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3 years ago

A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.

Physics

1 answer:

Alex73 [517]3 years ago

5 0

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

d' = d cos Ф

= 1430 cos 5.76°

= 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

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Rephrase the law of universal gravitation

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3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

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Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

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3 years ago

What is the maximum wavelength a photon can have if it is to possess sufficient energy to cause this dissociation?

Alona [7]

I attached the full question.
We need to calculate the enthalpy change first:
$\Delta H=$(247kJ)-(142.3kJ)=+105.2kJ$
Keep in mind this is the energy per mole, we need energy per atom:
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We can see that this reaction needs the energy to get started (enthalpy change is positive).
The energy of a photon is given with this formula:
$E=\frac{hc}{\lambda}$
We need this photon to have the same energy (or higher) as our enthalpy change:
$105.2=\frac{6.62\cdot 10^{-34}\cdot 3\cdot 10^ 8}{\lambda}\\ \lambda=\frac{6.62\cdot 10^{-34}\cdot 3\cdot 10^ 8}{105.2/(6.02\cdot 10^{23})} \\\lambda=1.138\mu m$
This wavelenght is clasified as near infrared.

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3 years ago