Is molten lava heterogeneous

Which unit of measurement is most appropriate for measuring the length of your computer screen?

Alika [10]

Answer:

Inches or Centimeters

Explanation:

Inches or centimeters are the most appropriate units of measurement for the computer screen because they are in the form of whole figures that can be easily comprehended. Usually the width of computer screens span from 10 to 20 inches.

Conversion of a 15inches computer to meters;

1inch=0.0254meter

15inches= ?

15inches*0.0254/1

=0.381meters

The above is a very small figure which might be hard for most people to fathom or visualize.

5 0

3 years ago

The number three ball moves off with a velocity of 5 m/s. What is the final velocity and direction of the eight ball?

Keith_Richards [23]

7 and a half millimeters long

6 0

3 years ago

on a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.8 cm and diameter 2.05 cm fr

emmainna [20.7K]

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.8 cm and diameter 2.05 cm from a storage room to a machinist, the weight of rod w is 23.41 N .

<h3>How is the weight calculated?</h3>

Given ,

density (d) = 7800 kg/m³

D = 2.05 cm = 0.0205 m

r = D/2 = 0.0205 / 2 =0.01025 m

h = 92.8 cm = 0.928 m

The weight is calculated by ,

W = m × g

But, to find mass (m)

m = d × V

The volume of the rod, because is cylindrical, is,

V = π × r² × h

∴ V = 3.14 × (0.01025)² × 0.928

∴ V = 3.14 × 1.051 × 10⁻⁴ × 0.928

∴ V = 3.0625 × 10⁻⁴

Now to calculate mass,

m = d × V

∴ m = 7800 × 3.0625 × 10⁻⁴

∴ m = 2.3887 Kg

Now, for calculating weight

W = m × g

∴ W = 2.3887 × 9.8

∴ W = 23.41 N

∴ The weight of the rod W assuming the free-fall acceleration is 23.41 N .

<h3>What is weight?</h3>

The weight of an object is the force exerted on it by gravity.
It is define weight as a vector quantity, the force of gravity acting on an object.
Some define weight as a scalar quantity that is the magnitude of gravity.

Can learn more about weight calculation from brainly.com/question/15685221

#SPJ4

6 0

1 year ago

In building a particle accelerator, you manage to produce a uniform electric field of magnitude 6.03 × 10 5 N/C in one 35.5 cm s

mamaluj [8]

Answer:

V = 2.14×10⁵ V.

W = 3.424×10⁻¹⁴ J.

Explanation:

Electric Potential: This can be defined as the work done in bringing a unit positive charge from infinity to that point, against the action of a field.

The S.I unit is V.

The expression containing electric potential, distance and electric field is given as,

V = E×r .............. Equation 1

Where V = Electric potential difference across the length of the accelerator's section, E = Electric Field, r = Length of the section.

Given: E = 6.03×10⁵ N/C, r = 35.5 cm = 0.355 m.

Substitute into equation 1

V = 6.03×10⁵×0.355

V = 2.14065×10⁵ V.

V ≈ 2.14×10⁵ V.

amount of Work required to move a proton through the section is given as,

W = qV ............... Equation 2

Where W = work required to move a proton through the section, q = charge on a proton V = Electric potential.

Given: V = 2.14×10⁵ V, q = 1.60 x 10⁻¹⁹ C.

Substitute into equation 2

W = (2.14×10⁵)(1.60 x 10⁻¹⁹)

W = 3.424×10⁻¹⁴ J.

6 0

3 years ago

A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant

OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

$a_t = ar$

where;

a is the angular acceleration and

r is the radius of the circular path

$a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2$

Determine time of the rotation;

$\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\$

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

$a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2$

The magnitude of total linear acceleration is given by;

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2$

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0

3 years ago