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Monica [59]
3 years ago
7

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v

elocity must an object be thrown upward (from a height of 3 meters) to reach a maximum height of 460 meters? (Round your answer to one decimal place.)
Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

<u>Vi = 94.64 m/s</u>

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vodomira [7]

Given parameters:

Mass on earth  = 50kg

Unknown:

Mass on planet Xenon = ?

Weight on planet Xenon = ?

Mass is the amount of matter contained in a particular substance.

Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.

         Weight  = mass x acceleration due to gravity

Planet Xenon has half the gravitational force of Earth.

 This translated gives \frac{9.8}{2}   = 4.9m/s²

Now, mass is always the same every where if the amount of matter in a substance does not change.

In this problem, mass = 50kg on planet xenon.

Weight =  mass x acceleration due to gravity  = 50 x 4.9  = 245N

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4 years ago
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal
kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both  spacecrafts, the velocities are described as follows;

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Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

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vB (0) = 0^2+ 3(0) + 10

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We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

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3 years ago
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jek_recluse [69]

i believe the answer is water vapor

hope this helps


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