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Keith_Richards [23]
2 years ago
11

you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?

Physics
1 answer:
laiz [17]2 years ago
8 0
At the same speed because it will slow down as it approaches the peak then speed up as it goes down again


it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something
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When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the fi
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Answer:

a)

125.6 rad/s

b)

25.12 rad/s²

Explanation:

a)

t = time required by the fan to get up to final operating speed = 5 sec

w = final operating rotational speed = 1200 rpm

we know that :

1 revolution = 2π rad

1 min = 60 sec

w = 1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}

w = \frac{1200\times 2\pi }{60}\frac{rad}{s}

w = 125.6 rad/s

b)

w₀ = initial angular speed = 0 rad/s

α = angular acceleration

using the equation

w = w₀ + α t

125.6 = 0 + α (5)

α = 25.12 rad/s²

5 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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