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3 years ago

I NEED HELP PLS ASAP IM LITERALLY CONFUSED??????????

Physics

2 answers:

Eduardwww [97]3 years ago

4 0

C. It pulls them together

Anastaziya [24]3 years ago

4 0

C: it pulls them together, gravity is constantly acting on everything and is pulling us to the ground

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Can someone help me figure out how to do a function formula for fx and fy

masha68 [24]

Answer: How to solve for FX and FY?

to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.

Explanation:

4 0

3 years ago

The diameter of a copper atom is approximately 2.28e-10 m. The massof one mole of copper is 64 grams. Assume that the atoms area

KIM [24]

1) Mass of one copper atom: $1.063\cdot 10^{-22} kg$

2) There are $9.33\cdot 10^{24}$ atoms in the cube

3) Mass of the cubical block: 992 kg

Explanation:

We are told here that the mass of one mole of copper is

$M=64 g$

for

$n=1 mol$ (number of moles)

We also know that the number of atoms inside 1 mole of substance is equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

This means that $N_A$ atoms of copper have a mass of M = 64 g. Therefore, we can find the mass of one copper atom by dividing the total mass by the avogadro number:

$m=\frac{M}{N_A}=\frac{64}{6.022\cdot 10^{23}}=1.063\cdot 10^{-22} kg$

We are told that the diameter of a copper atom is

$d=2.28\cdot 10^{-10} m$

We can assume that the atoms are arranged in a cube, and that they are all attached to each other; so the side of the cube can be written as size of one atom multiplied by the number of atom per side:

$L=Nd$

where

N is the number of atoms (rows) in one side of the cube

Since the side of the cube is

L = 4.8 cm = 0.048 m

We find N:

$N=\frac{L}{d}=\frac{0.048}{2.28\cdot 10^{-10}}=2.11\cdot 10^8$

This is the number of atom rows per side; therefore, the total number of atoms in the cube is

$N^3=(2.11\cdot 10^8)^3=9.33\cdot 10^{24}$

The total mass of the cubical block of copper will be given by the mass of one atom of copper multiplied by the total number of atoms, so:

$M= N^3 m$

where:

$N^3 = 9.33\cdot 10^{24}$ is the number of atoms in the cube

$m=1.063\cdot 10^{-22} kg$ is the mass of one atom

Therefore, substituting, we find:

$M=(9.33\cdot 10^{24})(1.063\cdot 10^{-22})=992 kg$

So, the mass of the cubical block is 992 kg.

Learn more about mass and density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

3 0

3 years ago

Equations to use: v= λ ∙ f v=d/t

Margarita [4]

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

$f=\frac{v}{2L}$

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

$v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s$

c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

$v=\frac{d}{t}$

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

$d=vt=(6,000 m/s)(3 s)=18,000 m$

3 0

3 years ago

Non examples of gravity ?

Afina-wow [57]

Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.

3 0

3 years ago

Read 2 more answers

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

3 0

3 years ago