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2 years ago

Who was the discoverer of the neutron

Physics

2 answers:

Pavel [41]2 years ago

6 0

Answer:

James Chadwick

Explanation:

In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron

Effectus [21]2 years ago

6 0

James Chadwick
He was a physicists and concluded in 1932 that the mass of an atom was located in its nucleus

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Although he did not present a mechanism, what were the key points of Alfred Wegener’s proposal for the concept of continental dr

valentinak56 [21]

Answer: Alfred Wegener provided some of the important points that supported the theory of continental drift. They are as follows-

The continents were once all attached together, and this can be proved by studying the coastlines of some of the continents that perfectly matches with one another.
The appearance of similar rock types and similar fossils (including both animals and plants) has also contributed much information that continents were once all together.

4 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar

Zolol [24]

Answer:

$T= 11.0003$ s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

$\omega ^2 r =Angular\ acc$

Considering the outer ring,

$\omega ^2 r = 9.8$

$\omega ^2= \frac{9.8}{30}$

$\omega = \sqrt{\frac{9.8}{30}}$

$\omega= 0.571 rad/s$

Therefore solving for Period T

Generally the equation for solving Period T is mathematically given as

$T= \frac{2\pi}{\omega}$

$T= \frac{2\pi}{0.571 rad/s}$

$T= 11.0003$ s

5 0

3 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0

3 years ago