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Arturiano [62]
2 years ago
12

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

ld allow astronauts to experience a sort of artificial gravity when walking along the inner wall of the station's outer rim. (a) Imagine one such station with a diameter of 121 m, where the apparent gravity is 2.60 m/s2 at the outer rim. How fast is the station rotating in revolutions per minute?
Physics
1 answer:
lyudmila [28]2 years ago
5 0

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration a_c, so:

a_c = w^2r

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

a_c = 2.6m/s2

So, Replacing the data, and solving for w, we get:

2.6m/s = w^2(60.5m)

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2\pi)= 1.976 rpm

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2 years ago
The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet
Naily [24]

Answer:

Speed of comet before collision is

v_{2_{i}}=-2.5\times10^{3}\quad km/h

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h

Case is considered as partially inelastic collision, by conservation of momentum

m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h

8 0
2 years ago
A soccer ball kicked with a force of 13.5 n accelerates at 6.5 m/s^2 to the right. what is the mass of the ball?
natita [175]

Answer:

2.08 kg

Explanation:

Newton's second law states that the acceleration of an object is proportional to the force applied to the object, according to the equation:

F=ma

where F is the force applied, m is the mass of the object and a its acceleration.

In this situation, the soccer ball is kicked with a force F=13.5 N and its acceleration is a=6.5 m/s^2, therefore its mass is

m=\frac{F}{a}=\frac{13.5 N}{6.5 m/s^2}=2.08 kg

6 0
2 years ago
Sebanyak 0,2 mol gas ideal berada dalam wadah yang volumenya 10 liter dan tekanan 1 atm . berapakah suhu gas tersebut
tamaranim1 [39]

The temperature of the gas is about 600 K

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u><em>Question (Translation):</em></u>

<em>A total of 0.2 moles of ideal gas are in containers with volume of 10 liters and a pressure of 1 atm. What is the temperature of the gas?</em>

<u>Given:</u>

number of moles = n = 0.2 moles

volume of gas = V = 10 liters

pressure of gas = P = 1 atm

gas contant = R = 0.0821 L.atm/mol.K

<u>Unknown:</u>

temperature of gas = T = ?

<u>Solution:</u>

PV = nRT

1 \times 10 = 0.2 \times 0.0821 \times T

10 = 0.01642 \times T

T = 10 \div 0.01642

T \approx 600 \texttt{ K}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

8 0
2 years ago
Which quantities decrease as the distance between a planet and the Sun increases? Check all that apply.
sweet-ann [11.9K]

<u>Answer</u>

Gravitational force

<u>Explanation</u>

Gravitational force is the force of attraction between two bodies of a given masses. It is calculated as follows:

Gravitational force = GM₁M₂/r²

From the formula it can be seen that gravitational force is inversely proportional to the r where are is the distance between the two bodies.

When the distance increases the gravitational force decreases.

5 0
2 years ago
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