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IrinaVladis [17]

2 years ago

5

A passenger on a jet airplane claims to be able to walk at a speed in excess of 500 mph. Can this be true?

1 answer:

alexdok [17]2 years ago

3 0

Answer:noo

Explanation:impossible

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If it takes 600 N to move a box 5 meters, how much work is done on the box?

Anika [276]

Answer:

<h2>3000 J</h2>

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force $\times$ distance

$= 600 \times 5$

Calculate the product

$= 3000 \:$ Joule

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

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A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 cm and a cross-sectional area of 15 cm2. T

stealth61 [152]

<span>A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?</span>

3 0

3 years ago

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Please select the word from the list that best fits the definition Which type of waves from the electromagnetic spectrum are use

dalvyx [7]

Visible light ultraviolet rays radio waves infrared waves

4 0

2 years ago

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What's the main function for the human digested system

tia_tia [17]

Answer:

The function of the digestive system is digestion and absorption. Digestion is the breakdown of food into small molecules, which are then absorbed into the body.

Explanation:I hope it helped

6 0

2 years ago

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A 28 kg mass suspends from a light rope 18 m long & is held to one side by the horizontal force, F, as shown below.

frutty [35]

Answer: 215.15 N

Explanation:

If we draw a free body diagram of the mass we will have the following:

$\sum{F_{x}}=-Tcos\theta + F=0$ (1)

$\sum{F_{y}}=Tsin\theta - mg=0$ (2)

Where $T$ is the tension force of the rope, $m=28 kg$ the mass, $g=9.8 m/s^{2}$ the acceleration due gravity and $mg$ is the weight.

On the other hand, we can calculate $\theta$ as follows:

$cos\theta=\frac{s}{l}$

$\theta=cos^{-1}(\frac{s}{l})$

Where $s=11.1 m$ and $l=18 m$

$\theta=cos^{-1}(\frac{11.1 m}{18 m})$

$\theta=51.9\°$ (3)

Now, we firstly need to find $T$ from (2):

$T=\frac{mg}{sin\theta}$ (4)

$T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}$

$T=348.69 N$ (5)

Substituting (5) in (1):

$F=Tcos\theta$ (6)

$F=348.69 N cos(51.9\°)$

Finally:

$F=215.15 N$

8 0

3 years ago