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2 years ago

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A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

e string is held so that it unravels without sliding as the disk is falling, what will be the wooden disk's acceleration? What would be the tension on string in the process?

1 answer:

Tju [1.3M]2 years ago

7 0

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

$a = \frac{2}{3}g$

Also from above equation the tension force in the string is

$T = \frac{1}{2}ma$

$T = \frac{mg}{3}$

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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2 years ago

You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel

Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

$T = 2 \pi \sqrt{\frac{m}{k}}$

squaring on both sides,

$T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}$

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

$y=T^2=0.0569x+0.0010$

the hanging mass M is x here, so comparing the equation we know that

$\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m$

b) In order to find the mass of the spring we make similar process, so comparing,

$\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g$

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2 years ago

An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the

laila [671]

Answer:

a) V = 1.866 10² V , b) V = 3.424 10⁵ V , c) v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

v = 0.027 c

v = 0.027 3 10⁸

v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

Em₀ = U = q V

final point. Electrons with maximum speed

Em_f = K = ½ m v2

Em₀ = Em_{f}

e V = ½ m v²

V = ½ m v² / e

let's calculate

V = ½ 9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 1.866 10² V

V = 1866 V

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 3.424 10⁵ V

V = 342402 V

c) this potential difference should give the protons the same speed as the electrons

v = 8.1 10⁶ m / s

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2 years ago

You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.

ankoles [38]

Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

Mass = 60 Kg

Energy = 5000 J

We need to calculate the specific heat

Using formula of energy

$Q= mc\Delta T$

$c =\dfrac{Q}{m\Delta T}$

Put the value into the formula

$c=\dfrac{5000}{60\times(37-13)}$

$c=3.47222\ J/kg^{\circ}C$

Hence, The specific heat is 3.47222 J/kg°C.

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2 years ago

a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what

mamaluj [8]

0.04m²

Explanation:

Given parameters:

Pressure = 250000Pa

Weight = 40000N

Unknown:

Area of each foot = ?

Solution:

Pressure is the force exerted per unit area of a body

Pressure = $\frac{force}{area}$

To find the area;

Area = $\frac{force }{pressure}$

Area = $\frac{40000}{250000}$ = 0.16m²

The force exerted by all the four feet is 0.16m²

the area of each feet = $\frac{0.16}{4}$ = 0.04m²

Learn more:

Pressure brainly.com/question/7139767

#learnwithBrainly

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2 years ago