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Svetllana [295]

3 years ago

7

Probably the deadliest aspect of a thunderstorm is _____. rain thunder lightning hail

2 answers:

max2010maxim [7]3 years ago

8 0

The answer to the given question is lightning.
Lightning is most probably the deadliest aspect of a thunderstorm. It is <span> a sudden </span>electrostatic discharge<span> during an </span>electrical storm. It happens <span> when there is an electrostatic discharge between </span>electrically charged<span> regions of a </span>cloud, which is referred as the i<span>ntra-cloud lightning or IC, between that cloud and another cloud or the CC lightning, or between a cloud and the ground (CG lightning).</span>

kipiarov [429]3 years ago

3 0

Probably the deadliest aspect of a thunderstorm is <u> lightning </u> .

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If F1 = 104 and F2 = 104, what will be the net force?

ladessa [460]

Answer:

208

Explanation:

add it together for the answer

3 0

3 years ago

A person hits a tennis ball with a mass of 0.058 kg against a wall.

horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

Let P is the average pressure on that area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{0.607\ N}{3\ m^2}$

P = 0.202 Pa

Hence, this is the required solution.

8 0

3 years ago

Determine the acceleration that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object

user100 [1]

For 3 kg mass: <em>F = $\lpha ma,a$ = $\frac{F}{m}$ = $\frac{12N}{3kg}$ = 4 $\lpha m$ / $s^{2}$ </em>

For 6 kg mass: <em>F = $\lpha ma,a$ = $\frac{F}{m}$ = $\frac{12N}{6kg}$ = 2 $\lpha m$ / $s^{2}$ </em>

Hope this helped, have a great day!~

3 0

3 years ago

Two balloons are charged with an identical quantity and type of charge: -6.25 x 10-6 C. They are held apart at a separation dist

steposvetlana [31]

Answer:

The electric force between them is 878.9 N

Explanation:

Given:

Identical charge $q = -6.25 \times 10^{-6}$ C

Separation between two charges $r = 0.02$ m

For finding the electrical force,

According to the coulomb's law

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

Here, force between two balloons are repulsive because both charges are same.

Where $k = 9 \times 10^{9}$

$F = \frac{9 \times 10^{9} \times (-6.25 \times 10^{-6} )^{2} }{4 \times 10^{-4} }$

$F = 878.9$ N

Therefore, the electric force between them is 878.9 N

8 0

3 years ago

Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a spe

yulyashka [42]

Explanation:

Mass of two soccer balls, $m_1=m_2=0.4\ kg$

Initial speed of first ball, $u_1=0$

Initial speed of second ball, $u_2=3.5\ m/s$

After the collision,

Final speed of the second ball, $v_2=0$

(a) The momentum remains conserved. Using the conservation of momentum to find it as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_1$ is the final speed of the first ball

$0.4\times 0+0.4\times 3.5=0.4v_1+0.4\times 0$

$0.4\times 3.5=0.4v_1$

$v_1=3.5\ m/s$

(b) Let $E_1$ is the kinetic energy of the first ball before the collision. It is given by :

$E_1=\dfrac{1}{2}mu_1^2$

$E_1=\dfrac{1}{2}\times 0.4\times 0$

It is at rest, so, the kinetic energy of the first ball before the collision is 0.

(c) After the collision, the second ball comes to rest. So, the kinetic energy of the second ball after the collision is 0.

Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers