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3 years ago

11

What is 7.4×10 to the second power

1 answer:

sleet_krkn [62]3 years ago

8 0

The correct answer is 740

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Consider the following statement:

vovangra [49]

Answer:

The statement "The magnetic field of a magnet comes out of the north pole and goes into the south pole" is imprecise

Explanation:

This is because the zero divergence equation (∇ · B = 0 ) is valid for any magnetic field, even if it is time dependent rather than static. Physically, it means that there are no magnetic charges otherwise we would have ∇ · B ∝ ρmag instead of ∇ · B = 0. Consequently, the magnetic field lines never begin or end anywhere in space; instead they form closed loops or run from infinity to infinity.

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3 years ago

Why do different stars have different life cycles

lana66690 [7]

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3 years ago

T or f :static electricity constantly flows in the same direction

Kazeer [188]

Answer:

False

because I got that question and I gotted right

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3 years ago

What is the speed of the sound wave if the frequency is 70 hertz and the wavelength is 0.4 meters

sesenic [268]

28 mps miles per second

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4 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

6 0

3 years ago