what (non living things) can symbolize the poem of invictus?
Provide a brief explanation why you chose this object ...

a. The Inverse Square Law states that the intensity or strength of a physical quantity decreases with the square of the distance from its source.

b. The values of E and r in the given circuit are: E = 4V (e.m.f. of the cell) and r = ∞

a. In the context of electrostatics, the Inverse Square Law describes the relationship between the electric field strength and the distance from a charged particle.

b. For the first scenario (V = 4V and I = 1A), we have:

4V = 1A * R1,

R1 = 4Ω.

For the second scenario (V = 2V and I = 2.5A), we have:

2V = 2.5A * R2,

R2 = 0.8Ω.

Total resistance in the circuit (when ammeter reads zero current) is given by:

= R1 + R2 + r.

Since the ammeter reads zero current, we have:

E = I * R_total,

4V = 0A * R_total,

R_total = ∞ (infinity).

Therefore, we can conclude that the internal resistance r of the cell is infinite (or very high compared to the resistances in the circuit).

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Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency \(f = ( \dfrac{1}{2 \pi}) \omega\)

where;

\(\omega = \sqrt{\dfrac{k}{m} }\)

Then;

\(f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )\)

However;

\(k = \dfrac{F}{x}\) and;

mass \(m = m_{car } + m_{person}\)

\(f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )\)

\(f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )\)

where;

\(F = m_{person}g\)

Then;

\(f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )\)

replacing the values;

\(f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )\)

\(\mathbf{f = 0.442 \ Hz}\)

Answer:

The last one - each  vector pointing towards the center of the circle must be the same length for uniform circular motion

The magnetic field in the cyclotron, is 0.292 T.

Angular velocity, ω = 2.8 x 10⁷rev/s

Charge of a proton, q = 1.6 x 10⁻¹⁹C

Mass of a proton, m = 1.67 x 10⁻²⁷kg

A cyclotron is a device that strongly accelerates the charge on charged particles or ions. The cyclotron amplifies the energy of the charged particles through the application of both magnetic and electric fields.

The expression for the angular velocity of the cyclotron is given by,

ω = qB/m

Therefore, the magnetic field in the cyclotron,

B = ωm/q

B = 2.8 x 10⁷x 1.67 x 10⁻²⁷/1.6 x 10⁻¹⁹

B = 0.292 T

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The linear speed will be the insect be 0.5759 meter/second carried collision with the near stationary photograph.

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

Given that an insect lands 0.1m from the center of the turn table.

Rotational speed of the turn table = 55 rev/min

= (55×2π/60) rad/second

= 5.759 rad/second.

Hence, the speed of the insect be = Rotational speed × length

= 5.759 rad/second × 0.1 M.

= 0.5759 meter/second.

Therefore, the speed of the insect be 0.5759 meter/second.

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Answer:

The distance between Jupiter and the Sun is 5.2 AU.

The answer would be A: 5.2 Au away from the sun (AU is the unit that is the same length as the distance between Earth and the Sun)

Answer:

i = 4.9 A

Explanation:

The expression for the magnetic force in a wire carrying a current is

          F = i L x B

bold letters indicate vectors.

The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition

            F - W = 0

            i L B = m g

They indicate the linear density of the cable λ = 0.2 kg / m

           λ = m / L

           m = λ L

we substitute

           i B = λ g

           i = \(\frac{ \lambda \ g}{B}\)

let's calculate

          i = 0.2 9.8 / 0.4

          i = 4.9 A

Complete Question

1 a A little girl pushes a 5.0 kg toy baby stroller at constant speed 7.0 m across the floor. She pushes on the handle with a force of 40 N at an angle of 30o with the horizontal.  How much work is done by the girl on the wagon?

1b  A farmhand pushes 20 ㎏ bale of hay 4m across the floor of the barn if she exerts a horizontal force of 60 N on the hay, how much work is done? (5 pts)

All parts are 4 points each

Answer:

1a

  \(W = 242.5 \ J\)

1b

    \(W = 240 \ J\)

Explanation:

Considering question a

From the question we are told that

       The mass of the toy baby stroller is  \(m = 5.0 \ kg\)

        The distance covered is    \(d = 7.0\ m\)

          The force the girl applies on the handle  is  \(F = 40 \ N\)

          The angle at which this force is applied is  \(\theta = 30^o\)

Generally the workdone is mathematically represented as

          \(W = F_x * d\)

Here  \(F_x\) is the force along the horizontal axis , this is mathematically represented as

           \(F_x = F cos (\theta )\)

=>         \(F_x = 40 * cos(30 )\)

=>         \(F_x = 34.64 \ N\)

So

              \(W = 34.64 * 7\)

=>           \(W = 242.5 \ J\)

Considering question b

From the question we are told that

       The mass of the toy baby stroller is  \(m = 20 \ kg\)

        The distance covered is    \(d = 4 \ m\)

          The force the girl applies on the handle  is  \(F = 60 \ N\)      

Generally the workdone is mathematically represented as

          \(W = F * d\)

=>     \(W = 60 * 4\)

=>     \(W = 240 \ J\)

The current in the circuit is 2.5 A.

Voltage across the wire AX, V = 4 - 2 = 2V

Resistance per unit length of wire AX, R/l = 2 Ω/m

Length of the wire, l = 0.4 m

Resistance of the wire,

R = R/l x R

R = 2 x 0.4 = 0.8 Ω

According to Ohm's law, the current in the wire AX,

I = V/R

I = 2/0.8

I = 2.5 A

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To better understand crash dynamics we have to look at "the laws of motion."

The laws of motion

The laws of motion were introduced by Sir Isaac Newton in 1687 in his book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), which defined the laws of motion, or three fundamental laws that govern the movement of bodies. The laws of motion, according to Newton, govern the motion of an object or a system of objects that interact.

It defines the concepts of force and mass, and the fundamental dynamics of motion.The following are the laws of motion:Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. The velocity of an object changes proportional to the force applied to it, and the acceleration of an object is proportional to both its force and its mass. For every action, there is an equal and opposite reaction.

Therefore, these laws are necessary to fully grasp crash dynamics because they explain how objects respond to outside forces that cause them to accelerate or decelerate.

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The rank of the forces in order of increasing magnitude is Fg < FB < FA.

option D is the correct answer.

The net force on an elevator moving upwards is determined by the force of gravity acting downwards and the normal force of the elevator acting upwards.

That is, the two forces acting on a person when he is moving in an elevator are:

When the two forces are of equal magnitude, the elevator will be static or moving with constant velocity.

When the magnitude of the two force are unequal, then the elevator will be accelerating upward or downward.

Since the elevator is moving upwards, it implies that the normal force is greater than the force of gravity acting downwards.

The box at the bottom will feel much heavier due to the weight of box and gravity acting downwards.

FA = FB + Fg

Thus, the force exerted on box A is the greatest, followed by the force on box B and then, the smallest is force of gravity.

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''The ability to use an applied force in order to make an object move'' is the  scientific definition of energy that relates it to work.

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Answer:

Relative density is the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.

Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

Answer: B) Methyl alcohol requires less than half as much energy per kg to evaporate than water does

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 kilo gram of liquid into its vapor state without change in its temperature.

Heat of vaporization is more for water than for methyl alcohol which means more heat is required to convert from liquid to vapour form.

As the Heat of vaporization for methyl alcohol  (1100) is almost half as that of Heat of vaporization for water (2257) , it means  Methyl alcohol requires less than half as much energy per kg to evaporate than water does.

Explanation:

Using the equation of motion S = ut+1/2gt² where;

S is the height of the table = 0.60m

u is the initial velocity = 0m/s

g is the acceleration due to gravity = 9.81m/s²

t is the time taken

Substitute

0.6 = 0+1/2(9.81)t²

0.6 = 4.905t²

t² = 0.6/4.905

t² = 0.1223

t = √0.1223

t = 0.3497

t≈0.35secs

Hence it will take 0.35seconds for a ball launched off the edge of these tables to reach  the floor

The final volume of the gas, when the pressure changes from 80.8 kPa to 101.3 kPa at constant temperature, is approximately 652.9 cm³.

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's Law can be represented by the equation: P₁V₁ = P₂V₂

Where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

Given:

Initial volume, V₁ = 817 cm³

Initial pressure, P₁ = 80.8 kPa

Final pressure, P₂ = 101.3 kPa

We need to find the final volume, V₂.

Using Boyle's Law equation, we can rearrange it to solve for V₂:

V₂ = (P₁V₁) / P₂

Plugging in the given values:

V₂ = (80.8 kPa * 817 cm³) / 101.3 kPa

Simplifying the expression:

V₂ ≈ 652.9 cm³

Therefore, the final volume of the gas, when the pressure changes from 80.8 kPa to 101.3 kPa at constant temperature, is approximately 652.9 cm³.

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The gauge pressure at point 2 is 98100 Pa or 9.81 x\(10^4\) Pa, which is equivalent to 6.97 x\(10^4\) Pa when rounded to two significant figures.

Step 1: Identification of the given data:

- Elevation at point 1 (h1) = 10.0 m

- Elevation at points 2 and 3 (h2 = h3) = 2.00 m

- Cross-sectional area at point 2 (A2) = 0.0480 \(m^2\)

- Cross-sectional area at point 3 (A3) = 0.0160 \(m^2\)

Step 2: Determination of the discharge rate:

As mentioned earlier, the discharge rate (Q) is given by Q = A2 * v2, and since the velocity at point 2 (v2) is negligible, the discharge rate will be 0.

Therefore, the discharge rate is 0 cubic meters per second.

Step 3: Determination of the gauge pressure at point 2:

To find the gauge pressure at point 2, we'll use Bernoulli's equation:

P1 + (1/2)ρ\(v1^2\) + ρgh1 = P2 + (1/2)ρ\(v2^2\) + ρgh2

Since the velocity at point 2 (v2) is negligible, the term (1/2)ρ\(v2^2\) can be ignored.

The equation simplifies to:

Patm + ρgh1 = P2 + ρgh2

We want to find the gauge pressure at point 2, so we'll subtract the atmospheric pressure (Patm) from P2:

\(P_g_a_u_g_e\) = P2 - Patm

Now let's substitute the given values into the equation:

\(P_g_a_u_g_e\) = (Patm + ρgh1) - Patm

\(P_g_a_u_g_e\) = ρgh1

Plugging in the values:

\(P_g_a_u_g_e\) = (1000 kg/m^3) * (9.81 \(m/s^2\)) * (10.0 m)

\(P_g_a_u_g_e\) = 98100 Pa

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Explanation:

It is given that,

The position of a particle is given by :

\(r(t)=(3t^2i+5j-6tk)\ m\)

(a) Velocity of a particle is given by :

\(v=\dfrac{dr(t)}{dt}\)

Putting values,

\(v=\dfrac{d}{dt}(3t^2+5-6t)\\\\v=(6ti-6k)\ m/s\)

The acceleration of the particle is given by :

\(a=\dfrac{dv}{dt}\\\\a=\dfrac{d}{dt}(6t-6)\\\\a=6i\ m/s^2\)

(b) At t = 0,

Velocity, v = 6k m/s

Acceleration, a = 6i m/s²

Answer:

ion channel

Explanation:

discussion of ion channel structure and function, followed by the classical description of the action potential and the ... Resistance and Conductance depend on the size and shape of the object that you are passing ... Intracellular and extracellular ion concentrations must ... applied to essentially any cell type.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

C) The book's inertia carried it forward.

When the bus stops suddenly, the book tends to remain in motion due to its inertia. The book was at rest on the seat of the bus, and when the bus stopped suddenly, the book continued moving forward with the same speed and direction it had before the bus stopped. As a result, the book slid off the seat and onto the floor.

Answer: As i know, please see the Explanation

Explanation:The cell contains 3% salt, so we can safely assume that it contains 100%−3%=97% water.

To calculate the mean add up the numbers you want to add and divide them by how many

Answer:

  e) about 10 m/s

Explanation:

Acceleration due to gravity is nominally* 9.8 m/s². That means the change in velocity each second is ...

  (9.8 m/s²)(1 s) = 9.8 m/s ≈ 10 m/s

_____

* This is the value expected to be used in the solution of many math and physics problems. The standard value of 'g' on Earth is defined as 9.80665 m/s². It varies from place to place and with altitude. At any given place, it may also vary with time as a result of changes in mass distribution within the Earth.

Answer:

yes

Explanation:

because it will help find who did the crime and it can also open jobs and opportunities for the people who likes sicnenae

Answer:

 If the inside light is not too bright, a person on the outside could not see inside because the light reflected from the outside is too bright.

Explanation:

some of the light bounces (or reflects) off the glass. The rest of the light keeps going through the glass object, but the light is bent (or refracted) as it moves from the air to the glass