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Yakvenalex [24]

2 years ago

8

A 26 cm object is 18 cm in front of a plane mirror. A ray of light strikes the object and is reflected off the mirror at a 42-de

gree angle. What is the angle of reflection? degrees What type of image is formed? What is the size of the image? cm

1 answer:

matrenka [14]2 years ago

8 0

Answer:

42 degrees, virtual image, same size as the object (26 cm)

Explanation:

The law of reflection states that:

- When a ray of light is incident on a flat surface (such as the plane mirror), the angle of reflection is equal to the angle of incidence

So, since in this case the angle of incidence is 42 degrees, the angle of reflection is also 42 degrees.

Moreover, the image formed by a plane mirror is always:

- Virtual (on the same side as the object)

- The same size as the object

So in this case, since the object's size is 26 cm, the image's size is also 26 cm.

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An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h

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Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

$K=\frac{1}{2}mv^{2}$

For the object thrown in the air:

$K=\frac{1}{2}.2.[v(t)]^{2}$

$K=(-9.8t+24)^{2}$

$K=96.04t^{2}-470.4t+576$

Kinetic energy of the object as a function of time: $K=96.04t^{2}-470.4t+576$

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

$U=mgh$

For the object thrown in the air:

$U=9.8.2.h(t)$

$U=9.8.2.(-4.9t^{2}+24t+60)$

$U=-96.04t^{2}+470.4t+1176$

Potential energy as function of time: $U=-96.04t^{2}+470.4t+1176$

Total kinetic and potential energy, also known as mechanical energy is

TME = $96.04t^{2}-470.4t+576$ + ( $-96.04t^{2}+470.4t+1176$ )

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

6 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

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Answer:

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

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

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

Explanation:

Part a

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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

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

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

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

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

<em>The void ratio is 0.77</em>

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

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

<em>Degree of Saturation is 0.43</em>

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.

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2 years ago

A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki

Anna71 [15]

Based on the direction of propagation compared to direction of vibration, waves are classified into:
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For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave

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2 years ago

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sladkih [1.3K]

Answer:

Explanation:

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v = k Q / R

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= 22.5 x 10⁵ V

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

= -9 x 10⁵

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= 13.5 x 10⁵ V

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

13.5 x 10⁵ V.

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Explanation:

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