All categories

3 years ago

Astronauts in orbit feel weightless because...

2 answers:

7 0

There is no gravity in orbit/space to pull them toward a certain area so they have full movement and that is why they float

Tatiana [17]3 years ago

4 0

You feel gravity when you feel things pressing together. Like your foot pressing against the floor, or your heart pressing against the inside of your chest. If everything around you is falling just like you are, then you never feel any things pressing together, and that feels like there's no weight.

You might be interested in

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

5 0

3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago

A coin is dropped off of a building landing on its side. It hits with a pressure of 400/2 It hits with a force of 0.1 Calculate

Arte-miy333 [17]

Complete Question:

A coin is dropped off of a building landing on its side. It hits with a pressure of 400 N/m². It hits with a force of 0.1N. Calculate the area of the coin?

Answer:

Area = 0.00025 m²

Explanation:

Given the following data;

Pressure = 400N/m²

Force = 0.1N

To find the area of the coin;

Pressure = Force/area

Area = Force/pressure

Substituting into the equation, we have;

Area = 0.1/400

Area = 0.00025 m²

5 0

2 years ago

A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1

kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

$\mu$ = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

$v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s$

Frequency is given by

$f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz$

Angular frequency is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s$

Maximum velocity of a particle is given by

$v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s$

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0

2 years ago

in which of the situations listed below is energy being transferred as heat to the system in order for the system to do work? In

In-s [12.5K]

The answer is option A, i think but i am not sure

5 0

3 years ago