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2 years ago

A car traveling at 12 m/s slows down with an acceleration of -1.5 m/s2. What is the

Physics

1 answer:

Alex777 [14]2 years ago

7 0

Answer:

using s=ut +0.5at^2

u=12m/s

a=-1.5m/s2

t=6 s

s=54m

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iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms

saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

$\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }$ where $V_{molar} = \frac{A}{\rho }$ an and A is the atomic mass of iron.

Therefore:

$\frac{2}{a^{3} } = \frac{N_{A} * p }{A}$

This implies that:

$A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }$

= $\frac{4}{\sqrt{3} }r$

Assuming that there is no phase change gives:

$\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }$

= 8.60 g/m³

3 0

3 years ago

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

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2 years ago

I am pretty sure its the last one.

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