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ohaa [14]
3 years ago
10

How to convert 8 feet to metres​

Physics
2 answers:
adelina 88 [10]3 years ago
8 0

Hello there,

We know that 1 foot = 0.305 metre.

So, 8 feet = 8 x 0.305

→ <u>2.438</u><u> </u><u>m.</u>

✍️ <em>By </em><em>Benjemin</em> ☺️

ohaa [14]3 years ago
7 0

Answer:

2.4384m

Explanation:

1 ft in m is 0.3048

8ft in m is x

therefore 8ft in m is

8 * 0.3048 = 2.4384

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strojnjashka [21]

Answer:

The volume of the block is equal to the volume of water displaced by the block.

Explanation:

Volume refers to the amount of space occupied by a given object (in this case the block). When an object such as the block is immersed in water, it displaces its own volume of water. This volume of water displaced is equal to the volume of the block. Hence we can write;

Final Volume of water - Initial Volume of water= Water Displaced = Volume of the block

Recall that the density of a body is given by;

Density= mass/volume

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8 0
3 years ago
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

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5 0
4 years ago
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