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3 years ago

What are solar flares

2 answers:

8 0

It's a brief of intense high energy radiation from the sun's surface. When it happens it disrupts the earth's communication and power line transmissions

crimeas [40]3 years ago

4 0

Solar flares are vast clouds of electrically charged subatomic particles that are continually blasted away from the Sun.

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Integrated Science- 8th grade science

padilas [110]

Answer:

A. a rigorously tested explanation

Explanation:

B. and D. are out - theories are not opinionated, they are factual
C. is out - not all theories are mathematical
A. is the best choice

3 0

2 years ago

Read 2 more answers

the kinetic energy of an object with mass m moving with a velocity of 5 m/s is 25 j what will be its kinetic energy when its vel

Dmitriy789 [7]

Answer:

100 J, 225 J

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity of the object

In this problem, the initial kinetic energy of the object is

K = 25 J

Then, the velocity is doubled, which means

v' = 2v

Therefore, the new kinetic energy will be

$K'=\frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2)=4K$

Therefore, the kinetic energy has quadrupled:

$K' = 4(25)=100 J$

Later, the velocity is tripled, which means

v'' = 3v

Therefore, the new kinetic energy will be

$K''=\frac{1}{2}m(3v)^2 = 9(\frac{1}{2}mv^2)=9K$

Therefore, the kinetic energy has increased by a factor of 9:

$K' = 9(25)=225 J$

4 0

3 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

3 years ago

Read 2 more answers

A normal walking speed is around 2.0 m/s . how much time t does it take the box to reach this speed if it has the acceleration 5

creativ13 [48]

Given:

u(initial velocity)=0

a=5.54m/s^2

v(final velocity)=2 m/s

v=u +at

Where v is the final velocity.

u is the initial velocity

a is the acceleration.

t is the time

2=0+5.54t

t=2/5.54

t=0.36 sec

6 0

3 years ago

Помогите плз!

Softa [21]

Все написано в скобках правильно

3 0

3 years ago