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kipiarov [429]
3 years ago
12

Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his

body or lift other objects.
True

False
Physics
2 answers:
just olya [345]3 years ago
4 0
The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.
Svetradugi [14.3K]3 years ago
4 0

Answer:

False

Explanation:

Every thing on the surface of Earth is at a state of rest i.e., it has a constant force of gravity acting on it which keeps things planted on the ground. Let us take an example of an elevator. When we are going down we experience a little bit of weightlessness. Now, imagine an elevator going at a faster rate down our bodies will feel complete weightlessness. This is the same phenomenon experienced by the astronauts in low Earth orbit.

So, everything experiences weightlessness so the astronauts need not apply much force to move or lift objects.

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a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one
prohojiy [21]

The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be 37^oC is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

f=\frac{v_T}{4L}

where, v_T=v\sqrt{\frac{T}{273} }

Since, T=37+273 K = 310 K

v = 331 m/s

\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s

Using this, we get:

f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz

Hence, the fundamental frequency is 367.42 Hz.

To learn more about Attention here:

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7 0
1 year ago
A cell phone is released from the top with the speed of 10ms what is the speed 3s after?
sergeinik [125]

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

6 0
3 years ago
Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

A_y=+15 m

\theta=32^{\circ}

We have to find the x-component of A and magnitude of A.

According to question

A_y=\mid A\mid cos\theta

Substitute the values then we get

15=\mid A\mid cos32

\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}

\mid A\mid=17.7m

tan\theta=\frac{perpendicular\;side}{Base}

tan32=\frac{A_x}{A_y}=\frac{A_x}{15}

0.62\times 15=A_x

A_x=9.3

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

6 0
3 years ago
Read 2 more answers
A real battery with internal resistance 0.460 Ω and emf 9.00 V is used to charge a 56.0-µF capacitor. A 21.0-Ω resistor is put i
Margaret [11]

Answer: 1.176×10^-3 s

Explanation: The time constant formulae for an RC circuit is given below as

t =RC

Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F

t = 56×10^-6 × 21

t = 1176×10^-6

t = 1.176×10^-3 s

4 0
3 years ago
Read 2 more answers
How many atoms are in SO<br> 3
kkurt [141]

Answer:

There are 6.87 x 1023 atoms in 1.14 mol SO3, or sulfur trioxide :

3 0
3 years ago
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