Hii Happy New year can you all please answer this

What did Thomson contribute to the model of atom

Ganezh [65]

<h3>Answer;</h3>

Electrons 

<h3>Explanation;</h3>

Thomson contributed to the model of an atom by discovery of electrons and thus proving the existence of sub-atomic particles in an atom.
Thomson used cathode ray tube, and demonstrated that cathode rays were negatively charged. According to his model normally known as the plum pudding in which he stated that an atom is composed of electrons as subatomic particles that are surrounded by positive charges to balance the electrons.

5 0

4 years ago

Read 2 more answers

Witch planet takes 84 earth year's to orbit the sun just once

alekssr [168]

Uranus takes 84 earth years to make a full rotation around the sun

8 0

3 years ago

Read 2 more answers

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(c) v = 0.31m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

4 years ago

Someone please help me will give BRAILIEST!!!!!

Alex777 [14]

Uranium is the right answer. Scientists use 5 percent of the uranium after the bomb is refused to create stronger and better nuclear bombs.

6 0

4 years ago

Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measure

chubhunter [2.5K]

Answer:

the smallest angle from the antennas is 47.3°

Explanation:

We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.

Therefore, the relation is:

λ = c /f

where

c is the speed of light constant
λ is the wavelength
f is the frequency

Thus,

λ = (3 × 10⁸ m/s) / (3.4 MHz)

= (3 × 10⁸ m/s) / (3.4 MHz)(10⁶ Hz/1 MHz)

= 88.235 m

Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as

θ = sin⁻¹(λ / d)

where

d is the distance between the two radio antennas

Thus,

θ = sin⁻¹(88.235 / 120)

θ = 47.3 °

Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is 47.3 °.

5 0

3 years ago

Hii Happy New year can you all please answer this​

Hii Happy New year can you all please answer this