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3 years ago

What will happen to the solar particle as the particles enter magnetosphere?

Physics

1 answer:

saul85 [17]3 years ago

8 0

Answer:

When the solar particles enters into the magnetosphere gets deflected.

Explanation:

When particles enter into the magnetosphere gets deflected because of the solar wind. After getting deflected they comes towards the earth's magnetic field. The Earth's outer space is live and also much dynamic. When solar wind as well as the plasma comes together in magnetosphere, electricity gets generated and also formation of electric field. The magnetosphere found to attract the charged particles only that are being emitted by the sun.

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You push a shopping cart for fun with a force of 60N. If the shopping cart has a mass of 12kg, what is the

posledela

Answer:

$\huge\boxed{\sf a = 5 \ ms^{-2}}$

Explanation:

<u>Given:</u>

Force = f = 60 N

Mass = m = 12 kg

<u>Required:</u>

Acceleration = a = ?

<u>Formula:</u>

F = ma

<u>Solution:</u>

Rearranging formula

a = F / m

a = 60 / 12

a = 5 ms⁻²

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>

4 0

3 years ago

A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt

ankoles [38]

Kinwtic energy = (40kg*2*2)/2=40*2=80J

4 0

3 years ago

Read 2 more answers

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$