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2 years ago

Which element is the main component of steel and the most widely used of all metals

Physics

2 answers:

Blizzard [7]2 years ago

8 0

Answer:

iron

Explanation:

Dafna1 [17]2 years ago

6 0

Iron

IRON is the main ingredient in various forms of iron and steel, but the various types of metals contain other elements as well.

BRAINLIEST??? HOPE IT HELPS!!!

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A hydrogen atom has a diameter of about 12.7 nm, express this diameter in centimeters.

Andrej [43]

1 meter = 1e9 nm
To get meters, divide nanometers by 1e9: 9.95nm / 1x10^9 = 9.95x10^-9 meters
Answer: 9.95e-9 meters

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2 years ago

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2 ways weak nuclear forces and gravity are alike <br><br> PLEASE HELP ):

shtirl [24]

They are attractive
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3 years ago

Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the poten

alukav5142 [94]

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

$V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}$

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

$V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}$

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

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3 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

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3 years ago

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3 years ago