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cluponka [151]
3 years ago
6

In the year 2055, a rocket was launched from a research laboratory on Mars. Mars has essentially no atmosphere. The test rocket

had an initial acceleration of 8.0 m/s2. Its engine burned for 7.00 seconds, and then the rocket coasted in a condition of free fall, under the influence of Mars's gravity, which is 3.71 m/s2. Find the maximum height reached by the rocket.
Physics
1 answer:
algol [13]3 years ago
4 0

Answer:

h = 618.64 m

Explanation:

First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:

h₁ = Vit + (1/2)at²

where,

h₁ = height gained during the burning of fuel

Vi = Initial Velocity = 0 m/s

t = time = 7 s

a = acceleration = 8 m/s²

Therefore,

h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²

h₁ = 196 m

Now we use 1st equation of motion to find final speed Vf:

Vf = Vi + at

Vf = 0 m/s + (8 m/s²)(7 s)

Vf =  56 m/s

Now, we calculate height covered in free fall motion. Using 3rd equation of motion:

2ah₂ = Vf² - Vi²

where,

a = - 3.71 m/s²

h₂ = height gained during free fall motion = ?

Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)

Vi = 56 m/s

Therefore,

(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²

h₂ = 422.64 m

So the total height gained will be:

h = h₁ + h₂

h = 196 m + 422.64 m

<u>h = 618.64 m</u>

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Answer:

250,000

Explanation:

<h2> </h2>

<h2>formula = ( F=ma </h2>

  • F=1500N
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3 years ago
A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto
lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

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5 0
3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force
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Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

7 0
3 years ago
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