Which of the following quantities are vectors?

The atomic number of a element if found by countingthe number of what in an atom?

S_A_V [24]

The Atomic number is found by counting the number of Protons found in the atom. Make brainliest if I helped best please :)

4 0

3 years ago

Draw a Lewis dot structure for Hf<br><br>

Serjik [45]

Answer:

There!!

Explanation:

mark me brainliest!!

3 0

2 years ago

.

fomenos

Answer:

The bulbs should be connected in parallel.

Explanation:

We want to find out a way to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other bulbs stays lit.

We must connect the two bulbs in parallel so that even when one bulb is burns out, it will have no effect on the other bulb and the 2nd bulb will keep on working. The current flowing in each bulb will depend upon the resistance of each bulb and the voltage will be same across each bulb.

On the other hand, if we use a series circuit then if one bulb burns out then the there is no flow of current in the circuit and therefore, the second bulb will not be operational.

The current flowing through each bulb is given by

I = V/R

The voltage across each bulb is given by

V = IReq

Where I is the current and Req is the equivalent resistance of the two bulbs connected in parallel and is given by

Req = (R₁*R₂)/(R₁+R₂)

The connection diagram is attached where two bulbs are connected in parallel and are power with a battery.

4 0

3 years ago

When an object is stationary, all of the forces acting on it are balanced

Leviafan [203]

With utmost clarity, that is truest of the truths

5 0

2 years ago

Read 2 more answers

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago