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2 years ago

A 2.0-cm-thick bar of soap is floating on a water surface

Physics

1 answer:

maxonik [38]2 years ago

5 0

Answer:

Explanation:

Given

thickness of soap bar t=2 cm

thickness underwater t'=1.5 cm

oil is poured in the water with specific gravity=0.6

Soap top surface is just level with the upper surface of oil i.e.

weight of soap is balance by buoyancy effect of oil and water

initially soap is balanced by water

$\rho _s\times A\times 2\times g=\rho _w\times A\times 1.5\times g$

$2\times \rho _s=1.5\rho _w$

if $\rho _s=density\ of\ soap$

$\rho _w=density of water$

$A=area\of\ cross-section$

$d=depth\ of\ soap\ in\ oil$

$\rho _s\times A\times 2\times g=0.6\rho _w\times A\times d\times g+\rho _w\times A\times (2-d)\times g$

$1.5\rho _w=0.6d\rho _w+\rho _w(2-d)$

$1.5=0.6d+2-d$

$0.4d=0.5$

$d=\frac{5}{4}$

$d=1.25\ cm$

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2 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

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2 years ago

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bekas [8.4K]

Answer:

I disagree.

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I hope this was the brainliest answer! Thank you for letting me help you.

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3 years ago

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