When working with a razor blade or exacto knife --

The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.

melamori03 [73]

If you subscribe I’ll answer QF Aotrx

8 0

2 years ago

If the angle of elevation of the cannon is decreased from 35 to 30 degrees, the vertical component of the ball's initial velocit

stealth61 [152]

Answer:

Determine the initial horizontal and vertical velocity for a dart launched with a ... an angle of 30 degrees. Vi=4m/s Vix Vicoso Mas cos 30 = = 30°. Viy=visine ... What is the resultant direction (angle) of the cannon ball (relative to a fixed frame)? ... Which of the following launch angles would result in the greatest time of flight iſ the.

Explanation:

I hope it helped

4 0

1 year ago

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

1 year ago

A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.

sladkih [1.3K]

Answer:

Explanation:

Potential due to a charged metallic sphere having charge Q and radius r on its surface will be

v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is

v = k Q / R

a ) On the surface of the shell , potential due to positive charge is

V₁ = $\frac{9\times10^9\times5\times10^{-6}}{6\times10^{-2}}$

On the surface of the shell , potential due to negative charge is

V₁ = $\frac{- 9\times10^9\times5\times10^{-6}}{6\times10^{-2}}$

Total potential will be zero . they will cancel each other.

b ) On the surface of the sphere potential

= $\frac{9\times10^9\times5\times10^{-6}}{2\times10^{-2}}$

= 22.5 x 10⁵ V

On the surface of the sphere potential due to outer shell

= $\frac{9\times10^9\times5\times10^{-6}}{5\times10^{-2}}$

= -9 x 10⁵

Total potential

=( 22.5 - 9 ) x 10⁵

= 13.5 x 10⁵ V

c ) In the space between the two , potential will depend upon the distance of the point from the common centre .

d ) Inside the sphere , potential will be same as that on the surface that is

13.5 x 10⁵ V.

e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.

5 0

1 year ago

When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe

ch4aika [34]

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0

1 year ago