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4 years ago

A material has a low resistivity. Which statement is the best prediction about

Physics

1 answer:

Ksivusya [100]4 years ago

8 0

Answer:A

Explanation:I’m taking the test rn and I got it correct

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The image shows a pendulum in simple harmonic motion. The pendulum starts at A and swings to E.

Genrish500 [490]

D: Velocity is decreasing. Acceleration is increasing.

A: Velocity is zero.

7 0

3 years ago

Read 2 more answers

At a pressure of one atmosphere methane boils at −161°C and freezes at −182.5°C. Consider a temperature scale where the boiling

lawyer [7]

Answer:

$-421.627906977\^{\circ}M$

Explanation:

Absolute zero in the Kelvin scale = -273.15 °C

Temperature difference in the boiling and freezing point

$-161-(-182.5)=21.5\ ^{\circ}C$

Change in 21.5°C = Change of 100°C in M

$1^{\circ}C=\dfrac{100}{21.5}=4.6511627907\ ^{\circ}M$

Difference in the Celsius scale

$-182.5-(-273.15)=90.65\ ^{\circ}C$

Change in $90.65\ ^{\circ}C$ corresponds to

$4.6511627907\times 90.65=421.627906977\ ^{\circ}M$

Therefore the absolute zero in the Methane scale is

$-421.627906977\ ^{\circ}M$

3 0

3 years ago

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.