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ololo11 [35]
3 years ago
8

What is the conductor's resistance if its length is 2m and has a cross-sectional area of 0.7m with a resistivity of4Ωm? a. 1.4 Ω

b. 0.08 2 11.42 Ω d. 0.35 c.

Physics
1 answer:
Firlakuza [10]3 years ago
3 0

Answer:

The answer is c. 11.42 Ohm

Explanation:

The conductor's resistance is calculated by the formula in the figure.

So, you have to replace the given values into the formula.

Resistance of a conductor is equal to the product of rho by the lengh of the conductor divided the cross-sectional area of the conductor.

R= 4 ohm.m . (2m/o.7m^{2} )\\R=11.42ohm

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Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

v=v_0+at\\\\v^2=v_0^2+2ad

Dividing the second equation by the first one, we obtain:

v=\frac{v_0^2+2ad}{v_0+at}

And, since v_0=0, then:

v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s

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We could use this data, plus any of the two initial equations, to determine the acceleration:

v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2

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3 years ago
Electromagnetic waves can transfer energy without a(n) _____________ (medium/electric field).
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Medium.
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If a wave is traveling at 60cm/second and has a wavelength of 15 cm what is the frequency?
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3 years ago
Read 2 more answers
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

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Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

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Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
2 years ago
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