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A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.

zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is ∝ v²

so we can say

$\frac{a2}{a1} = \frac{v2}{v1}$ .....................1

put here value

$\frac{2}{1} = \frac{v2}{22.1}$

solve it

v2 = $\sqrt{2 }$ × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0

3 years ago

While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.

lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0

3 years ago

A mechanical wave starts when matter is disturbed by a source of ______.

Gwar [14]

Answer:

MATERIAL MEDIUM

Explanation:

Wave is a disturbance that travels through a medium and transfer energy from one point to another without causing any permanent displacement of the medium itself. The two forms of wave are the mechanical wave and the electromagnetic waves.

Mechanical wave is a wave with requires MATERIAL MEDIUM for its propagation. This means that before wave can be propagated at times, material medium is needed e.g a ripple tank. A ripple tank is a mechanical device that generates waves using an instrument called stroboscope attached to it. This kind of wave requires an external source before it can propagate compared to electromagnetic waves that does not require material medium for its propagation.

5 0

3 years ago

One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and

taurus [48]

Answer:

(a) Initial volume will be 7.62 L

(b) Final temperature will be 303.85 K

Explanation:

We have given one mole of ideal gas done 3000 J

So work done W = 3000 J

Let initial volume is $V_1$ and initial pressure $P_1=1atm$ ( As pressure is constant )

Final volume $V_2=25L$ = 0.025 $m^3$

Number of moles n = 1

(B) From ideal gas of equation we know that $PV=nRT$

So $1.01\times 10^5\times0.025=1\times 8.31\times T$

T = 303.85 Kelvin

(B) For isothermal process work done is equal to

$W=nRTln\frac{V_2}{V_1}$

$3000=1\times 8.314\times 303.85\times ln\frac{0.025}{V_1}$

$ln\frac{0.025}{V_1}=1.1881$

$\frac{0.025}{V_1}=3.2808$

$V_2=0.00846m^3=7.62L$

So initial volume will be 7.62 L

5 0

3 years ago

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

3 years ago

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