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3 years ago

Which of the following periodic properties increases as you move from left to right across the periodic table?

Physics

2 answers:

Luda [366]3 years ago

8 0

Answer: B. electronegativity

Explanation: Electronegativity generally increases as you move of the periodic table and as you move to the right. All the other answers listed decrease as you move up a column of the periodic table.

andreyandreev [35.5K]3 years ago

7 0

Answer:

the answer choice is B

Explanation:

as we move from left to right , the atomic size decreases due to higher number of protons in the nucleus, which are able to attract the electrons more strongly. and so the electronegativity and electron affinity increases for the same reason. the nuclear charge increase due to more protons , and without an increase in inner electrons , there is less shielding effect. so effective nuclear charge increases.

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Why is the sky blue? The earth is surrounded by an atmosphere.

madreJ [45]

Answer:

The blue component of the spectrum of visible light has shorter wavelengths and higher frequencies than the red component. Thus, as sunlight of all colors passes through air, the blue part causes charged particles to oscillate faster than does the red part. The faster the oscillation, the more scattered light is produced, so blue is scattered more strongly than red. For particles such as air molecules that are much smaller than the wavelengths of visible light the difference is dramatic. The acceleration of the charged particles is proportional to the square of the frequency, and the intensity of scattered light is proportional to the square of this acceleration. Scattered light intensity is therefore proportional to the fourth power of frequency. The result is that blue light is scattered into other directions almost 10 times as efficiently as red light.

3 0

2 years ago

An apple drops from a tree and hits the ground in 1.5s what is its speed just before it hits the ground?

Zarrin [17]

Vf - Vo = gT

where

Vf = velocity of apple just before hitting the ground

Vo = initial velocity of apple = 0

T = time for apple to hit the ground = 1.5 sec (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

Vf - 0 = 9.8(1.5)

Vf = 14.7 m/sec.

8 0

3 years ago

You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of th

solmaris [256]

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

$v_x = u cos \theta$

where

u = 8.3 m/s is the initial velocity of the balloon

$\theta=27^{\circ}$ is the angle of projection

Substituting,

$v_x = (8.3)(cos 27^{\circ})=7.4 m/s$

The vertical component of the initial velocity of a projectile is given by

$u_y = u sin \theta$

where

u is the initial velocity

$\theta$ is the angle of projection

Here we have

u = 8.3 m/s

$\theta=27^{\circ}$

Substituting,

$u_y = (8.3)(sin 27^{\circ})=3.8 m/s$

The vertical component of the velocity of the balloon follows the suvat equation

$v_y = u_y - gt$

where

$v_y$ is the vertical velocity at time t

$u_y = 3.8 m/s$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

$v_y = 0$

So we get:

$0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s$

The maximum height of the balloon can be calculated using the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where

$u_y = 3.8 m/s$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

$s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m$

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

$t=\frac{2u_y}{g}$

where

$u_y$ is the initial vertical velocity

$g$ is the acceleration of gravity

Here we have

$u_y = 3.8 m/s$

$g=9.8 m/s^2$

Substituting,

$t=\frac{2(3.8)}{9.8}=0.78 s$

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

$d=v_x t$