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JulijaS [17]
2 years ago
7

Why do protons and neutrons need to be the same

Physics
1 answer:
qaws [65]2 years ago
3 0

Answer: A neutral atom much have an equal number of protons and electrons if not it would not work.

Explanation: (Have a Pleasurable day mate.)

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A 2.0-kg object moving at 5.0 m/s encounters a 30-Newton restive force
sveta [45]

The impulse experienced by the object is 3 N s.

<u>Explanation:</u>

Impulse is also termed as change in the momentum of the object. So, it is directly proportional to the force acting on the object and the time for which the force is acting on that object.

Thus, impulse experienced by an object is the product of force acting on the object for a given time period. So, it is the sudden influence of force on the given volume.

As the force is given as 30 N and the duration or the time is given as 0.1 seconds. Then, the impulse will be product of force with duration.

Impulse = Force × ΔTime = Force × Duration

Impulse = 30 × 0.1 = 3 N s.

Thus, the impulse experienced by the object is 3 N s.

6 0
3 years ago
What is the difference between a circuit and a current?
dimulka [17.4K]

Answer:

a circuit is a complete loop that carries a current in the form of electrons from negative to positive

Explanation:

The difference between current and circuit is that a circuit is a complete loop that carries a current in the form of electrons from negative to positive .

In a circuit, there are difference elements or components such as the battery, wire, resistor. The goal of a circuit arrangement is to completely carry current from one end to another.

The current is the quantity of charge that flows within the circuit per unit of time.

So, the battery supplies the electromotive force to move the current round the circuit.

3 0
2 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area= \sigma = 1.99*10^{-7} C/m^2

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates isE = \frac{\sigma}{\epsilon}

force acting on charge is F = q E

Work done by charge q id\Delta X =\frac{ q\sigma \Delta x}{\epsilon}

this work done is converted into kinectic enerrgy

\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}

solving for v

v = \sqrt{\frac{2q\Delta x}{\epsilon m}

\epsilon = 8.85*10^{-12} Nm2/C2

v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}

v = 1.15*10^{7} m/s

8 0
3 years ago
Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying
agasfer [191]

Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

V_d = \frac{l}{t} \rightarrow l = V_d t

The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

5 0
3 years ago
How do I ship a pringle in the mail without it breaking? This is a physics project.
vekshin1
Assuming you want it to be as small and lightweight as possible :

Cut a solid box roughly twice as big as the pringle. Put the pringle inside the box, and fill the remaining space with cotton, that will cushion the impacts. Be sure to apply the mention <em>FRAGILE</em> to the box, so that they'll take care of it properly.
5 0
3 years ago
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