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3 years ago

Farah claps his 600times in 60seconds. what is the friquency and periodb

Physics

1 answer:

rosijanka [135]3 years ago

5 0

Answer:

$\huge\boxed{\sf Frequency = 10\ Hz}$

$\huge\boxed{\sf Time\ Period = 0.1\ secs}$

Explanation:

<u>Given:</u>

No. of times = 600 times

Time = 60 seconds

<u>Required:</u>

Frequency = f = ?

Time Period = T = ?

<u>Solution:</u>

<h2>Frequency = No. of Times / Time</h2>

Frequency = 600 / 60

<u>Frequency = 10 Hz</u>

<h2>Time Period = 1 / Frequency</h2>

Time Period = 1 / 10

<u>Time Period = 0.1 secs</u>

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F = m*a, mass times acceleration.

F = 15*10 = 150 N

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Answer:

3.14946 rad/s

Explanation:

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2 years ago

You are on a train traveling east at speed of 18 m/s with respect to the ground. 1) If you walk east toward the front of the tra

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Answer:

19.2 m/s

Explanation:

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18 + 1.2 = 19.2

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3 years ago

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

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2 years ago

If the car’s speed decreases at a constant rate from 60 mi/hmi/h to 50 mi/hmi/h in 3.0 ss, what is the magnitude of its accelera

Kitty [74]

Answer:

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

Explanation:

Hi there!

The acceleration is defined as the change in velocity in a time:

a = Δv / Δt

Where:

a = acceleration.

Δv = change in velocity = final velocity - initial velocity.

Δt = elapsed time.

In this case:

Initial velocity = 60 mi/h

final velocity = 50 mi/h

elapsed time = 3.0 s

Let´s convert the time unit into h:

3.0 s · 1 h /3600 s = 1/1200 h

Now, let´s calculate the acceleration:

a = Δv / Δt

a = (50 mi/h - 60 mi/h) / 1/1200 h

a = -1.2 × 10⁴ mi/h²

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

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3 years ago

Farah claps his 600times in 60seconds. what is the friquency and periodb​

Farah claps his 600times in 60seconds. what is the friquency and periodb