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4 years ago

6

A 6.0-m board is leaning against a wall. A 28° angle is formed with the WALL What is the height above the ground where the board

makes contact with the wall? (In Meters)

1 answer:

FinnZ [79.3K]4 years ago

4 0

Answer:

the height above the ground where the board makes contact with the wall equals 5.297 meters above ground.

Explanation:

The wall and ladder are shown in the attached figure

In the figure we can see that

$cos(28^{o})=\frac{H}{L}\\\\cos(28^{o})=\frac{H}{6}\\\\\therefore H=6.0\times cos(28^{o})==5.297meters$

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During photosynthesis, what is the source of the carbon in<br> the sugar (CH20) that is created?

irakobra [83]

Answer:

Carbon dioxide

Explanation:

During photosynthesis cells use carbon dioxide and energy of sun to make sugar molecules and oxygen.

Hope i helped u...

8 0

3 years ago

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.

lorasvet [3.4K]

$\large{\underline{\underline{\pmb{\frak {\color {red}{Question:}}}}}}$

$\sf \red{When \: baking \: soda \: is \: mixed \: with \: lemon \: juice, \: bubbles \: are \: formed \: with \: the \: evolution \: of \: a \: gas.}$

$\large{\underline{\underline{\pmb{\frak {\color {blue}{Answer:}}}}}}$

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. The gas is formed in the reaction is Carbon dioxide. $CO_{2}$ is formed.

The change which happened in this reaction is a chemical change.

$\boxed{ \frak \green{Explanation}}$

Since, in chemical change we can't bring a substance to it's actual form how it was in earlier.

Examples: burning of paper is chemical, since we can't get the fine paper again after it is burnt.

Thus, the above reaction is also a chemical change, since we can't get back the lemon juice how it was earlier.

$\boxed{ \frak \red{Brainlysamurai}}$

5 0

2 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

4 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

Which sentence best describes a role of gravity in the formation of the

enyata [817]

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

8 0

3 years ago