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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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The river that flows through Rome is the Tiber.
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Read the paragraph below and answer the question that follows:
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B. Helium because it is constantly being made in the sun

Explanation:

From the given paragraph, we can conclude that helium is the most common element in a star such as the sun because it is constantly being made.

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8 0
3 years ago
Vectors are an important part of the language of science, mathematics, and engineering.
tensa zangetsu [6.8K]

Answer:

<u><em>a. True</em></u>

Explanation:

<em>Vectors are an important part of the language of science, mathematics, and engineering.</em>

4 0
2 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately
Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

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area  on which photons are received is given as

A = 4.80 cm^2 = 4.80 * 10^-4 m^2

now we can find the power received due to sunlight

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now we can say this power is due to photons that strikes on surface of earth

so here we can say

P = N\frac{hc}{\lambda}

given here that

\lambda = 510 nm

0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}

0.96 = N * 3.88 * 10^{-19}

N = \frac{0.96}{3.88*10^{-19}}

N = 2.47 * 10^{18}

so it will strike 2.47 * 10^18 photons on given area per second

3 0
3 years ago
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