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3 years ago

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A phone cord is 2.89 m long. The cord has a mass of 0.258 kg. A transverse wave pulse is produced by plucking one end of the tau

tcord. The pulse makes four trips down and back along the cord in 0.737 s. What is the tension in the cord

1 answer:

Aleksandr [31]3 years ago

8 0

Answer:

$T =87.87\ N$

Explanation:

Given,

Length , l=2.89 m

Mass , M=0.258 Kg

Now, displacement $d=4(2 \times 2.89) m$

$=23.12 m$

Velocity of waves

$v=\sqrt{T / \mu}$

$\mu=\frac{0.258}{2.89 }$

$=0.0893$

Now, velocity, $v=\frac{d}{t}=\frac{23.12}{0.737 s}$

$=31.37 m/s$

The tension

$T=v^{2} \mu=31.37 \times 31.37 \times 0.0893$

$T =87.87\ N$

Tension in the chord is equal to $T =87.87\ N$

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A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an

AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

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Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie

Katarina [22]

Answer:

18.63 N

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Assuming that the sum of torques are equal

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α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

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Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

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Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>please

REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

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t = 2 [s]

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<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

$25=5+2*t\\t = 10 [s]$

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

$v_{f}=4+10*2\\v_{f}=24 [m/s]$

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$v_{f}=5+8*10\\v_{f}=85 [m/s]$

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

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3 years ago

A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the

topjm [15]

<h2>5.3 km</h2>

Explanation:

This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

Let us denote each of the individual displacements by a vector. Consider the unit vectors $\vec{i}\textrm{ and }\vec{j}$ as the unit vectors in the direction of East and North respectively.

By simple calculations, we can derive the unit vectors $\vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2}$ in the directions North, $45^{o}$ South of West and $60^{o}$ North of West respectively.

So Total displacement vector = Sum of individual displacement vectors.

Displacement vector = $4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}$

Magnitude of Displacement = $|-4.25\vec{i}+3.165\vec{j}|=5.3km$

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scZoUnD [109]

Answer:

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