A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J

Answer:

Explanation:

For frequencies n generated in a string , the expression is as follows

n = 1 /2L√ ( T/m )

n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.

If T and m are constant , then

n x L = constant , hence n is inversely proportional to L or length of string.

Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times

Initial length of string is 65 cm .

x1 = 65 x 1 / 1.75 = 37.14 cm

x2 = 37.14 x 1/ 1.75 = 21.22 cm

x3 = 21.22 x 1 / 1.75 = 12.12 cm

x4= 12.12 x 1 / 1.75 = 6.92 cm

x5 = 6.92 x 1 / 1.75 = 3.95 cm

x6 = 3.95 x 1 / 1.75 = 2.25 cm

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n ≈ 14.0 g/mol

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

Answer:

no, all substances doesnot conduct heat

Answer:

No, all substances do not conduct heat easily because it depends on the nature of the substance. Some are good conductors of heat and some are bad. Therefore, it depends on their characteristics and their ability to conduct heat.

The bad conductors of heat are water, air, plastic, wood, etc.

Gold, Silver, Copper, Aluminium, Iron, etc. are good heat conductors as well as electrical conductors.

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

Answer:

[tex]\omega=109.67\ rad/s[/tex]

Explanation:

Given that,

The speed of the ball, u = 34 m/s

The ball is 0.310 m from the elbow joint.

We need to find the angular velocity (in rad/s) of the forearm.

We know that,

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]

So, the required angular velocity of the forearm is 109.67 rad/s.

Answer:

u can ask it to the person who give ot to u i dont no

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

Answer:

enchantment table language

Explanation:

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m[tex]_g[/tex]g = 0

where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m[tex]_g[/tex]g = 0

solve for m[tex]_g[/tex]

m[tex]_g[/tex] = ( Vρg - Mg ) / g

m[tex]_g[/tex] =  Vρg/g - Mg/g

m[tex]_g[/tex] =  ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m[tex]_g[/tex] / μ

where μ is the molecular mass of air

so

PV = (m[tex]_g[/tex]/μ)RT

solve for T

μPV = m[tex]_g[/tex]RT

T = μPV / m[tex]_g[/tex]R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T =  1405920 / 1979.442

T =  710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

The temperature required for the air to be heated is 710.26 K.

Given data:

The mass of a hot air-balloon is, m = 381 kg.

The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].

The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].

The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].

The condition where the hot air balloon is just about to take off is as follows:

[tex]F-mg - m'g =0[/tex]

Here,

m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,

[tex]F = V \times \rho \times g[/tex]

Solving as,

[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]

Now, apply the ideal gas law as,

PV = nRT

here, R is the universal gas constant and n is the number of moles and its value is,

[tex]n=\dfrac{m'}{M}[/tex]

M is the molecular mass of gas. Solving as,

[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]

Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,

[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]

Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.

Learn more about the ideal gas equation here:

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Answer:

I think it is of science is it true na i knew it bro dont take tension

Answer:

b. approaches zero.

Explanation:

The phenomenon is known as length contraction.

Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.

let the length of the train = L

Let the length observed when the train is in motion = L₀

Apply Einstein's special theory of relativity;

[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]

from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)

As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L).  (L₀ < L)

Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero.  (L₀ = 0)

Thus, the length of this object, as measured by a stationary observer approaches zero

Answer:

The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic? is your ansewer dont take tension

The string will be 108 cm long when it is stretched to the point where it becomes plastic.

Elasticity in physics and materials science refers to a body's capacity to withstand a force that causes distortion and to recover its original dimensions once the force has been withdrawn.

When sufficient loads are applied, solid objects will deform; if the material is elastic, the object will return to its original size and shape after the weights have been removed. Unlike plasticity, which prevents this from happening and causes the item to stay deformed,

Given parameters:

The elastic extensibility of a piece of string is 0.08.

The string is 100 cm long.

Hence,  it becomes plastic, after  it is stretched up to = 100 × 0.08 cm = 8 cm.  The string will be 108 cm long.

Learn more about elasticity here:

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Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s

Answer:

Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.

Explanation:

Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.

Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.

Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.

Answer:

Option (2) is correct.

The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Explanation:

An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.

The given problem is based on the concept and fundamentals of electromagnetic waves.  The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.

In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.

Learn more about the electromagnetic waves here:

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Nearly four hours every day, doing little to no physical activity.

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

Answer:

T₁ = 232.5 ºC

Explanation:

For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law

           V = i R

            R = V / i

i₀ = 1.28 A

            R₀ = 120 / 1.28

            R₀ = 93.75 ohm

i₁ = 1.229 A

             R₁ = 120 / 1.229

             R₁ = 97.64 or

Resistance in a metal is linear with temperature

            ΔR = α R₀ ΔT

where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹

            ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]

            ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]

            ΔT = 2,075 10² C

            ΔT = T₁-T₀ = 2,075 10²

            T₁ = T₀ + 207.5

             T₁ = 25+ 207.5

             T₁ = 232.5 ºC

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022

Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.