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otez555 [7]
2 years ago
5

A rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s2. Calculate the normal force (in N) acting on an a

stronaut of mass 90.4 kg, including her space suit.
Physics
1 answer:
vitfil [10]2 years ago
8 0

The normal force is 2820.48N in the negative y direction.

<h3>According to Newton's second law of motion, </h3>

Force = mass × acceleration

F = m×A

Note that rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s² .

The rocket accelerates upwards, hence the acceleration will be negative because it defies gravity's law (it keeps going into space without coming down)

Acceleration of the rocket = -31.2m/s²

Mass of the astronaut = 90.4kg

Normal force acting on the astronaut = -31.2 × 90.4kg

                                                                 = -2820.48N

Therefore, the normal force is 2820.48N in the negative y direction.

Learn more about normal force here:

brainly.com/question/13340671

#SPJ4

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light scattering by particles in a colloid or in a very fine suspension

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riadik2000 [5.3K]

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A) Both soap products were more effective than the hand sanitizer

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In the experiment shown below, which is greater, the force of gravity on the pith balls (Fg) or the electrostatic force between
Paraphin [41]

Answer:

Electronic force

Explanation:

Maybe because its warmer and may have more force

8 0
3 years ago
Plz plz help ...i really need steps for this answer
bija089 [108]

Answer:

3.9 seconds

Explanation:

Use constant acceleration equation:

y = y₀ + v₀ t + ½ at²

where y is the final position,

y₀ is the initial position,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

Given:

y = 0 m

y₀ = 15 m

v₀ = 15 m/s

a = -9.8 m/s²

Substituting values:

0 = 15 + 15t + ½ (-9.8) t²

0 = 15 + 15t − 4.9t²

0 = 4.9t² − 15t − 15

Solve with quadratic formula:

t = [ -b ± √(b² − 4ac) ] / 2a

t = [ 15 ± √((-15)² − 4(4.9)(-15)) ] / 2(4.9)

t = [ 15 ± √(225 + 294) ] / 9.8

t = (15 ± √519) / 9.8

t = -0.79 or 3.9

It takes 3.9 seconds for the stone to reach the bottom of the well.

The negative answer is the time it takes the stone to travel from the bottom of the well up to the top of the well.

6 0
3 years ago
Read 2 more answers
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

u =6.4013m/s

The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

h = 2.0906m

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

h = 2.0906+1.87

h = 3.9606m

4 0
4 years ago
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