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2 years ago

The diagram shows the molecular structure of propane. What is the chemical

Physics

2 answers:

ad-work [718]2 years ago

6 0

Answer:

Explanation:

Took the test

erica [24]2 years ago

4 0

Answer: B. C3H8

Explanation: Just took the test on A.PEX

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Pls how to solve this problem. help would be appreciated

Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

2 years ago

What is the rate of acceleration due to gravity on Earth?

Katena32 [7]

Answer:

Negative 9.8 meters per second squared

Explanation:

The negative is for the direction (down, towards the center of the earth). Often this can be estimated as -10 m/s^2 to make calculations easier.

3 0

2 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$