All categories

3 years ago

How would changing the proportions in an alloy change its properties in a 5th grade response plz.

1 answer:

6 0

Changing the proportions of substances in an alloy will change the physical properties of that alloy. The differing substances will change the metallic structure of the crystals of the metal alloy, and this alters the ductility, durability, hardness, tensile strength, toughness and other characteristics we assess are regards alloys.

You might be interested in

Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r

Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

$F_{c} = \dfrac{mv^{2}}{r}$

where, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0

3 years ago

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh

Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = $\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}$

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0

3 years ago

How many like does it take to get to the center of a tosipop

dmitriy555 [2]

252-364 licks

I'm not sure if it's correct or not

4 0

3 years ago

Read 2 more answers

What is the longitude of the middle of romano swamp?

Valentin [98]

37 degree West 47 degree North

5 0

3 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

3 years ago