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3 years ago

13

When powdered iron is left exposed to the air, it rusts. explain why the mass of the rust is greater than the mass of the powder

ed iron?

1 answer:

Stells [14]3 years ago

7 0

There is more surface area to rust even though microscopic. which is why there is more rust than iron

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Why do flames go upwards

BabaBlast [244]

Answer:

Easy search it on g o o g l e

4 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate

kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid $R=1.06 m$

Emf induced $E=8.50\times 10^{-6} V/m$

no of turns per meter n=450

we know Induced EMF is given by

$\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A$

Magnetic Field is given by

$B=\mu _0ni$

thus $\frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}$

Area of cross-section

$A=\pi R^2$ where

solving integration we get

$E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2$

where r=distance from axis

R=radius of Solenoid

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s$

4 0

3 years ago

Answer fast , is si unit newton/meter square or pascal or both

Readme [11.4K]

Answer:

Both

Explanation:

The S.I. unit of pressure is newton/meter square or pascal as both represent the same dimensional value.

8 0

3 years ago

What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

yuradex [85]

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s

8 0

3 years ago