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Dmitry [639]
3 years ago
5

How could rescue workers use squeezing or compressing to get energy to their flashlights during rescue missions?

Physics
1 answer:
Neko [114]3 years ago
3 0

Answer:

Explanation:

During rescue missions, different types of energy can be devices for flashlight, this could be human powered energy such as squeezing or compressing. In flashlight electrical energy is converted to light and thermal energy.

A squeezing or compressing to get energy for flashlight can be regarded as "DYNAMO PROCESS" it involves spinning of "fly wheels" into the flashlight through consistent squeezing ,which is connected to a dynamo(Dynamo supply electrical current). Hence the needed light is seen on the bulb of the flashlight.

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These are intense, rotating convectional systems that develop over warm ocean areas in the tropics and subtropics, primarily dur
asambeis [7]

Answer:

True.

Explanation:

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4 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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Answer:

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