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swat32
3 years ago
14

How long will a plane have to fly continuously with 900 miles per hour in order to cover the same distance as that from Earth to

the next star, Proxima Centauri?
Physics
1 answer:
hammer [34]3 years ago
4 0

I'm not sure about the distance to the nearest star, but it's probably about        4 light-years (L-y).

1 L-y = 1.86 * 10E5 mi/sec * 3600 sec/hr * 24 hr/day * 365 day/yr

1 L-y = 5.9 *10E12 mi   and 4 L-y = 2.3 *10E13 mi    distance to star

2.3 * 10E13 mi / 900 mi/hr = 2.6 * 10E10 hr       hours to star

2.6 * 10E10 hr / (24 hr/day) = 1.1 * 10E9 day     days to star

1.1 * 10E9 day / 365 day/yr = 3 * 10E6 yr = 3 million years to star

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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

brainly.com/question/2562955

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

8 0
3 years ago
What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?
Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

120 V = (0.5 A) R

R = 240 Ω

6 0
3 years ago
Read 2 more answers
Which of the following is a subatomic particle?
mr Goodwill [35]

Answer:

neutron.

Explanation:

subatomic particles include,

neutron.

proton.

electron.

hope it helps. :)

4 0
3 years ago
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci
Alecsey [184]

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

Answer:  

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

8 0
3 years ago
What makes steel durable
sdas [7]

It is durable because it is one of the strongest metals and doesn't corrode easily.

4 0
3 years ago
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