Need help! Need help! Need help! Need help! Need help! Need help!

Answer:

The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".

Explanation:

As the outer spherical shell is conducting, so there is no electric field in side from

⇒ [tex]r_b_1 < r < r_b_2[/tex].

So the electric potential at all points inside the conducting shell that from

⇒ [tex]r_b_1<r<r_b_2[/tex]  

and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:

⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]

Thus the above is the right solution.

Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

Answer:

 f = v / 4L

the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Explanation:

In wind instruments the wave speed must meet

          v = λ f

          λ = v / f

from v is the speed of sound that depends on the temperature

          v = v₀ [tex]\sqrt{1+ \frac{T [C]}{273} }[/tex]

where I saw the speed of sound at 0ºC v₀ = 331 m/s  the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation

            (F -32) 5/9 = C

            76ºF = 24.4ºC

            45ºF = 7.2ºC

With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside

at T₁ = 24ºC               v₁ = 342.9 m / s

at T₂ = 7ºC                 v₂ = 339.7 m / s

To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change

              λ / 4 = L

              λ= 4L

              v / f = 4L

              f = v / 4L

Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Answer:

b

Explanation:

Answer:

Explanation:

From the question we are told that:

Number of turns [tex]N=10[/tex]

Area [tex]a=0.23m^2[/tex]

Magnetic field [tex]B=0.947T[/tex]

Generally the equation for maximum flux is mathematically given by

 [tex]\phi=NBa[/tex]

 [tex]\phi=10*0.047*0.23[/tex]

 [tex]\phi=0.1081wbi[/tex]

Therefore induced emf

 [tex]e= \frac{d\phi}{dt}[/tex]

Since

 [tex]t=0[/tex]

Therefore

 [tex]e=0[/tex]

Explanation:

Given:

[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]

[tex]v_1[/tex] = 40 cm/s

[tex]v_2[/tex] = 90 cm/s

[tex]A_2[/tex] = ?

a) The continuity equation is given by

[tex]A_1v_1 = A_2v_2[/tex]

Solving for [tex]A_2[/tex],

[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]

[tex]= 2\:\text{cm}^2[/tex]

b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of

[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]

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

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:

Option (c).

Explanation:

Let the mass of each cart is m and the force is F.

Time for cart A is 2t and for cart B is t.

Work done is given by the

W= force x displacement

As the distance is given by

S= u t +0.5 at^2

So, when the time is doubled the distance is four times.

So, WA = F x 4 S

WB = F x S

WA= 4 WB

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.

We have that the angle is

[tex]\theta=32.53[/tex]

From the Question we are told that

E⃗ =5.00×105ı^N/C

Generally the equation for Tension   is mathematically given

[tex]W=Tcos\theta[/tex]

Where

[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]

[tex]\theta=32.53[/tex]

For more information on this visit

https://brainly.com/question/20746649?referrer=searchResults

Answer:

The spring constant of the spring is 10.3 N/m.

Explanation:

Given that,

Mass of a bungee jumper, m = 59 kg

The period of oscillation, T = 0.25 min = 15 sec

We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :

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

Where

k is the spring constant

[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]

So, the spring constant of the spring is 10.3 N/m.

Answer: When a pathogen enters their colony, ants change their behavior to avoid the outbreak of disease. In this way, they protect the queen, brood and young workers from becoming ill. These results, from a study carried out in collaboration between the groups of Sylvia Cremer at the Institute of Science and Technology Austria (IST Austria) and of Laurent Keller at the University of Lausanne, are published today in the journal Science.

Explanation: search for it.

Answer:

4.926 L Y 7.389 L

Explanation:

first you calculate the tank volume

V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]

then you convert to liters

[tex]12315 cm^{3}[/tex] = 12.315 l

then you calculate the liters of water

2/5(12.35 l) = 4.926 l

finally we calculate the amount without water

12.315 l - 4.926 l = 7.389 l

HERE IS MORE INFORMATION ON THE SUBJECT. THEY REMOVED THE

ENGLISH SITE BUT YOU CAN USE TRANSLATOR

LINK: https://gscourses.thinkific.com/courses/fisicai

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:

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.

Explanation:

pls write the full question

Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

Answer:

The correct option is (e) "none of the above".

Explanation:

Given that,

A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.

If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.

We know that the time period can be calculated as :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

Where

l is the length

g is acceleration due to gravity

It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.

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:

none of those are right, its technically 96.5%. so i would say 97% is your best bet because thats closest and it just rounds up :)

Explanation:

[tex]Hello[/tex] [tex]There![/tex]

[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]

The answer is...

C. Work and time.

[tex]HopeThisHelps!![/tex]

[tex]AnimeVines[/tex]

Answer:

Distance traveled = 3 * 10E8 * 17.2 * 10E-9 = 5.16 m

.81 / 3 * 10E8 = 2.7 * 10E-9    normal time thru glass

(20.8 - 17.2) * E10-9 = 3.6 * 10E-9   additional time due to glass

c tg = c n ta      where tg and ta are the times spent in glass and air

(Note you can also write Va = n Vg  or  D / ta = n D / tg)

n = tg / ta = 3.6 / 2.7 = 1.33 the index of refraction of the glass

Wavelength (air) = Wavelength (glass) * n

Wavelenght = 560 nm / 1.33 = 421 nm

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:

https://brainly.com/question/28790459

#SPJ5

Answer:

Current

I think The choose (B)

B. Electric current

Answer:

No

Explanation:

You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.

Answer:

 E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]

Explanation:

For this exercise we use that the electric field is a vector, so the resulting field is

          E_total = E₁ + E₂                      (1)

since the field has the same direction in the space between the planes

Let's use Gauss's law for the electric field of each plate

Let's use a Gaussian surface that is a cylinder with the base parallel to the plate, therefore the normal to the surface and the field lines are parallel and the angle is zero so cos 0 = 1

          Ф  = ∫ .dA = [tex]q_{int}[/tex] /ε₀

if we assume that the charge is uniformly distributed on the plate we can define a charge density

         σ = q_{int} A

as the field exists on both sides of the plate on the inside

          E A = A σ / 2ε₀

          E = σ / 2ε₀

we substitute in equation 1

         E = σ /ε₀

for the complete plate

         σ = Q / A = Q / L²

we substitute

         E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]

[tex]\implies F_1 < F_2[/tex]

[tex] \implies F_{net} = F_2 - F1[/tex]

[tex]\implies F_{net} = 42 -15[/tex]

[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]

The net force on the 4.0 kg box is 27 N towards EAST.