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2 years ago

12

The nuclear particle carrying a charge of +1e is the

2 answers:

babunello [35]2 years ago

6 0

The nuclear particle is a proton

aalyn [17]2 years ago

6 0

Inside the nucleus only two particles are present

1. Neutrons

2. Protons

These are also known as nucleons as combined

neutrons are chargeless and having mass same as that of protons while protons are positively charged particles and its charge is same in magnitude as that of electrons

SO the charge given here is +1e which means the magnitude of charge is same as the magnitude of electron but it is a positive charge

<u>so it must be proton</u>

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Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

Light is emitted by electrons when they drop from one energy level to a lower level. Which transition results in the emission of

natulia [17]

Answer:

Level 4 to level 2

Explanation:

Electrons in an atom are contained in specific energy levels (1, 2, 3, and so on) having different distances from the nucleus. When light is emitted by electrons from one energy level to a lower level, level 4 to level 2 has the greatest energy.

Hence, the correct option is "Level 4 to level 2".

3 0

3 years ago

The atomic nucleus contains two subatomic particles, the proton and the neutron. Atoms of different elements have different numb

9966 [12]

The answer is option A .

7 0

3 years ago

Read 2 more answers

A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th

Stolb23 [73]

Answer:

The current is $I_b = 400 \ A$

Explanation:

From the question we are told that

The length of the segment is $l = 2.50 \ m$

The current is $I_a = 1000 \ A$

The force felt is $F = 4.0 \ N$

The distance of the second wire is $d = 5.0 \ cm = 0.05 \ m$

Generally the current on the second wire is mathematically represented as

$I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4 \pi * 10^{-7} \ N/A^2$

=> $I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }$

=> $I_b = 400 \ A$

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2 years ago

If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh

Gala2k [10]

the correct answer is c

4 0

3 years ago