properties of elements within a ________ on the periodic table change in a predictable way from one side of the table to another

Est ce qu’un cricket c’est un consommateur

Sholpan [36]

Answer:

Oui

Explanation:

7 0

3 years ago

We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength

san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

$\theta=1.22\frac{\lambda}{b}$

θ = 50.0*10^-7 rad

λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

$b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm$

hence, the smallest diameter objective lens is 13.4cm

8 0

3 years ago

Read 2 more answers

If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$

3 0

3 years ago

a wave is described by where x is in meters, y is in centimeters and t is in seconds. The angular frequency is

Sergeeva-Olga [200]

Complete Question

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave frequency is

Answer:

The value is $w = 10 \ rad /s$

Explanation:

From the question we are told that

The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)

Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

$y(x,t) = Asin(kx + wt )$

Where w is the angular frequency

Now comparing this equation with that given we see that

$w = 10 \ rad /s$

7 0

3 years ago

A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density

Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

mole fraction (H2SO4)= 0.034
mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

n = m / MW
= ∑ AWT
mole fraction = moles of A component / total moles of solution
ρ = m /v

Where:

m = mass
n = moles
MW = molecular weight
AWT = atomic weight
ρ = density
v = volume

Information about the problem:

m solute (H2SO4) = 17.75 g
v(solution) = 100 ml
ρ (solution)= 1.094 g/ml
AWT (H)= 1 g/mol
AWT (S) = 32 g/mol
AWT (O)= 16 g/mol
mole fraction(H2SO4) = ?
mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)= 98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)= 18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092 mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092 mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092 mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

Solvent
Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

8 0

2 years ago