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3 years ago

What type of charge does an electron have?

Physics

2 answers:

motikmotik3 years ago

6 0

An electron has a negative charge! I hope you do good on your test! Please make me brainiest.

GarryVolchara [31]3 years ago

4 0

An electron has a negative charge. Hope this helps.

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The amount of light lost within a fiber-optic system is known as

densk [106]

Attenuation is the correct answer.

6 0

3 years ago

Which of the following has the fewest calence electrons?

dlinn [17]

Answer:

both magnesium and strontium have 2 valence electrons

Explanation:

potassium has 8 valence electrons, and nitrogen has 5.

8 0

2 years ago

Suppose A=BnCm, where A has dimensions LT, B has dimensions L2T-1, and C has dimensions LT2. Then the exponents n and m have the

julsineya [31]

Explanation:

The expression is :

$A=B^nC^m$

A =[LT], B=[L²T⁻¹], C=[LT²]

Using dimensional of A, B and C in above formula. So,

$A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}$

Comparing the powers both sides,

2n+m=1 ...(1)

2m-n=1 ...(2)

Now, solving equation (1) and (2) we get :

$n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}$

Hence, the correct option is (E).

5 0

3 years ago

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its

True [87]

Answer:

The angular velocity is $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

The mass of wooden gate is $m_g = 4.5 kg$

The length of side is L = 2 m

The mass of the raven is $m_r = 1.2 kg$

The initial speed of the raven is $u_r = 5.0m/s$

The final speed of the raven is $v_r = 1.5 m/s$

From the law of conservation of angular momentum we express this question mathematically as

Total initial angular momentum of both the Raven and the Gate = The Final angular momentum of both the Raven and the Gate

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is zero

Note: This above is the generally formula for angular momentum of square objects

The final angular velocity of the Raven is $m_r * v_r * \frac{L}{2}$

The final angular velocity of the Gate is $\frac{1}{3} m_g L^2 w$

Substituting this formula

$m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

Substituting value

$\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

$6w = 4.2$

$w = \frac{6}{4.2}$

$w = 1.43\ rad/sec$

5 0

3 years ago

A circuit has a current of 2.4 A. The voltage is increased to 4 times its original value, while the resistance stays the same. H

9966 [12]

Answer:

Resistance will become 4 times the previous value

Explanation:

We have given current in the circuit i = 2.4 A

According to ohm's law current in the circuit is given by $I=\frac{V}{R}$

So $2.4=\frac{V}{R}$ ............eqn 1

Now voltage is increased to 4 times so new voltage = 4 V

And current in the circuit is same as 2.4 A

We have to fond the resistance so that after increasing voltage current will be same

So $2.4=\frac{4V}{R_{unknown}}$ ..........eqn 2

Dividing eqn 1 and 2

$1=\frac{V}{R}\times \frac{R_{unkown}}{4V}$

$R_{unknown}=4R$

So resistance will become 4 times the previous value

6 0

3 years ago

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