All categories

3 years ago

Astronauts experience weightlessness because they are in?? what

2 answers:

4 0

The weightlessness that astronauts experience is sometimes refered to as 'zero gravity'. This statement isn't true however, because their is gravity is space, just not as much. The answer would be because the astronauts are in low gravity. :D

andreev551 [17]3 years ago

3 0

Astronauts experience weightlessness because they are in space and it has zero Gravity which means u have no weight in space so u can obviously float in space Hope this helps:) mark as best please:)

You might be interested in

The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton

Brrunno [24]

Answer:

A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C) $\frac{F_e}{F_g}$ = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = $6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }$

F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

F = $k \ \frac{q_1q_2}{r^2}$

let's calculate

F = $9 \ 10^9 \ \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}$

F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

$\frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }$

= 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0

2 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.

marishachu [46]

Answer:

a) the magnitude of the force is

F= Q( $\frac{kqs}{r^3}$ ) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b) the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = $\frac{kq}{r^{2} }$

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = $\frac{kp}{r^{3} }$ , where p = q × s

E(r) = $\frac{kqs}{r^{3} }$ where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q( $\frac{kqs}{r^3}$ )

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0

2 years ago

Mickey had three substances. He hit each of them with a hammer. A red

Verizon [17]

Answer:

A & D

Explanation:

6 0

2 years ago

What type of compound forms Molecules?

guajiro [1.7K]

Answer:

Elements that are made from only one type of atom

Explanation:

7 0

3 years ago

n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us

barxatty [35]

Answer:

The answer is " $(1.265 \pm 0.010) \ s \ and \ 0.709 \%$ "

Explanation:

In point i:

$T_{theo}= 2\pi \sqrt{\frac{l}{g}}$

$=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s$

If error in the theoretical time period :

$\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}$

$= 0.010 \ s$

$T_{theo} = (1.265 \pm 0.010) \ s$

In point ii:

$\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100$

<h3> $= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%$ </h3>

5 0

2 years ago