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3 years ago

10

In most cases, how many electrons does it take to completely fill the

1 answer:

Lyrx [107]3 years ago

4 0

Answer: 2

Explanation:

Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.

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Help please I don't understand this

BARSIC [14]

It's either staying there or is going at the same pace

4 0

3 years ago

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)

7 0

3 years ago

A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.

Nadusha1986 [10]

<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E = 27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = $\frac{1}{2}$ m v²

= $\frac{1}{2}$ x 27 x ( 6.1 )² = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = $\frac{3290}{14}$ = 235 N

3 0

3 years ago

Calculate the potential energy of a 4 kg cat crouched 3 meters off the ground

Tom [10]

Gravitational potential energy = mass × gravity × height

Ep = (4)(9.81)(3)

Energy = 117.72 Joules

= 1.2x10^2 Joules

7 0

3 years ago

Read 2 more answers

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e

Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

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

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

5 0

3 years ago