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3 years ago

A rod is given a charge of +16.7 UC, how many extra protons does It contain?

Physics

1 answer:

satela [25.4K]3 years ago

3 0

Answer:

We know that the electronic charge is 1.6 E-19 coulombs

Therefore 1 coulomb of charge contains 1.6 E19 electrons

A charge of 16.7 E6 coulombs therefore contains

N = 16.7 E6 * 1.6 * E-19 = 2.67 E-12 electrons

Also, the charge on the proton equals that on the electron

You might be interested in

Which sentence states Newton’s third law?

hammer [34]

<span>If two objects collide, each object exerts a force equal to and in the opposite direction of the other.</span>

3 0

3 years ago

Read 2 more answers

When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.

Readme [11.4K]

F=K*X,
F=M*a

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m

8 0

4 years ago

Two 2.0 g plastic buttons each with + 40 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on

EleoNora [17]

Answer:

a. There are three potential energy interaction. b. 2.16 m/s c. 2.16 m/s d. 0 m/s

Explanation:

a. There are three potential energy interaction.

Let the charges be q₁ = +40 nC, q₂ = +250 nC and q₃ = + 40 nC and the distances between them be q₁ and q₂ is r, the distance between q₂ and q₃ is r and the distance between q₁ and q₃ is r₁ = 2r respectively. So, the potential energies are

U₁ = kq₁q₂/r, U₂ = kq₁q₃/2r and U₃ = kq₂q₃/r

U = U₁ + U₂ + U₃ = kq₁q₂/r + kq₁q₃/2r + kq₂q₃/r (q₁ = q₃ = q and q₂ = Q)

U = kqQ/r + kq²/2r + kqQ/r = qk/r(2Q + q/2)

b. To calculate the final speed of the left 2.0 g button, the potential energy = kinetic energy change of the particle.

ΔU = -ΔK

0 - qk(2Q + q/2)/r = -(1/2mv² - 0). Since the final potential at infinity equals zero and the initial kinetic energy is zero.

So qk(2Q + q/2)/r = -1/2mv²

v = √[2qk(2Q + q/2)/mr] where m = 2.0 g r = 2.0 cm

substituting the values for the variables,

v = √[2 × 40 × 10⁻⁹ × 9 × 10⁹(2 × 250 × 10⁻⁹ + 40 × 10⁻⁹/2)/2 × 10⁻³ × 2 × 10⁻²]

v = √[360(500 × 10⁻⁹ + 20 × 10⁻⁹)/2 × 10⁻⁵]

v = √[720(520 × 10⁻⁹)/4 × 10⁻⁵] = 2.16 m/s

c. The final speed of the right 2.0 g button is also 2.16 m/s since we have the same potential energy in the system

Since the net force on the 5.0 g mass is zero due to the mutual repulsion of the charges on the two 2.0 g masses, its acceleration a = 0. Since it starts from rests u = 0, its velocity v = u + at.

Hence,

v = u + at = 0 + 0t = 0 m/s

8 0

4 years ago

What happens when a spinning ice skater draws in her outstretched arms? Group of answer choices Her moment of inertia increases

kotegsom [21]

Answer:

Her moment of inertia decreases causing her spin to speed up. The physics law behind this phenomenon is the conservation of angular momentum.

Explanation:

<em>Theory</em>

<u>The Law of conservation of angular momentum</u>

The angular momentum of a rotating body or a system remains constant unless it is acted upon by an external unbalance force.

Angular momentum = moment of inertia × angular velocity

The moment of inertia = mass×[perpendicular distance from axis of rotation]²

___________________________________________________________

⇒When skater draws in her outstretched arms the mass distribution get concentrated towards the axis of rotation so the moment of inertia of the body decrease.

But angular moment should conserve so angular velocity increases (spin increases)

5 0

3 years ago

A current of 17.0 mA is maintained in a single circular loop of 2.00 \mathrm{~m} circumference. A magnetic field of 0.800T is di

poizon [28]

The magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$ .

We have current carrying single circular loop placed in a magnetic field

parallel to the plane of the loop.

We have to determine the Magnetic moment of the loop.

<h3>What is the formula to calculate the magnetic moment of loop?</h3>

The formula to calculate the magnetic moment of the loop is -

M = NIA

where -

N - Number of turns

I - Current in the loop

A - Area of the loop

According to the question, we have -

I = 17mA = 0.017A

Circumference (C) = 2 meters

B = 0.8 T

Now -

C = 2

2 $\pi$ r = 2

$\pi$ r = 1

r = $\frac{1}{\pi }$

r = 0.32

Using the formula -

M = NIA

M = $\pi$ NI $r^{2}$

M = 3.14 x 1 x 0.017 x 0.32 x 0.32

M = 3.14 x 4 x 0.017

M = 5.5 x $10^{-3}$ A $m^{2}$

Hence, the magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$

To solve more questions on Magnetic moments, visit the link below-

brainly.com/question/13933686

#SPJ4

3 0

1 year ago