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Elina [12.6K]
3 years ago
6

An extension cord is used with an electric weed trimmer that has a resistance of 17.9 Ω. The extension cord is made of copper (r

esistivity 1.72 x 10-8 ohm m) wire that has a cross-sectional area of 1.71 x 10-6 m2. The combined length of the two wires in the extension cord is 86.3 m. (a) Determine the resistance of the extension cord. (b) The extension cord is plugged into a 120-V socket. What voltage is applied to the trimmer itself?
Physics
1 answer:
Naddika [18.5K]3 years ago
8 0

Answer:

(a) R_{c}=0.87ohms

(b) V_{T}=114.44V

Explanation:

Part (a)

The total length of copper cord L=86.3 m

The cross sectional area A=1.71×10⁻⁶m²

The resistivity of copper p=1.72×10⁻⁸Ω

Thus the resistance of extension cord is

R_{c}=p\frac{L}{A}\\R_{c}=(1.72*10^{-8} )\frac{86.3}{1.71*10^{-6}}\\R_{c}=0.87ohms

Part (b)

The resistance of trimmer Rt=17.9 ohms

When voltage of 120V is applied then the current I is passing through series circuit is

I=\frac{120V}{R_{c} +R_{T} }\\I=\frac{120V}{0.87 +17.9 } \\I=6.4A

Thus the voltage across the trimmer is:

V_{T}=IR_{T}\\V_{T}=(6.4)*(17.9)\\V_{T}=114.44V

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Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

                v = 0,     s = 84 m,     \mu = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

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Also,    

               v^{2} = u^{2} + 2as

              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0
3 years ago
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained
marta [7]

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

  • <em>mass of the metal hammer, m = 1.0 kg</em>
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  • <em>specific heat capacity of iron, 450 J/kg⁰C</em>

The increase in temperature of the metal hammer is calculated as follows;

Q = K.E\\\\mc \Delta T = \frac{1}{2}  mv^2\\\\\Delta T = \frac{v^2}{2 c}

where;

<em>c is the </em><em>specific heat capacity</em><em> of the metal hammer</em>

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Assuming the metal hammer is iron, c = 450 J/kg⁰C

\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

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Please help with the working of this question
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Answer:

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Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in
maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

v^{2} = \frac{dr}{dt}^{2}  +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2} - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

\frac{r}{z}=tan(45)

or:

z = r cot(45) - (2)

and:

\frac{dz}{dt} = \frac{dr}{dt} cot(45)

replacing (2) in (1) we obtain:

v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}  - (3)

Now the kinetic energy is given as:

T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}) - (4)

And the potential energy is given by:

V = -mgz = -mgr cot(45)

So the Langrangian is given by:

L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)

And the equations of motion are:

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\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c

For r

\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

mr\frac{d\theta}{dt}=c

Which is the magnitude of the angular momentum

7 0
3 years ago
The Hubble Space Telescope (HST) orbits 569,000m above Earth’s surface. Given that Earth’s mass is 5.97 × 10^24 kg and its radiu
Soloha48 [4]
Refer to the diagram shown below.

M = 5.97 x 10²⁴ kg, mass of the earth
h = 5.69 x 10⁵ m, height of HST above the earth's surface
R = 6.38 x 10⁶ m, radius of the earth

Note that
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R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m

The force between the earth and HST is
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Let v = tangential velocity of the HST.

The centripetal force acting on HST is equal to F.
Therefore
m*[v²/(R+h)] = (GMm)(R+h)²
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v = 7.5699 x 103 m/s

Answer
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3 years ago
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