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4 years ago

What is a magnetosphere?

1 answer:

4 0

A magnetosphere is the region of space surrounding an astronomical object in which charged particles are controlled by that object's magnetic field. The magnetic field near the surface of many astronomical objects resembles that of a dipole.

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Cellus

WINSTONCH [101]

Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

$d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m$

So, its initial position was 471 m.

5 0

4 years ago

A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary

riadik2000 [5.3K]

Answer:

Radius between electron and proton $= 6.804\times 10^{-10}m$

Explanation:

The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.

The electrostatic force is given as = $\frac{kq_1q_2}{r^2}$

Where,

k = coulomb's law constant (9×10⁹ N-m²/C²)

q₁ and q₂ = charges = 1.6 × 10⁻¹⁹ C

r = radius between the proton and the electron

Also,

Centripetal force on the moving electron is given as:

= $\frac{m_eV^2}{r}$

where,

$m_e$ = mass of the electron (9.1 ×10⁻³¹ kg)

V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)

Now equating both the formulas, we have

$\frac{kq_1q_2}{r^2}$ = $\frac{m_eV^2}{r}$

⇒ $r = \frac{kq_1q_2}{m_eV^2}$

substituting the values in the above equation we get,

$r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}$

⇒ $r = 6.804\times 10^{-10}m$

3 0

3 years ago

A balloon filled with helium gas has an average density of rhob = 0.22 kg/m3. The density of the air is about rhoa = 1.23 kg/m3.

MissTica

Answer:

$a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

45.03681 m/s²

Explanation:

$F_b$ = Buoyant force

W = Weight of the balloon

$\rho_a$ = Density of air = 1.23 kg/m³

$\rho_b$ = Density of balloon = 0.22 kg/m³

$v_a$ = Volume of air

$v_b$ = Volume of balloon

$F_b=\rho_av_bg$

$W=\rho_bv_bg$

g = Acceleration due to gravity = 9.81 m/s²

The net force acting on the balloon is

$F=F_b-W\\\Rightarrow F=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=v_bg(\rho_a-\rho_b)\\\Rightarrow a=\frac{g}{\rho_b}(\rho_a-\rho_b)\\\Rightarrow a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

The equation is $a=g\left(\frac{\rho_a}{\rho_b}-1\right)$

$a=g\left(\frac{\rho_a}{\rho_b}-1\right)\\\Rightarrow a=9.81\times \left(\frac{1.23}{0.22}-1\right)\\\Rightarrow a=45.03681\ m/s^2$

The acceleration of the balloon is 45.03681 m/s²

8 0

3 years ago

What processes most directly helps create soil from rocks

Neko [114]

The process that most directly helps create soil from rocks is Weathering.

4 0

3 years ago

You drop a 30 g pebble down a well. You hear a splash 2.7 s later. Ignoring air resistance, how deep is the well? Assume g = 9.8

lys-0071 [83]

I will assume here that the well is sufficiently short so that the time the sound takes to come from the bottom of the well to our ear is negligible.

Since the pebble moves by uniformly accelerated motion, the distance it covers is given by
$S= \frac{1}{2}gt^2$
where
$g=9.81 m/s^2$ is the gravitational acceleration
$t=2.7 s$ is the time the pebble takes to reach the bottom of the well

Therefore, the depth of the well is
$S= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2 = 35.7 m \sim 36 m$
and the correct answer is B.

8 0

3 years ago