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rewona [7]
3 years ago
6

During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m

uch charge is transferred from the cloud to the earth during such a strike?
Physics
1 answer:
Nezavi [6.7K]3 years ago
3 0

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current I = 2.5 \times 10^{4} A

Time t = 40 \times 10^{-6} sec

We know that the current is the rate of flow of charge.

From the formula of current,

<h3>  I = \frac{Q}{t}</h3>

Where Q = charge transfer between cloud and earth.

 Q =I t

Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}

Q = 1 C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

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A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed
Marina86 [1]

Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a  ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

a=\dfrac{qVB\ sin\theta}{m}

a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2

a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2

a=0.212 m/s²

4 0
3 years ago
Someone help im boredd
Oduvanchick [21]

Answer:

ok what is the question you need help with

Explanation:

:)

6 0
2 years ago
Read 2 more answers
Stio
Ainat [17]

The true statements about magnetic fields and forces will be A,D and E.

<h3>What is a magnet?</h3>

An iron piece,alloy, or other substance with its constituent atoms arranged in such a way that it shows magnetism qualities,

The function of the magnet is attracting other iron-containing objects or aligning itself in a magnetic field.

There are two poles of the magnet;

1. North Pole.

2. South Pole.

The same poles repel each other, while the opposite poles attract each other. In a sense, south-south and north-north repel. While the north-south and the south-north attract each other.

The correct statements are;

(A). The north pole attracts the south pole of a magnet.

(D)Forces caused by magnetic fields are weaker farther from the magnet.

(E)Magnetic forces can act on an object even if the object isn't touching the magnet.

Hence, the true statements about magnetic fields and forces will be A,D and E.

To learn more about the magnet, refer to the link;

brainly.com/question/13026686

#SPJ1

3 0
2 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
What is the mass of a ball that has 29j of potential energy and is lifted 2.0m?​
Salsk061 [2.6K]

Answer:

1.48kg

Explanation:

Here,

potential energy (P.E) = 29j

height (h) = 2m

acceleration due to gravity(g) =

9.8m {s}^{ - 2}

mass(m) = ?

we know,

P.E = mgh

or, 29 = m×9.8×2

or, 29/19.6 = m

or,m = 1.48kg

6 0
2 years ago
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