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3 years ago

DESPERATE WILL GIBE BRAINLIST AND THANKS

Physics

1 answer:

const2013 [10]3 years ago

4 0

Answer:

True

Radiometric dating is a process of identifying the age of a material based on known half-lives of decaying radioactive materials found in both organic and inorganic objects. Radiometric dating is often used to determine the age of rocks, bones, and ancient materials.

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The muzzle velocity of a rifle bullet is 800. m⋅ s −1 m⋅s−1 along the direction of motion. If the bullet weighs 27.0 g g , and t

11Alexandr11 [23.1K]

Answer:

1.98 × 10⁻³³m

Explanation:

It is given that,

Mass of the bullet, m = 27 g = 0.027 kg

Velocity of bullet, v = 800 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

$p=mvp=0.027\times 800 = 21.6\ kg-m/s$

Uncertainty in momentum is,

$\Delta p=0.2\%\ of\ 21.6\\\Delta p=0.0432$

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x\geq \dfrac{h}{4\pi}\Delta x=\dfrac{h}{4\pi \Delta p}\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.0432}\Delta x=1.98\times 10^{-33}\ m$

4 0

4 years ago

This lab is investigating the relationship between mass, ________, and momentum.

Sav [38]

This lab is investigating the relationship between mass, Speed , and momentum.

Momentum is manufactured from the mass and speed of an object. it's miles a vector quantity, owning a significance and a direction. If m is an object's mass and v is its speed, then the object's momentum is p.

Momentum in an easy way is a quantity of movement. right here amount is measurable because if an item is moving and has mass, then it has momentum. If an object no longer flows then it has no momentum. however, in regular existence, it has an important but many people didn't understand it.

Momentum gives the connection between the mass, pace, and direction of an object. Any exchange in momentum results in pressure. So, an exchange in momentum is used to determine the force appearing upon the item.

Learn more about momentum here:-brainly.com/question/1042017

#SPJ1

7 0

2 years ago

A 2kg object is moving horizontally with a speed of 4m/s 2. How much net force is required t keep the object moving at this spee

dezoksy [38]

To continue moving at constant speed in a straight line requires NO net force. Zero. Nada. If there IS any net force on the object, then its speed or direction will change.

8 0

3 years ago

Which of the following properties did Rutherford use in his experiment?

Ronch [10]

Answer:

(C) The positive charge of the alpha particles and the negative charge of the electrons are the properties Rutherford used in his experiment

Explanation:

In his scattering experiment, scientist Ernest Rutherford used the property of positively charged alpha particles and negatively charge electrons. He performed this experiment by passing some of the alpha particles through a gold foil. The result was that the some alpha particles scattered while some passed through the gold foil without collision.

He concluded that the alpha particles are centrally positively charged and needed a large amount repelling force. This experiment of Rutherford is also known as Rutherford model of atom. This experiment helped him in doing so many other discoveries.

3 0

4 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago