PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?

A) 8.03 x 10^16 nuclei

B) 4.01 x 10^16 nuclei

C) 2.02 x 10^16 nuclei

D) 1.61 x 10^17 nuclei

Answer:

Explanation:

Answer:

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Explanation:

The formula for calculating the elastic potential energy is:

[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]

By rearrangement and using (K) as the subject;

[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]

[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]

k = 2.3 × 10⁻² N/m

Now; the formula used to calculate the frequency of the small oscillation is:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

where;

m = mass of each atom

assuming

m = 1.66 × 10⁻²⁶ kg

Then:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Answer:

B) Blackbody radiation

MARK THIS AS BRAINLIEST PLEAWESSS :)

Answer:

Black body radiation

Explanation:

Answer:

A. Archaea

Explanation:

Living organisms have been taxonomically classified into a highest ranking taxa called DOMAIN. The domains are as follows: Eukarya, Prokarya (bacteria) and Archaea. Although both Domains Bacteria and Archaea are unicellular, the Archaea is specifically characterized by their ability to survive in harsh weather conditions like the thermal vents tiny organisms were collected from.

Also, members of the domain Archaea have a unique cell wall different from other organisms like bacteria. Hence, based on the matching characteristics of the collected tiny organisms, they would be found in the Domain ARCHAEA.

Answer:

uhm oxygen tank, a suit (like an astronaut suit) uh food, and a space ship

Explanation:

Answer:

A) vectors: veloicty, force

scalar:  speed, work

B)  t = 1.75 s,  C)   v = - 17 2 m / s

Explanation:

We answer each part separately

A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude

vector quantities: the speed of a car number is the magnitude and direction is where it goes

Force, the number is the magnitude and above that applies gives direction

Scalar magnitude: how quickly the number of the speedometer of the car

Temperature, work

B) I = 15 m height to the soil and get to calculate time = 0

        y = y₀ + v₀ t - ½ g t²

as the ball is loose its initial velocity is zero

       0 = 0 +0  - ½ g t²

       t = [tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]

       t = 1.75 s

C) the velocity to the reach the floor

      v = vo - g t

      v = 0 - g t

      v = - 9.8 1.74

       v = - 17 2 m / s

The negative signt iindicates that the speed goes down

Answer:

total internal reflection

Answer:

The bird appears farther

Explanation:

This is because as the light from the bird travels into the water which has a higher refractive index than air, light rays from the kingfisher bird bend towards the normal at the water surface and thus enter the eye of the scuba diver. Now, if we project the light rays from the eyes of the scuba diver into the air, we see that they appear to come from a point farther than that of the actual kingfisher bird perched on the branch.

So, the bird appears to the diver to be farther from the surface than the actual bird

Answer:

T = 56.11 N

Explanation:

Given that,

The equation of a wave is :

y = 0.08 sin(469t – 28.0x),

where x and y are in meters and t is in seconds

The linear mass density of the wave = 0.2 kg/m

The speed of wave is given by :

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Also,

[tex]v=\dfrac{\omega}{k}[/tex]

We have,

[tex]k=469\ and\ \omega=28[/tex]

Put all the values,

[tex]\dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}\\\\(\dfrac{\omega}{k})^2=\dfrac{T}{\mu}\\\\T=(\dfrac{\omega}{k})^2\times \mu[/tex]

Put all the values,

[tex]T=(\dfrac{469}{28})^2\times 0.2\\\\T=56.11\ N[/tex]

So, the tension in the string is 56.11 N.

Answer:

1. 2160 J

2. 2940 J

3. 64 J

Explanation:

1. Determination of the work done.

Weight (W) = 180 N

Height (h) = 12 m

Workdone =?

Wd = W × h

Wd = 180 × 12

Wd = 2160 J

Thus, the Workdone is 2160 J

2. Determination of the work done.

Mass (m) = 50 Kg

Height (h) = 6 m

Acceleration due to gravity (g) = 9.8 m/s²

Workdone =?

Wd = mgh

Wd = 50 × 9.8 × 6

Wd = 2940 J

Thus, the Workdone is 2940 J

3. Determination of the work done.

Force (F) = 16 N

Distance (d) = 4 m

Workdone =?

Wd = F × d

Wd = 16 × 4

Wd = 64 J

Thus, the Workdone is 64 J

Answer:

[tex]\boxed {\boxed {\sf 2 \ Newtons}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F= m \times a[/tex]

The mass of the stone is 0.2 kilograms and the acceleration is 10.0 meters per square second.

Substitute the values into the formula.

[tex]F= 0.2 \ kg * 10.0 \ m/s^2[/tex]

Multiply.

[tex]F=2 \ kg*m/s^2[/tex]

Convert the units.

[tex]F= 2 \ N[/tex]

The force is 2 Newtons.

Answer:

1 cal/s =4.184w

p=50 cal/s =2093w

v=12v

P = V*I

I =P/V

I = 17.43 A

P =1²*R

R = P/I²

Answer:

Aceleración, a = 2 m/s²

Explanation:

Dados los siguientes datos;

Velocidad inicial = 108 km/h

Tiempo = 10 segundos

Velocidad final = 36 km/h

To find the average acceleration;

Conversión:

36 km/h to meters per seconds = 36*1000/3600 = 10 m/s

108 km/h to meters per seconds = 108*1000/3600 = 30 m/s

I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;

[tex] V = U + at[/tex]

Dónde;

V es la velocidad final.

U es la velocidad inicial.

a es la aceleración.

t es el tiempo medido en segundos.

Sustituyendo en la fórmula, tenemos;

[tex] 30 = 10 + a*10 [/tex]

[tex] 30 = 10 + 10a [/tex]

[tex] 10a = 30 - 10 [/tex]

[tex] 10a = 20 [/tex]

[tex] Aceleracion = \frac{20}{10}[/tex]

Aceleración, a = 2 m/s²

Answer:

The electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]

The magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]

Explanation:

We are given:

[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m[/tex]

The equation used to calculate the electric field follows:

[tex]\vec{E}=\frac{kq\hat{r}}{r^2}[/tex]

OR

[tex]\vec{E}=\frac{kq}{r^2}\frac{\vec{r}}{|\vec{r}|}[/tex]

We know:

[tex]|\vec{r}|=r[/tex]

[tex]q=1.6\times 10^{-19}C[/tex]

So, the equation becomes:

[tex]\vec{E}=\frac{kq(\vec{r})}{r^3}[/tex]            .....(1)

Putting values in equation 1, we get:

[tex]\vec{E}=\frac{(9\times 10^9}(1.6\times 10^{-18})(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})}{\left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{E}=(1.84\times 10^{18}(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})\\\\\vec{E}=1.65\times 10^{9}\hat{i}+3.68\times 10^{8}\hat{j}[/tex]

Hence, the electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]

The magnetic field due to moving charge is given by:

[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \hat{r}}{r^2}[/tex]

OR

[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \vec{r}}{r^3}[/tex]              .....(2)

We are given:

[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m\\\\\vec{v}=-3\times 10^6\hat{i}[/tex]

Putting values in equation 2, we get:

[tex]\vec{B}=\frac{\mu_o\times (1.6\times 10^{-19})\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]}{4\pi\times \left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{B}=(20.42)\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]\\\\\vec{B}=-1.23\times 10^{-14}\hat{k}[/tex]

Hence, the magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]

Answer:

the man has returned from his trip

Answer:

just did by typing this lol

Answer:

66,7 seconds

Explanation:

the formula for height/distance is : S=v.t

Answer:

The wire is made of silver (ρ =  1.59×10⁻⁸ ohms/m)

Explanation:

Applying,

R = ρL/A................. Equation 1

Where R = Resistance length of the wire, ρ = Resistivity of the wire, L = Length of the wire, A = crosssectional area of the wire

make ρ the subject of the equation

ρ = RA/L............. Equation 2

From the question,

Given: R = 2.53×10⁻³ ohms, A = 3.14×10⁻⁶ m², L = 0.5 m

Substitute these values into equation 2

ρ = (2.53×10⁻³)(3.14×10⁻⁶)/0.5

ρ = 1.59×10⁻⁸ ohms/m

Hence from the resistivity chart, the wire is made of silver

Answer:

0.035 J

Explanation:

Applying,

W = ke²/2.............. Equation 1

Where W = workdone by the stretching the spring, k = spring constant, e = extension.

make k the subject of the equation

k = 2W/e²............... Equation 2

From the question

Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m

Substitute these values into equation 2

k = (2×2)/(0.15²)

k = 177.78 N/m

Hence, work need to stretch the spring from 33 cm to 35 cm

therefore,

e = 35-33 = 2 cm = 0.02 m

Substitute into equation 1

W = 177.78(0.02²)/2

W = 0.035 J

Answer:

Things you should do for your family

things you shouldn't

Answer:

Explanation:

In a reversible reaction which has reached dynamic equilibrium , rate of forward reaction is equal to rate of backward reaction .

Following is a reversible chemical reaction .

A + B = C + D

Rate of forward reaction = k₁ x [ A ] x [ B ]

Rate of backward reaction = k₂ x [ C ]  x [ D ]

k₁ x [ A ] x [ B ] =  k₂ x [ C ]  x [ D ]

[ A ] x [ B ] = k₂ / k₁  [ C ]  x [ D ]

[ A ] x [ B ] = k   [ C ]  x [ D ]

The products of the forward and backward reactions remain

constant at equilibrium.

Hence option A and D are correct statement .

Answer:

kinetic energy

Explanation:

hop it is helpful

mark me brainlist

The television has weight (8 kg) g = 78.4 N, and the magnitude of the normal force between the table and television would be the same, 78.4 N. This mean the maximum magnitude of static friction between the table and television is

0.48 (78.4 N) ≈ 37.6 N ≈ 38 N

and this is the minimum required force needed to get the television to slide.

Answer:

Explanation:before the phase change the substance is a particle.

OPTION D is the correct answer.

Refer to the attachment for complete calculation...

Answer:

Force B is zero.

Explanation:

If the plane is able to move forward then force D is greater than B. In this case because the plane is moving in the direction of D, then B is assume to be zero.

The force B in the given image will be 0. The correct option is A.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force.

An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. It is not possible for an object to contain a force.

If the plane may go forward, force D must be greater than force B. In this instance, B is taken to be zero because the plane is travelling in the direction of D.

Thus, the correct option is A.

For more details regarding force, visit:

https://brainly.com/question/13191643

#SPJ2

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

Answer:

In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid. In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid. 

Explanation:

this is the correct

Answer:

λ  = 4.023 10⁻⁷ m

Explanation:

The double-slit interference phenomenon is described by

          d sin θ = (m + ½) λ          destructive interference

          d sin θ = m λ                  constructive interference

we can use trigonometry

         tan θ = y / L

how these experiments occur for small angles

         tan θ = sin θ/cos θ = sin θ

         sin θ = y / L

we substitute

        d y / L = (m + ½) λ        destructive interference

        d y / L = m λ                 constructive interference

with the expression for constructive interference we look for the separation of the slits

        d = m λ L / y

        d = 1 603 10⁻⁹ 3 /4.84 10⁻³

        d = 3.738 10⁻⁴ m

Now let's analyze the case where the distance for constructive and destructive interference occurs at the same point y = 4.84 mm = 4.84 10⁻³m

          d y / L = (m + ½)  λ

          λ = [tex]\frac{ d \ y}{L\ (m+ 1/2) }[/tex]

the first strip is for m = 1  

let's calculate

         λ  = [tex]\frac{3.738 \ 10^{-4} 4.84 \ 10^{-3} }{ 3 \ ( 1 + 0.5) }[/tex]

         λ  = 4.023 10⁻⁷ m

Answer:

crystallography.......

Answer:

a)   μ = 0.0136, b)   F = 22.8 N

Explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

        p₀ = m v₀

final instant. After inelastic shock

        p_f = (m + M) v

the moment is preserved

        p₀ = p_f

        m v₀ = (m + M) v

        v = [tex]\frac{m}{m + M} \ v_o[/tex]

We look for the speed of the block with the bullet inside

        v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]

        v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

         W = ΔK

          W = ½ (m + M) [tex]v_f^2[/tex]  - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

         W = - fr x

we substitute

           - fr x = 0 - ½ (m + M) vo²

           fr = ½ (m + M) v₀² / x

the friction force is

          fr = μ N

          μ = fr / N

equilibrium condition

          N - W = 0

          N = W

          N = (m + M) g

we substitute

         μ = ½ v₀² / x g

we calculate

          μ = ½ 0.448 ^ 2 / 0.75 9.8

          μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

          W = ΔK

initial block velocity is zero vo = 0

         F x₁ = ½ M v² - 0

         F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]

         F = ½ 2.50 0.448² / 0.0110

         F = 22.8 N