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2 years ago

Use the periodic table from the lesson to answer the following question. The atomic mass of F is ___. 9 10 19 55

Physics

2 answers:

Basile [38]2 years ago

7 0

18.9 but round it and you get 19

almond37 [142]2 years ago

4 0

<h2><u>What is the atomic mass of the element F?</u></h2><h3>A. 9</h3><h3>B. 10</h3><h3>C. 19</h3><h3>D. 55</h3><h3><u>Answer</u></h3>

In this case if given those choices your answer would be (C. 19).

However in more complicated Chemistry rules you would learn that the periodic table only showcases 1 of the many possible masses of the element.

Hope this help! :D

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A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

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1 year ago

An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the ma

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Answer:

Explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

because of negative charge on electron .

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2 years ago

Where will the spacecraft be when the gravitational forces acting on it are equal?

klio [65]

It would be not be able to move yet it would be in the air

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2 years ago

An object undergoes two successive displacements:

Gre4nikov [31]

The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

$d_1=79 m$ at $16.9^{\circ}$

So its components are

$d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m$

2nd displacement is:

$d_2=16.7 m$ at $31.1^{\circ}$

So its components are

$d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m$

Therefore, the x- and y-components of the net displacement are:

$d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m$

Therefore, the magnitude of the final displacement is:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m$

Learn more about displacement:

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2 years ago

Why echo sound cannot be heard in small room

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They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.

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2 years ago