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olga nikolaevna [1]
3 years ago
6

The nucleus of an atom can have either a positive or negative charge. please select the best answer from the choices provided t

f
Physics
2 answers:
Airida [17]3 years ago
5 0
F - False.

The nucleus of an atom is positively charge.
Ivahew [28]3 years ago
3 0

Answer:

FALSE

Explanation:

Nucleus of an atom consist of neutrons and protons and it is the smallest part of atom.

So the total charge on the nucleus is due to charge of protons as we know that neutrons are neutral in nature and they do not have any charge on it.

So we will have

Q_{proton} = + 1.6 \times 10^{-19} C

Q_{neutron} = 0 C

Q_{electron} = - 1.6 \times 10^{-19} C

so here we can say that total charge on a nucleus must be positive charge which is only due to protons present inside the nucleus.

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A coaxial cable carries a current I =0.02A through a wire of radius Rw=.2mmin one direction and an identical current through the
Anna [14]

Answer:

a)  B = 10⁻¹ r , b)  B = 4 10⁻⁹ / r , c) B=0

Explanation:

For this exercise let's use Ampere's law

            ∫ B. ds = μ₀ I

Where I is the current locked in the path. Let's take a closed path as a circle

          ds = 2π dr

          B 2π r = μ₀ I

          B = μ₀ I / 2μ₀ r

Let's analyze several cases

a) r <Rw

Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density

          j = I / A

For this case

         j = I /π Rw² = I’/π r²

         I’= I r² / Rw²

The magnetic field is

        B = (μ₀/ 2π) r²/Rw²   1 / r

        B = (μ₀ / 2π) r / Rw²

calculate

        B = 4π 10⁻⁷ /2π   r / 0.002²

        B = 10⁻¹ r

b) in field  between   Rw <r <Rs

In this case the current enclosed in the total current

      I = 0.02 A

      B = μ₀/ 2π   I / r

      B = 4π 10⁻⁷ / 2π  0.02 / r

      B = 4 10⁻⁹ / r

c) the field outside the coaxial Rs <r

In this case the waxed current is zero, so

       B = 0

7 0
3 years ago
4. Why is this a double replacement reaction?<br> HCl+Fes- FeCl+H2S
My name is Ann [436]

Answer:

the H from the HCl switched places with the Fe from the FeS.

6 0
3 years ago
In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

7 0
3 years ago
11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
Marat540 [252]

a =  \frac{v - u}{t}

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

approximately 3.33 m/s^2

the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

5.0 m/s^2

5.0 > 3.33 \\ so \:  the \: answer  \: is \: no

6 0
3 years ago
Can the big bang's sound still be heard?
AleksandrR [38]
There is no sound in space because there is no medium to move it. However i think the big bang can still been seen in the form of background radiation but i am unsure. 
4 0
3 years ago
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