What is in between the nucleus and the electrons in an atom?

raketka [301]

Hello!

Since the two weights are off the table, the block will move towards letter F.

I hope this helps :))

5 0

3 years ago

A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill

LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

$\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s$

(ii) The linear speed of the small metal object is calculated as;

$v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s$

6 0

3 years ago

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is:

$\bar F = \frac{m\cdot (v_{2}-v_{1})}{\Delta t}$

$\bar F = 12000\,N$

The average impact force is 12000 newtons.

5 0

3 years ago

In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i

pantera1 [17]

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

Δx = 4.68 x 10⁻³ m = 4.68 mm

5 0

3 years ago

You need to build a prototype with machined parts that withstand a saline corrosive environment and temperatures above 200 degre

kondor19780726 [428]

Answer:

Stainless steel

Explanation:

I will try to order the solutions from the least correct to the most correct.

Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, Polycaprolactone is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.

On the other hand we have Untreated aluminum, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.

We have to compare the two steels.

The Mild Steel has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.

Finally, we have stainless steel, which, as the name implies, contains in some of its variations chromium, zinc or magnesium in its alloys, which makes it highly resistant to corrosion.

In addition its melting point is above 1500 ° c.

The best choice is stainless steel.

5 0

3 years ago