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2 years ago

9

What pair of atoms forms ionic bonds

1 answer:

WARRIOR [948]2 years ago

8 0

K,F potassium , fluorine form ionic bond,

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Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the

kykrilka [37]

Answer:

A) -2N

B) Left

C) -0.5

Explanation:

A) -12 + 10

B) More force is acted on in that direction

C) Net force/Mass (-2/4)

5 0

2 years ago

Help me with this physics math?

solong [7]

Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes

(6.4)^2 x 10^12

= 40.96 x 10^12 .

Now it's just a matter of mashing out the fraction.

The 'mantissa' (the number part) is

6/40.96 = 0.1465

and the order of magnitude is

10^24 / 10^12 = 10^12 .

Put it all together and you've got

1.465 x 10^11 .

4 0

3 years ago

Which of the following elements is commonly found in the Earth's crust, living matter, oceans, and atmosphere? A. carbon B. argo

jekas [21]

The answer is carbon.

8 0

3 years ago

Read 2 more answers

In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s

serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is s = 0.9689 m

8 0

3 years ago

Leoni is participating in four drawing competitions. If the probability of her losing any

Alex17521 [72]

Answer:

(a) $Probability = 0.7599$

(b) $Probability = 0.2646$

Explanation:

Represent losing with L and winning with W.

So:

$L = 0.7$ --- Given

$n = 4$

Probability of winning would be:

$W = 1 - L$

$W = 1 - 0.7$

$W = 0.3$

The question illustrates binomial probability and will be solved using the following binomial expansion;

$(L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

So:

Solving (a): Winning at least 1

We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4

So, we have:

$Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

Substitute value for W and L

$Probability = 4 * 0.7^3*0.3 + 6*0.7^2*0.3^2 + 4*0.7*0.3^3 + 0.3^4$

$Probability = 0.7599$

<em>Hence, the probability of her winning at least one is 0.7599</em>

Solving (a): Wining exactly 2

We look at the above and we list out the terms where the powers of W is exactly 2

So, we have:

$Probability = 6L^2W^2$

Substitute value for W and L

$Probability = 6*0.7^2*0.3^2$

$Probability = 0.2646$

<em>Hence, the probability of her winning exactly two is 0.2646</em>

6 0

2 years ago