Ask question

Ask question

All categories

3 years ago

6

The uncertainty in position of a proton confined to the nucleus of an atom is roughly the diameter of the nucleus. If this diame

ter is d = 7.8×10−15 m, what is the uncertainty in the proton's momentum?

1 answer:

Oksanka [162]3 years ago

4 0

Answer:

0.135E-19kgm/s

Explanation:

Using the uncertainty principle, we find

Dp = h / (4π Dx)

= (6.63×10-34Js)/4π(3.90×10-15m) = 0.135×10-19kg m/s

You might be interested in

What are some examples of irreversible processes in nature?

castortr0y [4]

Answer:

Burning. When you burn something, it turns into ash you can't make that thing turn into what it was before.

8 0

3 years ago

A pendulum has 576 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

Natalka [10]

Ideally, 576 J because energy is conserved.
In the real world, a tiny tiny tiny tiny bit less than 576 J ,
because we live in a world with friction and air resistance.

3 0

3 years ago

The density of silver is 10.5 g cm3. a piece of silver with a mass of 61.3 g would ovvupy a volume of ?

natali 33 [55]

The relationship between mass m, volume V and density d is:
$d= \frac{m}{V}$
The silver has density $d=10.5 g/cm^3$ , and the mass of the piece of silver is $m=61.3 g$ . Therefore we can calculate its volume using the previous formula:
$V= \frac{m}{d}= \frac{61.3 g}{10.5 g/cm^3}=5.84 cm^3$

7 0

2 years ago

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$

$\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}$

4 0

2 years ago

I NEED HELP PLZZZZZZ

schepotkina [342]

Mostly GPE and a little KE since the ball is high up (GPE) and it's also moving (KE) but not as much as it had when you first threw it

4 0

2 years ago