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deff fn [24]
3 years ago
13

A mass m at the end of a spring vibrates with a frequency

Physics
1 answer:
Wittaler [7]3 years ago
6 0

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

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Vesnalui [34]

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A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball
lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.  

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

M_{1} u_{1} +M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

Since initial velocity for M1 ball is zero, then

M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

and

M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

So, on solving all the above equation, we get an equation for velocity and that is

\frac{2M_{2}u_{2} }{(M_{1}+M_{2}  }=final velocity of ball with mass 2.5 kg

v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is \frac{1}{2}*2.5*(4.67)^{2}  =27.09 J

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0
3 years ago
Pravat exerts a force of 30 N to lift a bag of groceries 0.5 m. How much work did Pravat do on the bag?
Dahasolnce [82]

Answer:

15 J

Explanation:

Work = Force x Distance

15= 30 x 0.5

Have a blessed day!

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frosja888 [35]

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It is sometimes described as the amount of work needed to move a test charge between two sites, expressed as a unit of charge. Volt is the potential difference's SI unit (V). We only take into account the charge between the locations P and Q when current moves between them in an electric circuit. Electric potential difference between two sites is referred to as voltage, also known as electric pressure, electric tension, or (electric) potential difference. an electric field that is static.

Vh = I*Rn

Vh = 39/5.476*5.40v

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brainly.com/question/13521443

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5 0
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Answer:

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Then the throw will slow down because of the gravitational pull.

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If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.

So to do so you need to throw the bananas with a speed of at least 9,81m.s

Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.

I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.

kind regards

4 0
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