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2 years ago

A(n) has charge but negligible mass, whereas a(n) has mass but no charge.

Physics

2 answers:

Dahasolnce [82]2 years ago

7 0

Answer:

An electron has charge but negligible mass, whereas a neutron has mass but no charge.

Explanation:

Atoms are made of equal number of protons and electrons. Almost all the mass of an atom is in the nucleus of the atom. The nucleus is made of protons and neutrons (except hydrogen, which has no neutron)

The charge of an electron is $-1.6 \times 10^{-19}$ C

The charge of a proton is $+1.6 \times 10^{-19}$ C

The charge of a neutron is 0 C

The mass of a proton is $1.6726 \times 10^{-27}$ kg or 1.007276 u

The mass of a neutron is $1.6749 \times 10^{-27}$ kg or 1.008664 u

The mass of an electron is $9.1 \times 10^{-31}$ kg or 0.00054858 u

As we can see the proton is 1,838 times heavier than an electron and the neutron is 1,840 times heavier than an electron.

So the electron has a negligible mass and a negative charge but the neutron has no charge but it is heavier than electron and proton

vfiekz [6]2 years ago

6 0

Electron;Neutron is the correct answer.

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

docker41 [41]

1) At x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) At x = 6.6 cm, $E_y=0$

3) At x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) At x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field is zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

The electric field of an infinite sheet of charge is perpendicular to the sheet:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.

So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).

So,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

And the negative sign indicates that the direction is to the right.

We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).

So the total field along the y-direction is zero.

This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), $dE_y$ , is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis, $-dE_y$ . Therefore, the net field along the y-direction is always zero.

Here it is similar to part 1), but this time the point is located at

x = 1.45 cm

so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is

$E=E_1-E_2$

Substituting the same expressions of part 1), we find

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative sign indicates that the direction is to the left.

This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.

Here we note that the slab is conductive: this means that the charges in the slab are free to move.

We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).

The net charge per unit area of the slab is

$\sigma=+64\mu C/m^2$

And this the average of the surface charge density on both sides of the slab, a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Also, the infinite sheet located at x = 0, which has a negative charge $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the left surface of the slab, so

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

Here we want to calculate the value of the x-component of the electric field at

x = 3.34 cm

We notice that this point is located inside the slab, because its edges are at

a = 2.9 cm

b = 4.0 cm

But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero

We calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is $\sigma_a = +65.25 \mu C/m^2$

As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is

2.9 cm < x < 4 cm

So the correct answer is

"none of these region"

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0

2 years ago

Show your work and resoning for the below requirement.

Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

- $T_{y}$ L + W₂ L + W₁ L / 2 = 0

T_{y} = W₂ + W₁ / 2

T_{y} = 120 + 150/2

T_{y} = 195 lb

we use trigonometry to find the cable tension

sin 30 = T_{y} / T

T = T_{y} / sin 30

T = 195 / sin 30

T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

T < 500 lb

4 0

2 years ago

A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if

saul85 [17]

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

F = 0.15×10×2×sin0

F = 0

Goodluck

7 0

2 years ago

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

Anna007 [38]

Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

3 0

2 years ago

All scientists try to base their conclusions on _____.

seraphim [82]

Either theory or evidence

8 0

2 years ago

Read 2 more answers

A(n) ________ has charge but negligible mass, whereas a(n) ________ has mass but no charge.

A(n) has charge but negligible mass, whereas a(n) has mass but no charge.