Kinetic energy is the energy an object has due to its
die hele

Answer:

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

Answer:

sorry I don't know I am only in 7th grade

Answer:

The electric field at the cylinder surface = 80.19 kN/C

Electric flux via the cylinder [tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]

Explanation:

Given that:

For the filament

The length = 4.95 m

The charge = 2.00  µC

The charge per unit length for the filament can be computed as:

[tex]\lambda = \dfrac{q}{l}[/tex]

[tex]\lambda = \dfrac{2}{4.5}\mu C/m[/tex]

Using Gauss's law:

[tex]\phi_E = \oint E^{\to}*dA^{\to}[/tex]---- (1)

where;

electric flux = [tex]\phi_E[/tex]

permittivity of free space = [tex]\varepsilon_o[/tex]

electric field = E

surface area = dA

However, the electric flux [tex]\phi_E[/tex] via the cylinder can be expressed as:

[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]

Equation (1) can now be rewritten as:

[tex]\dfrac{\lambda l'}{\varepsilon_o} = \oint E^{\to}*d(2 \pi rl')[/tex]

[tex]|E| =(\dfrac{\lambda l'}{\varepsilon_o 2 \pi rl' })[/tex]

replacing the values into the above equation:

[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \mu C/m) (1.65 \ cm)}{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (10 \ cm )(1.65 cm) })[/tex]

[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \times 10^{-6} C/m) }{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (0.1 \ m ) })[/tex]

[tex]\mathbf{|E| =80.19 \ kN/C}[/tex]

Thus, the electric field at the cylinder surface = 80.19 kN/C

The electric flux now is calculated using the said formula:

[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]

[tex]\phi_E = \dfrac{(\dfrac{2}{4.5}\times 10^{-6} \ C)(0.0165 \ m)}{8.85 \times 10^{-12} \ C^2/Nm^2}[/tex]

[tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]

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:

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:

Josiah Spiers in 1867 was when scripture union was founded

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:

c

Explanation:

Answer:

Explanation:

9ooooo

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

Answer:

C. 5730 years

Explanation:

N(t) = N(0)e^-kt

The half-life is T = 5730 years,

e^-kT = 1/2

→ k = - ln(1/2) / T

→ - ln(1/2) / 5730

→ 1.209681 x 10^-4 years^-1

The amount present dropped to 50%.

Then one half-life has elapsed, so the age is 5730 years.

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: See explanation

Explanation:

Celsius and Fahrenheit are the scales that are used in the measurement of temperature. Celcius is also refered to as centigrade. The relation that exist between Celsius and Fahrenheit is typically proportional.

The conversion from Celsius to Fahrenheit is expressed as:

F = (9/5 × C) + 32.

The conversion from Fahrenheit to Celsius is expressed as:

C = 5/9(F - 32)

For example to convert 100°C to Fahrenheit will be:

F = (9/5 × C) + 32.

F = (9/5 × 100) + 32

F = 180 + 32

F = 212°F

Answer:

To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)

Explanation:

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:

The fish gobbles the mosquito at height 18 cm.

Explanation:

Initial velocity, u = 3.7 m/s

horizontal distance, d = 28 cm

Angle, A = 39 degree

Let the time is t.

Horizontal distance = horizontal velocity x time

d =  u cos A x t

0.28 = 3.7 cos 39 x t

t = 0.097 s

Let the height is h.

Use the second equation of motion

[tex]h =u t-0.5 gt^2\\\\h= u sin A t - 0.5 gt^2\\\\h= 3.7 sin 39 \times 0.097 - 0.5\times 9.8\times 0.097\times0.097\\\\h =0.226 -0.046 \\\\h=0.18 m=18 cm[/tex]

Answer:

For (a): The chemical shift is [tex]2.08\delta[/tex]

For (b): The chemical shift is [tex]9.85\delta[/tex]

For (c): The chemical shift is [tex]7.5\delta[/tex]

Explanation:

To calculate the chemical shift, we use the equation:

[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]

Given value of spectrometer frequency = 200 MHz

Given peak position = 416 Hz

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]

Given peak position = [tex]1.97\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]

Given peak position = [tex]1.50\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]

Answer:

0.217seconds is the answer

Answer: Choice D) 19.6 J

=======================================

Work Shown:

m = 2 kg = mass

g = 9.8 m/s^2 = acceleration of gravity (approximate)

h = 1 m = height

---------

PE = potential energy

PE = m*g*h

PE = 2*9.8*1

PE = 19.6 J

The vase has approximately 19.6 Joules of potential energy.

We didn't have to make any conversions because the unit Joule is equivalent to the more complicated unit of kg*m^2/s^2 so it only involves kilograms, meters and seconds. If however the mass was given in say grams (instead of kg), then you'd need to convert to kg.

Statements imply it is thrown with velocity 30cos30° horizontally and 30sin30° vertically.

Vertically:

Total time taken = 2 x time to go up

= 2(v - u)/a

= 0 - 30sin30°)/(-g)

= 30/g

Therefore, it would travel 30/g sec in horizontal direction as well.

Horizontally :

Distance = horizontal speed x time

= 30cos30° (30/g)

= 450√3 /g

If g = 10, distance is 45√3 m.

Vertically,

Distance = vert. speed x (time of flight/2)

= 30sin30° x (30/g)/2

= 90 m.

Time taken = 30/g = 3 sec

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]

OPTION C is the correct answer.

The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.

The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.

The isotope decay of an atom is given by the equation:

ln [A] = -kt + ln [A]₀

The rate constant, k is:

k = ln 2 / Half life

k = ln 2 / 4.96 x 10³

k = 1.40 × 10⁻⁴ s⁻¹

t = 1.98 x 10⁴

[A]₀ = 3.21 x 10¹⁷

ln [A] = -1.40 × 10⁻⁴  ×  1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538

[A] = 2.02 x 10¹⁶ nuclei

Thus the correct option is C.

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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:

The carbon rod in batteries is used as an inert electrode

Explanation:

A battery is considered as a power source that consists of one or more electrochemical cells having an external connections to provide power to the electrical devices such as the lights, bulbs, fans, mobile phones, etc.

It contains a positive terminal and a negative terminal.

The carbon rod in the battery does not help in the electrochemical reactions. It acts as an inert electrode and helps to flow the electrons only.

Thus the true statement is :

The carbon rod in batteries is used as an inert electrode.

I recommend that they chill out. After that, they can do a web search on the phrase "human rights." They will learn that it describes each particular person's political objectives, at least those who claim to be morally superior to everyone else.

Answer:

17 seconds

Explanation:

Given that:

The mass attached to the spring (m) = 0.30 kg

The spring constant (k) =  2.00 N/m

The damping constant (b) = 0.025 kg/s

The initial distance [tex]x_o[/tex] = 5.0 cm

The initial final amplitude [tex]A_f[/tex] = 2.5 cm and not 2.5 m, please note the mistake, if it is 2.5 m, our time taken will be -93.7 sec, and we do not want a negative time value.

To start with the angular frequency damping using the formula:

[tex]\omega_{\gamma}= \dfrac{b}{2m}[/tex]

[tex]\omega_{\gamma}= \dfrac{0.025 \ kg/s}{2(0.3 \ kg)}[/tex]

[tex]\omega_{\gamma}=4.167 \times 10^{-2} \ s^{-1}[/tex]

In the absence of damping, the angular frequency is:

[tex]\omega_o = \sqrt{\dfrac{k}{m}}[/tex]

[tex]\omega_o = \sqrt{\dfrac{2 \ N/m}{0.3 kg}} \\\\\omega_o = 2.581 \ s^{-1}[/tex]

The initial amplitude oscillation can be computed by using the formula:

[tex]A_i = e^{-\omega_{\gamma}t} x_o \sqrt{\dfrac{\omega_o^2}{\omega_o^2-\omega_f^2}}[/tex]

[tex]A_i = e^{-\omega_{\gamma}0} (5.0 \ cm) \sqrt{\dfrac{2.581^2}{2.581^2-(4.167*10^{-2})^2}}[/tex]

[tex]A_i = 5.0006 \ cm \\ \\ A_i = 5.001 \ cm[/tex]

The final amplitude, as well as the initial amplitude, can be illustrated by using the relation:

[tex]A_f = e^{-\omega_{\gamma}t}A_i\\ \\ e^{-\omega_{\gamma}t} = \dfrac{2. 5 \ cm}{5.001 cm}\\ \\ = 0.4999\\ \\ \implies -\omega_{\gamma}t_f = \mathsf{In (0.4999)} \\ \\ t_f = \dfrac{\mathsf{-In (0.4999)}}{4.167*10^{-2} \ s^{-1}} \\ \\[/tex]

[tex]t_f = 16.64 \ sec \\ \\ \mathbf{t_f \simeq 17 sec}[/tex]

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

The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.

When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.

For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.

In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.

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By George, you've nailed it, Stacy !

That's a fact, uh huh.

Truer words were never written.

Your statement is one of unquestionable veracity.

The pure truthiness of it cannot be denied.

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