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defon
3 years ago
5

(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the per

iod of its rotation?
Physics
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

\huge\boxed{T = 0.025\ seconds}

Explanation:

<u>Given:</u>

Frequency = f = 40 Hz

<u>Required:</u>

Time period = T = ?

<u>Formula:</u>

\sf T = 1 / f

<u>Solution:</u>

T = 1 / 40

T = 0.025 seconds

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
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A 10.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is su
inna [77]

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

Substitute these values into equation 2

e = 10/395

e = 0.025 m

7 0
3 years ago
The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7
My name is Ann [436]

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

I=\dfrac{CB^2}{2\mu_0}

I=\dfrac{CE^2}{2\mu_0 C^2}

E=\sqrt{{2I\mu_0 C}}

E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}

E=930.84 N/C

Therefore answer is 930.84 N/C

4 0
3 years ago
A string wound around a cylinder of 10.0 cm
laila [671]
A string wound around a cylinder of 10 cm<span> radius has a 150 gram mass attached. When released, the mass accelerates at 50 </span>cm/s2<span>.</span>
7 0
4 years ago
Which of the following pieces of home-made equipment is appropriate for use?
denis-greek [22]

Answer:

your neighbors biycyle inner tubes

Explanation:

hope this helps :)

4 0
2 years ago
The net external force on the propeller of a 3.0 kg model airplane is 6.8 N forward.What is the acceleration of the airplane? An
solong [7]

The correct answer to the question is 2.27 m/s^2 i.e the acceleration of the body is 2.27 m/s^2 along the forward direction.

CALCULATION:

As per the question, the net external force on the propeller of model airplane F =  6.8 N.

The mass of the model air plane m = 3.0 kg

We are asked to calculate the acceleration of the air plane.

From Newton's second law of motion, we know that the net external force acting on a body is equal to the product of mass with acceleration of that body.

Mathematically force F = m × a

                                  ⇒  a=\ \frac{F}{m}

                                          =\ \frac{6.8\ N}{3.0\ kg}

                                          =\ 2.27\ m/s^2                  [ans]

The direction of acceleration is along the direction of force. Hence, the acceleration of the propeller is 2.27 m/s^2 along forward direction.

8 0
4 years ago
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