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3 years ago

12

As temperature increases, ________. Group of answer choices the resistance of a conductor remains the same the resistance of a c

onductor increases the resistance of a conductor decreases the voltage of the conductor decreases

1 answer:

amm18123 years ago

6 0

Answer:

resistance of a conductor increases

Explanation:

The resistance of conductors is directly proportional to the temperature of the conductor. This implies that when the temperature of the conductor is increased, the resistance of the conductor increases likewise.

This is applied in the resistance thermometer. Resistance thermometers are useful for accurate temperature measurements at very high or very low temperatures.

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If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t

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Answer:

The velocity is $4.6 m/s^2$

Explanation:

Given:

Force = 500N

Distance s= 0

To find :

Its velocity at s = 0.5 m

Solution:

$\sum F_{x}=m a$

$F\left(\frac{4}{5}\right)-F_{S}=13 a$

$500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a$

$400-(500 s)=13 a$

$a = \frac{400 -(500s)}{13}$

$a = (30.77 -38.46s) m/s^2$

Using the relation,

$a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}$

$a=v \frac{d v}{d s}$

$v d v=a d s$

Now integrating on both sides

$\int_{0}^{v} v d v=\int_{0}^{0.5} a d s$

$\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s$

$\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}$

$\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]$

$\left[\frac{v^{2}}{2}\right]=[15.385-4.807]$

$\left[\frac{v^{2}}{2}\right]=10.578$

$v^{2}=10.578 \times 2$

$v^{2}=21.15$

$v = \sqrt{21.15}$

$v = 4.6 m/s^2$

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4 years ago