Ask question

Ask question

All categories

2 years ago

7

How does earths magnetic field protect us from solar flares

2 answers:

Vinvika [58]2 years ago

5 0

Answer:

Earth's magnetic field serves to deflect most of the solar wind,

whose charged particles would otherwise strip away the ozone layer that

protects the Earth from harmful ultraviolet radiation.

Alex787 [66]2 years ago

4 0

Answer:

Earth's magnetic field serves to deflect most of the solar wind, whose charged particles would otherwise strip away the ozone layer that protects the Earth from harmful ultraviolet radiation.

You might be interested in

A star has a surface temperature of 30,000 K. At what wavelengrh does the peak radiation occour

stellarik [79]

The peak radiation will occur at 97 nm.

3 0

3 years ago

when an electric current is flowing through a wire which of the following is also produced energy domain field force

qaws [65]

When an electric current is flowing through a wire, there is a magnetic <em>field</em> produced. The lines of magnetic force are circles around the wire.

4 0

4 years ago

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

7 0

3 years ago

The Higher The Value Of Coefficient Of Friction The _____ The Resistance To Sliding.

Vadim26 [7]

Lower the resistance to sliding.

7 0

3 years ago

Hey stob it.<br> Please help me.<br> Cmon help me.<br> Plz.

Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

6 0

3 years ago