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Elden [556K]
2 years ago
12

A space station consists of two donut-shaped living chambers, A and B, that have the radii shown in the drawing. As the station

rotates, an astronaut in chamber A is moved 2.30 102 m along a circular arc. How far along a circular arc is an astronaut in chamber B moved during the same time?

Physics
1 answer:
tensa zangetsu [6.8K]2 years ago
6 0
The angle subtended by the arc of movement of both chambers is the same. Thus:
S = r∅, for chamber A:

2.3 x 10² = 3.2 x 10² ∅
∅ = 0.72 rad

∅ is the same for B so:
S = 1.1 x 10³ x 0.72
S = 7.92 x 10² m
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Hope you can answer this: A Student Visits A Farm And Makes These Notes In Her Journal.
Strike441 [17]

Answer:

Common Sense

Explanation:

The chick has probably seen other chicks get caught by the Hawk and knows not to go near, or saw a giant bird flying straight towards it and used common sense to identify it as danger and run away. Although if this is for a test or a grade or something, please do not use the answer, it is most likely incorrect. This is honestly my best answer.

8 0
1 year ago
8. The resistance of a bagel toaster is 14 Ω. To prepare a bagel, the toaster is operated for one minute from a 120-V outlet. Ho
IgorC [24]

Answer: The energy delivered to the toaster is 264.490KJ

Explanation:

Here is the complete question:

The resistance of a bagel toaster is 14 ?. To prepare a bagel, the toaster is operated for one minute from a 120-V outlet. How much energy is delivered to the toaster?

Step-by-step explanation:

Please see attachment below

7 0
2 years ago
Read 2 more answers
If the net force acting on an object increases by 50 percent, then the acceleration of the object will
ki77a [65]
<span>If the net force acting on an object increases by 50 percent, then
the acceleration of the object will also increase by 50 percent.

This answer is not offered among the list of choices.

So the correct response is "D. none of the above"</span>
8 0
2 years ago
(b) Figure 4 shows a car travelling on a motorway.
Alik [6]

Answer:

To calculate anything - speed, acceleration, all that - we need <em>data</em>. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to <em>measure </em>it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring <em>time </em>and <em>distance</em>, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

  1. Press start on the stopwatch right as you pass a lamp post
  2. Each time you pass another lamp post, press the lap button on the stopwatch
  3. Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
  4. Record your data
  5. Calculate the time intervals for passing each lamp post using the lap data
  6. Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

5 0
1 year ago
For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a
jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

R=\frac{L}{A\sigma }

R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

i = \frac{V}{R}

i = \frac{0.060}{0.0178}

i = 3.4 A

(c)

Current density is given as

J = \frac{i}{A}

J = \frac{3.4}{5.3\times10^{-6}}

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

E = \frac{J}{\sigma }

E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}

E = 9.01 x 10⁻³ V/m

5 0
2 years ago
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