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Helen [10]
3 years ago
8

A skateboard is moving to the right with a velocity of 8 m/s after a steady Gust of wind that last 5 seconds the skateboarder is

moving to the right with a velocity of 5 m/s assuming the acceleration is constant what is the acceleration of the skateboard during the five second time.
Physics
1 answer:
Rzqust [24]3 years ago
6 0

The acceleration of the skateboard is -0.6 m/s^2 (to the left)

Explanation:

The motion of the skateboard is a uniformly accelerated motion, so we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the skateboard in this problem, choosing right as positive direction, we have:

u = +8 m/s is the initial velocity

v = +5 m/s is the final velocity

t = 5 s is the time

Solving for a, we find the acceleration:

a=\frac{v-u}{t}=\frac{5-8}{5}=-0.6 m/s^2

And since the sign is negative, the direction is to the left.

Learn more about acceleration here:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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Answer:

The first harmonic is: 250Hz, second harmonic 500Hz, third harmonic 750Hz.

Explanation:

Use the frequency f, speed v, and wavelentgh L relationship:

v = f\cdot L\implies f = \frac{v}{L}

We are given the speed v=400 m/s. The base wavelength on a string of length 80cm is twice the length of the string (a "half wave" along the full length of the string), so:

f = \frac{400\frac{m}{s}}{2\cdot0.8 m}= 250\frac{1}{s}=250 Hz

The fundamental frequency (first harmonic) is 250 Hz

The second harmonic is produced by one full wave across the string (adding one node in the middle), so L=80cm in this case, therefore the second harmonic frequency is: f2 = 2*250=500Hz

the third harmonic add another node (and a half wave) to the pattern and the wavelength will be 2/3 of 80cm, so f3=3*250Hz = 750Hz


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A tension force of 175 N inclined at 20.0° above the horizontal is used to pull a 40.0 - kg packing crate a distance of 6.00 m o
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Answer:

(a) Work done by the tension force is 987 J

(b) Coefficient of kinetic friction between the crate and surface is 0.495

Explanation:

(a)

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W=Fd\cos (\Theta )

W=175 \times 6 \times  \cos (20 )J=987J

Thus work done by the tension force is 987 J

(b)

Normal force on the crate is given by

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Since crate is moving with constant speed . Therefore using Newtons second law .

Fcos(\Theta ) -\mu_k N=0

Where \mu_k=coefficient of kinetic friction

\therefore \mu_k=\frac{F\cos (\Theta )}{N}

=>\mu_k=\frac{175\times \cos 20}{332}

=>\mu _k= 0.495

Thus coefficient of kinetic friction between the crate and surface is 0.495

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