Question 5 of 10
Which of the following best describes a property of water?
O A. Dissolving very few substances
B. Remaining liquid over a narrow range of temperatures
C. Having low surface tension
D. Requiring a lot of energy to change temperature

Answer:

a) the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Explanation:

Given the data in the question;

separation  between two slits  d = 0.36 mm = 0.00036 m

Separation between two adjacent fringes β = 5.5 mm = 0.0055 m

Distance of screen from slits D = 1.6 m

n = 2

a) the wavelength the light emits;

Using the formula;

β = (nD/d)λ

To find wavelength, we make λ the subject of formula;

βd = nDλ

λ = βd / nD

so we substitute

λ = ( 0.0055 m × 0.00036 m ) / ( 2 × 1.6 m )

λ = 0.00000198 / 3.2

λ = 6.1875 × 10⁻⁷ m

Therefore, the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes;

To find the distance between the two n = 1 dark fringes, we use the following formula;

y[tex]_m[/tex] = 2nλD / d

given that n = 1, we substitute

y[tex]_m[/tex] = ( 2 × 1 × ( 6.1875 × 10⁻⁷ m ) × 1.6 m ) / 0.00036 m

y[tex]_m[/tex] = 0.00000198 / 0.00036

y[tex]_m[/tex] = 0.0055 m

y[tex]_m[/tex] = 5.5 × 10⁻³ m

Therefore, the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Answer:

lungs. hopes this helps

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.

It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex]    and   [tex]$\phi_{B,f} = 0$[/tex]

Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]

  [tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]

  = 30.27 V

Therefore, the emf generated is 30.27 V.

Hi there!

[tex]\large\boxed{1560 kgm/s}[/tex]

Recall that:

P = m · v, where:

P = momentum

m = mass (kg)

v = velocity (m/s)

Thus:

P = 156 · 10

P = 1560 kgm/s

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

[tex]tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0[/tex]

The drag force D = Wsinθ

[tex]\dfrac{1}{2}C_v \rho AV^2 = W sin \theta[/tex]

where;

[tex]\rho = 1.23 \ kg/m^3[/tex]

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient [tex]C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}[/tex]

[tex]C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}[/tex]

[tex]C_D =0.9378[/tex]

Answer:  I think that its b, they would have difficulty sprinting in a race

Explanation:

a) v = 1.94 × 10^5 m/s

b) Ms = 2.09 × 10^24 kg

Explanation:

Given:

m = 0.257M (M = mass of earth = 5.972×10^24 kg)

= 1.535×10^24 kg

r = 7.18×10^9 m

T = 2.7 days × (24 hr/1 day) × (3600 s/1 hr)

= 2.3328×10^5 s

a) To find the orbital speed of the planet, we need to find the circumference of the planet's orbit first:

C = 2×(pi)×r

= 2(3.14)(7.18×10^9m)

= 4.51×10^10 m

The orbital speed v is then given by

v = C/T

= (4.51×10^10 m)/(2.33×10^5 s)

= 1.94 × 10^5 m/s

b) We know that centripetal force Fc is given by

Fc = mv^2/r

where v = orbital speed

r = average orbital radius

m = mass of planet

We also know that the gravitational force FG between the star K2-116 and the planet is given by

FG = GmMs/r^2

where m = mass of planet

Ms = mass of star K2-116

r. = average orbital radius

G = universal gravitational constant

= 6.67 × 10^-11 m^3/kg-s^2

Equating Fc and FG together, we get

Fc = FG

mv^2/r = GmMs/r^2

Note that m and one of the r's get cancelled out so we are left with

v^2 = GMs/r

Solving for the mass of the star Ms, we get

Ms = rv^2/G

=(7.18 × 10^9 m)(1.94 × 10^5 m/s)^2/(6.67 × 10^-11 m^3/kg-^2)

= 2.09 × 10^24 kg

Answer:

0.0872

Explanation:

the solution to the problem can be found in the attachment section. Please go through and feel free to ask your doubts.

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

Solution :

The conditions for the maximum in the Young's experiment is :

d sin θ = m λ,     where m = 0, 1, 2, 3, .....

The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,

d sin θ =  λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]

Given : d = 100 λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]

[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]

 [tex]$=0.573^\circ$[/tex]

  = 0.01 rad

Answer:

a) size of the bright spot is proportional to the hollow size

b) as the size of the hole increases, the circular point decreases.

Explanation:

a) In this case the diameter of the hole is much greater than the wavelength, as the size of the hole is many orders greater than the wavelength we are in the part of geometric optics,

Consequently the size of the bright spot is proportional to the hollow size.

Consequently the size increases

b) in this case the hole diameter d = 100 10⁻⁶m and the wavelength that for the green color is lam = 500 nm  = 5 10⁻⁷ m

We see that angles are very small so the wavelength of the office is greater than the wavelength, but you can observe the effects of diffraction

          d sin θ = 1.22 m λ

the numerical constant appears by solving the equation in polar coorθdinates, because the hole is circular

the first zero occurs for m = 1

        sin θ = 1.22 λ / d

In these experiments the angles are small

        sin θ = θ

we substitute

       θ = 1.22 λ/ d

       θ = 1.22 500 10⁻⁹ / 100 10⁻⁶

       θ = 6.1 10⁻³

without the hole diameter increases by 20%            

       d’ = 1.2 d

we substitute

    θ'= 1.22 λ / d'

    θ’ = 1.22 λ /1.2 d

    θ‘= 1.22 λ /d  [tex]\frac{1}{1.22}[/tex]

    θ ’=  θ  0.83

    θ ’= 6.1 10⁻³ 0.83

    θ' = 5 10⁻³ rad

Therefore, the answer is that as the size of the hole increases, the circular point decreases.

Answer:

a)   w = 31.4 rad / s,  b)  a = 118.4 m / s²

Explanation:

a) let's reduce to the SI system

   w = 5 rev / s (2pi rad / 1 rev)

   w = 31.4 rad / s

b) the expression for the centripetal acceleration is

      a = v² / r

linear and angular variables are related

      v = w r

    we substitute

     a = w² r

     a = 31.4² 0.120

     a = 118.4 m / s²

Explanation:

F=w/s

F=40,000j/330m

F=121.212N

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

[tex]time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m[/tex]

Therefore, the distance between the submarine and the ocean floor is 11,250 m

c. a natural process

It is a natural process

Answer:

[tex]12.35\:\mathrm{m/s}[/tex]

Explanation:

We can use the work-energy theorem to solve this problem. The work-energy theorem states that the work done on an object will be equal to that object's change in kinetic energy. Thus, we have the following equation:

[tex]W=\Delta KE,\\W=\frac{1}{2}mv^2-0,\\122=\frac{1}{2}\cdot 1.6\cdot v^2,\\v^2=152.5,\\v\approx \boxed{12.35\:\mathrm{m/s}}[/tex]

Answer:

30 newtons

explanations

data given

mass=6kg

acceleration=5

f=m×a

6×5=30

Answer:

dino :)

Explanation:

Answer:

3 pls give me brainliest

Explanation:

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = [tex]\frac{x-x_1}{t}[/tex]

we substitute the values

             v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = [tex]\frac{6600 - x_1}{4}[/tex]

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = [tex]\frac{6600 -128 a}{4}[/tex]

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]

or,

    [tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]

On substituting the values, we get

      [tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]

      [tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)

On the Y-axis,

⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]

                        [tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]

                    [tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]

From equation 1, we get

           [tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]

                        [tex]T_1+3T_1=278.4 \ N[/tex]

                                [tex]4T_1=278.4 \ N[/tex]

                                  [tex]T_1=\frac{278.4}{4}[/tex]

                                       [tex]=69.6 \ N[/tex]  

Answer:

69.58

Explanation:

Complete question:

For a galvanic cell to generate an electric current flowing from anode to cathode, what must be true?

(a) Electrons flow from the anode to the cathode

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode

(c) Electrons flow from higher potential energy to lower potential energy

(d) All of the above are true.

Answer:

(d) All of the above are true.

Explanation:

A galvanic or Voltaic cell is a primary type of electrochemical cell that is used to generate electrical energy from the chemical reactions that take place in it.

It consists of a positive electrode (cathode) and a negative electrode (anode) for the movement of charges.

(a) Electrons flow from the anode to the cathode. TRUE

Anode is the negative electrode and for electron current, electrons flow from negative electrode to positive electrode.

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode. TRUE

Based on electron current flow.

(c) Electrons flow from higher potential energy to lower potential energy. TRUE

The driving force of the electron flow is the potential difference. Electrons must flow from higher potential to lower potential.

All the options are correct, so we select option "D"

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.

Given-

velocity of the ice skater is 12 m/ sec.

Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.

[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]

The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,

[tex]W_{f}=\bigtriangleup KE +0[/tex]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,

[tex]W_{f} =u_{f} mgx[/tex]

Here,  [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .

Equate the value of kinetic energy and work done of friction for further result, we get,

[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]

[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]

[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]

[tex]u_{f} =0.077[/tex]

Hence, the coefficient of kinetic friction between the ice and skates is 0.077.

For more about the friction, follow the link below-

https://brainly.com/question/13357196

Answer:

a PNP transistor used as an amplifier

Explanation:

The diagram show a PNP transistor used as an amplifier.PNP transisitor having one 2 P type and the 1 N type of semiconductor.

This bipolar PNP junction transistor is made up of three layers of semiconductor material, two of which are P-type & one of which is N-type. It consists of three terminals.

The transistor's emitter allows it to supply the majority of charge carriers. In relation to the ground, the emitter always is forward biased.

As a result, the base receives the vast majority of charge carriers. A transistor's emitter is strong and of modest size.

The collector collects the vast majority of the charge carrier delivered by the emitter. Reverse bias is always present at the collector-base junction.

The charge collector region is moderately mixed and capable of collecting the charge.

The diagram shows a transistor used in a circuit.

A signal with a small change in voltage is input into a circuit that includes a transistor connected to 4 resistors and 2 capacitors. The signal output from the transistor has a much larger change in voltage than the input.

The diagram show a PNP transistor used as an amplifier.

Hence option C is correct.

To learn more about the PNP transistors refer to the link;

https://brainly.com/question/1492057

Answer:

f = 1.255 cm

Explanation:

The Radius of Curvature:

The radius of that hollow sphere, whose part is the spherical mirror, is known as ‘The Radius of Curvature’ of  mirror.

Focal Length:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’

The focal length is basically equal to the half of the radius of curvature of the mirror:

[tex]f = \frac{r}{2}[/tex]

where,

f = focal length = ?

r = radius of curvature = 2.51 cm

[tex]f = \frac{2.51\ cm}{2}[/tex]

f = 1.255 cm

Answer:

person that you know and like (not a member of your family), and who likes you

As we know larger the area of contact lesser the pressure. So, in order to reduce the pressure heavy vehicles have broad tyres to increase the area of contact with the ground. Heavy vehicles have broad tyres because broad tyres have large area of contact and less pressure on the ground.

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