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3 years ago

Based on the trend in valence electrons across periods for main-group

Physics

2 answers:

Andreyy893 years ago

6 0

Answer: 4

Explanation: a p e x

Sergeeva-Olga [200]3 years ago

4 0

Answer: 4

Explanation:

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A ball rolls downhill with a constant acceleration of 4m/s squared. If it started from rest,it’s velocity at the end of 3 meters

vladimir2022 [97]

Answer:

4.9 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (it starts from rest)

$a=4 m/s^2$ is the acceleration

s = 3 m is the distance covered

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(4)(3)}=4.9 m/s$

3 0

3 years ago

What causes an atom to be neutral?

Ugo [173]

Answer:

Explanation:

If there are the same number of protons (+) as electrons (-) they will cancel out and make the atom neutral.

4 0

2 years ago

How much force to move 18 kg at a rate of 3m/s

Stolb23 [73]

Answer:

54N

Explanation:

F = ma

F = (18kg)(3m/s)

F = 54N

4 0

2 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

2 years ago

E20E20: The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to

Lubov Fominskaja [6]

The correct answer (sample response) is:

The image seems to be behind the mirror, but nothing is really there.

Include the following in your response:

The image appears to be behind the mirror.

If someone looks behind the mirror, there is no image there.

$|Huntrw6|$

5 0

2 years ago

Read 2 more answers