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3 years ago

What do all elements have in common?

1 answer:

7 0

The common<span> feature is that the atoms of </span>all elements<span> consist of electrons, protons, and neutrons. Hope this helps!</span>

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Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If

frez [133]

Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2

v2 = v1×A1/A2

and

r2 = 2×r1

then:

A2 = 4×A1

so,

v2 = (v1)/4

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)

= 9.75×10^4 Pa

= 9.8×10^4 Pa

Therefore, the pressure in the wider section is 9.8×10^4 Pa

5 0

3 years ago

Two iron weights, one twice the mass of the other, are dropped from the top of a building. Compared with the lighter weight, the

Whitepunk [10]

The twice as heavy weight will hit the ground with more force, or impact.

4 0

3 years ago

Read 2 more answers

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

5 0

3 years ago

To get an idea of the size of magnetic fields at the atomic level, consider the magnitude of the magnetic field due to the elect

artcher [175]

Answer:

The magnetic field is 14.08 T.

Explanation:

Given that,

Speed of electron $v=2.2\times10^{6}\ m/s$

Distance $d =0.5\times10^{-10}\ m$

Suppose we need to find the magnetic field at the location of the proton

We need to calculate the magnetic field

Using formula of magnetic field

$B = \dfrac{\mu_{0}I}{2r}$ ....(I)

Using formula of current

$I=\dfrac{q}{T}$

Using formula of time

$T= \dfrac{2\pi r}{v}$

Put the value of current and time in equation (I)

$B=\dfrac{\mu \times q\times v}{2\pi r\times2r}$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times2.2\times10^{6}}{4\pi\times(0.5\times10^{-10})^2}$

$B=14.08\ T$

Hence, The magnetic field is 14.08 T.

3 0

3 years ago

I really need the answer to this question please

Sliva [168]

I believe the answer is option A

5 0

3 years ago