Ask question

Ask question

All categories

SCORPION-xisa [38]

1 year ago

10

An atom of argon has a radius of 71.pm and the average orbital speed of the electrons in it is about ×3.9107/ms. calculate the l

east possible uncertainty in a measurement of the speed of an electron in an atom of argon. write your answer as a percentage of the average speed, and round it to 2 significant digits.

1 answer:

Anna11 [10]1 year ago

7 0

Answer: 2.1 %

Explanation:

The radius of the Argon atom, r = 71 pm = 7.1 × 10 ⁻¹¹ m

Average orbital speed of electrons, v = 3.9 × 10⁷ m/s

From uncertainty principle:

Δx m Δv ≥ h/4π

mass of electron, m = 9.1 ×10⁻³¹ kg

Δx = radius of the argon atom = 7.1 × 10 ⁻¹¹ m

$\Rightarrow \Delta v = \frac {6.626 \times 10^{-34} m^2kg/s}{4\times 3.14 \times 7.1 \times 10^{-11} m \times 9.1 \times 10^{-31} kg}$

$\Delta v = 8.2 \times 10^5 m/s$

Percentage uncertainty:

$\frac{\Delta v}{v} \times 100\% = \frac {8.2 \times 10^5 m/s}{3.9 \times 10^7 m/s} \times 100 \%= 2.1 \%$

You might be interested in

A mass is oscillating horizontally on a spring. At the locations A, B, C, D, and E, photogates are used to measure the speed of

svetoff [14.1K]

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is $PE_{elastic} = 50J$

The kinetic energy at point D is $KE = 75J$

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is $E_ {m} = 100J$

So at point B

$E_{m} = PE_{elastic} +KE$

$100 = PE_{elastic} + KE$

KE at point B is 50J

So $PE_{elastic} = 100 - 50 = 50J$

Now at point D

$E_{m} = PE_{elastic} +KE$

$PE_{elastic}$ at point D is 25J

So $KE = 100 - 25 = 75J$

4 0

1 year ago

An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight

EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

so here we have

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0

2 years ago

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

2 years ago

A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi

tekilochka [14]

This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N

The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
= 1721.63 N</span>

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely, so we can say that the horizontal component is zero:

Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
= 811.66 N</span>

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0

2 years ago

The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in

solmaris [256]

Answer: the correct answer is 7.8026035971 x 10^(-13) joule

Explanation:

Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get

m = 4.00260 u + 222.01757 u = 226.02017 u .

The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is

Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,

which is equivalent to an energy change of

Delta E = (0.00523 u)*(931.5MeV/1u)

Delta E = 4.87 MeV

Converting 4.87 MeV to Joules

1 joule [J] = 6241506363094 mega-electrón voltio [MeV]

4 mega-electrón voltio = 6.40870932 x 10^(-13) joule

4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule

5 0

2 years ago