Explain why water, with its high specific heat capacity, is utilized for heating systems such as hot-water radiators.

Answer:

The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:

1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.

2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.

3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.

Explanation:

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω ,  R₃ = 1.72 10³ Ω ,  R₄ = 5.4 10³ Ω    1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

   V = i R

let the voltage

1st resistance

         V = i R

         R₁ = V / i

         R₁ = 14.7 / 1 10⁻³

         R₁ = 14.7 10³ Ω

power is

        P = V i

        P = 14.7 1 10⁻³

        P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

          R₂ = 8.18 / 1 10⁻³

          R₂ = 8.18 10³ Ω

Power

          P = 8.18 1 10⁻³

          P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

          V₁ + V₂ + V₃ = 24.6

          V₃ = 24.6 - V₁ -V₂

          V₃ = 24.6 - 14.7 - 8.18

          V₃ = 1.72 V

          R₃ = 1.72 / 1 10⁻³

          R₃ = 1.72 10³ Ω

power

          P = Vi

          P = 1.72 10⁻³

          P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

           30 = V₁ + V₂ + V₃ + V₄

           V₄ = 30 - V₁ -V₂ -V₃

           V₄ = 30 -14.7 - 8.18 - 1.72

           V₄ = 5.4 V

          R₄ = 5.4 / 1 10⁻³

          R₄ = 5.4 10³ Ω

Power

         P = V i

         P = 5.4 10⁻³

         P = 0.0054 W

⅛ W resistance

The values ​​of these resistance are commercially

Let's check the consumption of the circuit

  R_total = R₁ + R₂ + R₃ + R₄

  R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

   R_total = 30 10³

the current circulating in the circuit is

     i = V / R_total

     i = 30/30 10³

     i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

Answer:

4.5x 10^ -9m

Explanation:

See attached file

Answer:

The radius is  [tex]r_{min} = 0.00226 \ m[/tex]

Explanation:

   From the question we are told that

      The  elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]

      The length is  [tex]L = 4.0 \ m[/tex]

      The weight of the commercial sign is    [tex]F_s = 8000 \ N[/tex]

       The maximum extension of the wire is  [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]

Generally the elastic limit of an alloy (stress) is is mathematically represented as

            [tex]\sigma = \frac{ F_s }{ A }[/tex]

Where A is the cross-sectional area of the wire which is mathematically represented as

         [tex]A = \pi r^2[/tex]

here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign

So

          [tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]

=>       [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]

substituting values

             [tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]

           [tex]r_{min} = 0.00226 \ m[/tex]

Answer:

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Explanation:

Given:

Frequency = 91.5 MHz

Power = 11.5 Kw = 11,500 J/s

Find:

Number of photons emit per second

Computation:

Total energy with frequency (E) = hf

Total energy with frequency (E) = 6.626×10⁻³⁴  × 91.5×10⁶

Total energy with frequency (E) = 6.06×10⁻²⁶ J

Number of photons emit per second = 11,500 / 6.06×10⁻²⁶

Number of photons emit per second = 1897.689 × 10²⁶

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]

[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]

Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]

Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Answer:

cost_cost = $ 96

Explanation:

In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system

         performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)

         perfomance= 8,927 km / l

now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km

          #_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters

Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos

            cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638

             cost_gasoline = #_gasoline   cost_litro

             cost_gasoline = 225.72   0.638

             cost_gasoline = $ 144

This cost is for the one way trip, the total round trip cost is

             cost_total = 2 cost_gasoline

             cost_total = $ 288

Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is

                cost_person = total_cost / #_people

                cost_person = 288/3

                cost_cost = $ 96

Answer:

10.8rev

Explanation:

Using

Wf²-wf = 2 alpha x theta

0²- 56.36x56.36/ 2(-20.13) x theta

Theta = 68.09 rad

But 68.09/2π

>= 10.8 revolutions

Explanation:

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

1,860 g

Explanation:

In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R²   ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

g' = 1.09 m/s²

Answer:

Explanation:

Surface of earth,

[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]

free fall acceleration at height h,

[tex]g_h = \frac{GM}{(R+h)^2}[/tex]

where

G = gravitational constant

R = Radius of earth

M = mass of earth

therefore,

[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]

Where height h = 2R

[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]

For more information, visit

https://brainly.com/question/17162343

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

Answer:

8 seconds

160 meters

Explanation:

Given in the y direction:

Δy = 0 m

v₀ = 40 m/s

a = -10 m/s²

Find: t

Δy = v₀ t + ½ at²

0 m = (40 m/s) t + ½ (-10 m/s²) t²

0 = 40t − 5t²

0 = 5t (8 − t)

t = 0 or 8

Given in the x direction:

v₀ = 20 m/s

a = 0 m/s²

t = 8 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²

Δx = 160 m

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      [tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

Answer:

Explanation:

Formula for calculating the relationship between  the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:

[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]  where;

Vs and Vp are the emf in the secondary and primary coil respectively

Ns and Np are the number if turns in the secondary and primary coil respectively

Ip and Is are the currents in the secondary and primary coil respectively

Since the are all equal to each other, then we can equate any teo of the expression as shown;

[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

Given parameters

Np = 500-turns

Ns = 2000-turns

Is = 3.0Amp

Required

Current in the primary coil (Ip)

Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]

[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]

Hence the current in the primary coil is 12Amp

Answer:

The pressure is [tex]P = 583333 \ N/m^2[/tex]

Explanation:

From the question we are told that

  The area of the edge is  [tex]A = 0.72 cm^2 = 0.72 *10^{-4}\ m[/tex]

    The  force is [tex]F = 42 \ N[/tex]

The pressure is mathematically represented as

            [tex]P = \frac{F}{A}[/tex]

substituting values

           [tex]P = \frac{42}{0.72*10^{-4}}[/tex]

           [tex]P = 583333 \ N/m^2[/tex]

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

Answer:

0.273m/s

Explanation:

first find out the meaning of 0.90×10−4m3/s

literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s

1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?

cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²

so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²

they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s

Answer:

The projectile strikes the tower at a height of 354.824 meters.

Explanation:

The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:

Horizontal motion

[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]

Vertical motion

[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.

[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

The time spent for the projectile to strike the tower is obtained from first equation:

[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]

If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:

[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]

[tex]t \approx 7.071\,s[/tex]

Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])

[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]

[tex]y \approx 354.824\,m[/tex]

The projectile strikes the tower at a height of 354.824 meters.

Answer:

The  angular acceleration is  [tex]\alpha = 3.235 \ rad/s ^2[/tex]

Explanation:

From the question we are told that

    The moment of inertia is  [tex]I = 0.034\ kg \cdot m^2[/tex]

     The  net torque is  [tex]\tau = 0.11\ N \cdot m[/tex]

Generally the net torque is mathematically represented as

           [tex]\tau = I * \alpha[/tex]

Where [tex]\alpha[/tex] is the angular acceleration so  

        [tex]\alpha = \frac{\tau }{I}[/tex]

substituting values

         [tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]

        [tex]\alpha = 3.235 \ rad/s ^2[/tex]

Answer:

The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

[tex]v=\dfrac{c}{\mu}[/tex]

Put the value into the formula

[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]

[tex]v=1.06\times10^{8}\ m/s[/tex]

We need to calculate the time of short pulse of light beam

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]

[tex]t=2.37\times10^{-9}\ sec[/tex]

Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Answer:

Electric field is zero at point 4.73 m

Explanation:

Given:

Charge place = 50 cm  = 0.50 m

change q1 = 5 µC

change q2 = 4 µC

Computation:

electric field zero calculated by:

[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]

Where electric field is zero,

First distance = x

Second distance = (x-0.50)

So,

E1 = E2

[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]

[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]

x = 0.263 or x = 4.73

So,

Electric field is zero at point 4.73 m

Answer:

The maximum force that can be lifted by this system is  51,478.4 kg

Explanation:

Given;

maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²

diameter of the piston, d = 65 cm = 0.65 m

The maximum gauge pressure of the piston is given as;

[tex]Hp = \frac{F}{A}[/tex]

Where;

F is the maximum force of the piston

A is the area of the piston

[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]

F = Hp x A

F = 1.52 x 10⁶N/m² x 0.3319m²

F = 504488 N

Force is given as;

F = mg

m = F/g

m = 504488/9.8

m = 51,478.4 kg

Therefore, the maximum force that can be lifted by this system is  51,478.4 kg

Answer:

Number of turns of wire(N) = 3,036 turns (Approx)

Explanation:

Given:

Diameter = 13 Cm

emf = 5.6 v

Note:

The given question is incomplete, unknown information is as follow.

Magnetic field increases = 0.25 T in 1.8 (Second)

Find:

Number of turns of wire(N)

Computation:

radius (r) = 13 / 2 = 6.5 cm = 0.065 m

Area = πr²

Area = (22/7)(0.065)(0.065)

Area = 0.013278 m²

So,

emf = (N)(A)(dB / dt)

5.6 = (N)(0.013278)(0.25 / 1.8)

5.6 = (N)(0.013278)(0.1389)

N = 3,036.35899

Number of turns of wire(N) = 3,036 turns (Approx)

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

Answer:

1. Berenice  = 0.67 D

2. Avishka  = 1.50 D

3. Francesca  = 2.00 D

4. Edouard = 2.75 D

Explanation:

The farsighted people are those with near point greater than 25 cm.

They include Avishka, Berenice, Edouard and Francisca.

A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative

From the lens formula, 1/f = 1/v  1/u; but v is negative

Therefore, 1/f = 1/u - 1/v

But, power of a lens = 1/f in meters.

Therefore,  P =  1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

P = 1/ 0.25 - 1/0.4 = 1.5 D

For Berenice, u = 0.25 m, v = 0.3 m

P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v = 0.80 m

P = 1/0.25 - 1/0.80 =2.75 D

For Francesca, u = 0.25 m, v = 0.50 m

P = 1/0.25 - 1/0.5 = 2.0 D

When The farsighted people are those with a near point greater than 25cm.

Berenice is = 0.67 D

Avishka is = 1.50 D

Francesca is = 2.00 D

Edouard is = 2.75 D

When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.

Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.

Also, From the lens formula, 1/f = 1/v 1/u; but v is negative

Thus, 1/f = 1/u - 1/v

But, when the power of a lens = 1/f in meters.

Thus, P = 1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

After that, P = 1/ 0.25 - 1/0.4 = 1.5 D

Then, For Berenice, u = 0.25 m, v = 0.3 m

Now, P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v is = 0.80 m

Then, P = 1/0.25 - 1/0.80 is =2.75 D

Now, For Francesca, u = 0.25 m, v is = 0.50 m

Therefore, P = 1/0.25 - 1/0.5 = 2.0 D

Find more information about Hyperopic Vision here:

https://brainly.com/question/25676535

Answer:

in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.

Answer:

At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.

Explanation:

The number of particles (N)  counts are inversely proportional to the distance between the source and the detector.  

By using the below formula we can find the number of counts.

[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]

The number of count per second, when the distance is 3 meters.

[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]

Number of count per second when the distance is 10 meters.

[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

Answer:

λ = 667.85 nm

Explanation:

Let f be the frequency detected by the observer

Let v be the speed at which the observer is moving.

Now, when the direction at which the observer is moving is away from the source, we have the frequency as;

f = f_o√((1 - β)/(1 + β))

From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;

λ = λ_o√((1 + β)/(1 - β))

Where, β = Hr/c

H is hubbles constant which has a value of 0.0218 m/s • ly

c is speed of light = 3 × 10^(8) m/s

r is given as 2.40 x 10^(8) ly

Thus,

β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))

β = 0.01744

Since we are given λ_o = 656.3 nm

Then;

λ = 656.3√((1 + 0.01744)/(1 - 0.01744))

λ = 667.85 nm

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

[tex]E = Blv[/tex]

where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

[tex]v[/tex] is the velocity of the fluid through the field = ?

[tex]B[/tex] is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]

2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]

[tex]v[/tex] = 2.88/1.44 = 2 m/s