\theta=(\frac{0+30}{2})(1.36)=20.4 rad$

And since $1 rev = 2 \pi rad$ , we can convert into number of revolutions completed:

$\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev$

Learn more about angular motion:

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3 years ago

What is the frequency of a photon with an energy of 4.56 10-19 j?

Brrunno [24]

Give me some answer choices and i will be happy to help

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3 years ago

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a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

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In the Atlantic Ocean

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D I believe it is hope it woke

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