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2 years ago

Identify the labeled parts in the figure.

Physics

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

I. a, c, f and h

II. e

III. b, d, g and i

IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

II. Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e

III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

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Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

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y = -20 cm

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$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

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The distance the beetle travels on the ground is 0.3677181864 m

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3 years ago

The springs of a 1500 kg car compress 5.00 mm when its 68 kg driver gets into the driver's seat. Part A If the car goes over a b

elena-s [515]

Answer:

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Explanation:

Given;

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extention of the spring, x = 5 mm = 5 x 10⁻³ m

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The weight of the driver is calculated as;

F = mg

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The spring constant, k, is calculated as;

k = F/m

k = (666.4 N) / (5 x 10⁻³ m)

k = 133,280 N/m

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ω = 2πf

f = ω / 2π

f = (9.426) / (2π)

f = 1.5 Hz

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