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masya89 [10]
3 years ago
11

Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.

Physics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

The horizontal distance covered by the firework will be \frac{1876.8}{g}

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

v_{y}=u_{y}+(-g) t

u_{y}= u\sin \phi

and v_{y}=0

therefore \frac{u\sin \phi }{g} =t

writing equation of motion in horizontal direction:

s_{x}=u_{x}t

u_{x} = u\cos \phi

therefore the equation becomes s_{x}=\frac{u^{2}   \sin \phi  \cos \phi}{g}

therefore horizontal distance traveled =\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}

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Answer:

No.

Explanation:

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6 0
2 years ago
If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.
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3 years ago
The hydrogen stored inside a large weather balloon has a mass of 13.558 g. What is the volume of this balloon if the density of
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6 0
3 years ago
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7 0
4 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

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Read more about vertical weight here:

brainly.com/question/15244771

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5 0
2 years ago
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