According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is -6048 N
<h3>
What does Newton's third law say ?</h3>
The law state that in every action, there will be equal and opposite reaction.
Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.
Let us first calculate the force involved in the acceleration of the rocket by using the formula
F = ma
Where mass m = 87.4 kg, acceleration a = 69.2 m/s2
Substitute the two parameters into the formula
F = 87.4 x 69.2
F = 6048.08 N
According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.
Therefore, the normal force acting on the astronaut is -6048 N approximately
Learn more about forces here: brainly.com/question/12970081
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Answer:
Differences between freefall and weightlessness are as follows:
<h3>
<u>Freefall</u></h3>
- When a body falls only under the influence of gravity, it is called free fall.
- Freefall is not possible in absence of gravity.
- A body falling in a vacuum is an example of free fall.
<h3>
<u>Weightlessness</u></h3>
- Weightlessness is a condition at which the apparent weight of body becomes zero.
- Weightlessness is possible in absence of gravity.
- A man in a free falling lift is an example of weightlessness.
Hope this helps....
Good luck on your assignment....
Answer:
A major challenge in the drug delivery field is to enhance transport of therapeutics across biological barriers such as the blood brain barrier (BBB), the small intestine, nasal, skin and the mouth mucosa.
Answer:
Explanation:
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Answer:
H / R = 2/3
Explanation:
Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.
Initial. Lowest point
Em₀ = K = 1/2 m v²
Final. In the sought height
= U = mg h
Energy is conserved
Em₀ =
½ m v² = m g h
v² = 2 gh
Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass
Initial. Lower
Em₀ = K = ½ I w²
Final. Heights sought
Emf = U = m g R
Em₀ =
½ I w² = m g R
The moment of inertial of a cylinder is
I =
+ ½ m R²
I= ½
+ ½ m R²
Linear and rotational speed are related
v = w / R
w = v / R
We replace
½
w² + ½ m R² w² = m g R
moment of inertia of the center of mass
= ½ m R²
½ ½ m R² (v²/R²) + ½ m v² = m gR
m v² ( ¼ + ½ ) = m g R
v² = 4/3 g R
As they indicate that the linear velocity of the two points is equal, we equate the two equations
2 g H = 4/3 g R
H / R = 2/3