Which of the following scientists won a Nobel Prize for pioneering work in the
study of the evolution of stars?
A. Christian Doppler
B. Warren Washington
C. Charles Kuen Kao
-D. Subrahmanyan Chandrasekhar

Answer:

[tex]E_0=0.021v/m[/tex]

Explanation:

From the question we are told that:

Length [tex]l=1.73m[/tex]

Voltage [tex]V=36.0mV[/tex]

Frequency [tex]F=500Hz[/tex]

Generally the equation for maximum Electric Field  is mathematically given by

 [tex]E_0=\frac{\triangle}{r}[/tex]

 [tex]E_0=\frac{36*10^{-3}}{1.73}[/tex]

 [tex]E_0=0.021v/m[/tex]

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:

4

Explanation:

Answer:

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

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         [tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex]            (you has an mistake in the formula)

         [tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]

         [tex]\frac{1}{C_{eq1}}[/tex] = 0.1   10⁶

         [tex]C_{eq1}[/tex] = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          [tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]

          [tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]

          [tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶

          [tex]C_{eq2}[/tex] = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]

          U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]

          U₄ = 0.5 J

Answer:

 [tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029

Explanation:

The refractive index is defined

         n = c / v

         v = c / n

the speed of light per se wave is constant, so we can use the relations of uniform motion

         v = x / t

         t = x / v

we substitute

         t = x n / c

let's calculate the time

vacuum

        t₁ = 1000 1/3 10⁸

        t₁ = 3.333333 10⁻⁶ s

air

        t₂ = 1000 1.00029 / 3 10⁸

        t2 = 3.3343 10⁻⁶ s

the relationship between these times is

       t₂ / t₁ = 3.3343 / 3.3333333

       t₂ / t₁ = 1.00029

Complete Question

Complete Question is attached below

Answer:

[tex]q=1.558*10^{-9}c[/tex]

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength [tex]E_l=784.75N/m[/tex]

Right field strength [tex]E_r=776.38 N/m[/tex]

Front field strength [tex]E_f=725.5 N/m[/tex]

Back field strength [tex]E_b=749.54 N/m[/tex]

Top field strength [tex]E_t=944.95 N/m[/tex]

Bottom field strength [tex]E_{bo}=1082.58 N/m[/tex]

Generally, the equation for  Charge flux is mathematically given by

[tex]\phi=EAcos\theta[/tex]

Where

Theta for Right,Left,Front and Back are at an angle 90

[tex]cos 90=0[/tex]

Therefore

[tex]\phi =0[/tex] with respect to Right,Left,Front and Back

Generally, the equation for  Charge Flux is mathematically also given by

[tex]\phi=\frac{q}{e_o}[/tex]

Where

[tex]Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2[/tex]

Therefore

[tex]Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t[/tex]

[tex]Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)[/tex]

[tex]Q_{net}=176N/C m^2[/tex]

Giving

[tex]q=\phi*e_0[/tex]

[tex]q=176N/C m^2*1.558*10^{-12}c[/tex]

[tex]q=1.558*10^{-9}c[/tex]

Answer:

C

Explanation:

Similarity

Answer: [tex]1.79\ rad/s^2[/tex]

Explanation:

Given

Initial angular speed is [tex]\omega_1=26.2\ rad/s[/tex]

Final angular speed is [tex]\omega_2=36.5\ rad/s[/tex]

Time period [tex]t=5.75\ s[/tex]

Magnitude of the fan's acceleration is given by

[tex]\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}[/tex]

Insert the values

[tex]\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2[/tex]

Thus, fan angular acceleration is [tex]1.79\ rad/s^2[/tex]

Answer:

The angular acceleration is given by 1.8 rad/s^2.

Explanation:

initial angular speed, wo = 26.2 rad/s

final angular velocity, w = 36.5 rad/s

time, t = 5.75 seconds

The first equation of motion is

[tex]w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2[/tex]

Answer:

8000 N

Explanation:

Applying

P = F×V.............. Equation 1

Where P = Power of the sport car, F = resistive force acting on the sport car, V  = Speed of the sport car.

make F the subject of the equation

F = P/V............ Equation 2

From the question,

Given: P = 200 kW = 200000 W, V = 25 m/s

Substitute these values into equation 2

F = 200000/25

F = 8000 N

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:

Current = 0.000294 A

Explanation:

Below is the given values:

Given the charge = 9.00 C

Time = 8.50 h

Use the below formula to find the current:

Current = Q / t

Now plug the values:

Current = 9 / (8.5 x 3600)

Current = 0.000294 A

Answer:

W = 2523.67 J

Explanation:

Given that,

The mass of surfer, m = 77 kg

He starts with a speed of 1.3 m/s

It drops through a height of 1.65 m and ends with a speed of 8.2 m/s.

We know that, the work done is equal to the change in kinetic energy. So,

[tex]W=\dfrac{1}{2}m(v_2^2-v_1^2)\\\\=\dfrac{1}{2}\times 77\times (8.2^2-1.3^2)\\W=2523.67\ J[/tex]

So, the required work done is equal to 2523.67 J.

Answer:

A. 16 J

B. 16 J

C. 8 N

Explanation:

A. Determination of the kinetic energy.

Mass (m) = 0.5 Kg

Velocity (v) =. 8 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 8²

KE = ½ × 0.5 × 64

KE = 0.5 × 32

KE = 16 J

B. Determination of the Workdone by the force.

Kinetic energy (KE) = 16 J

Workdone =.?

Workdone and kinetic energy has the same unit of measurement. Thus,

Workdone = kinetic energy

Workdone = 16 J

C. Determination of the force.

Workdone (Wd) = 16 J

Displacement (s) = 2 m

Force (F) =?

Wd = F × s

16 = F × 2

Divide both side by 2

F = 16 / 2

F = 8 N

Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)

= 250 x (340/( 340-15))

= 261.54 Hz

Answer:

no parallel is the correct answer

Answer:

c). It wouldn't change.

Explanation:

[tex]{ \bf{F = \frac{ \ \gamma _{o}NI }{l} }}[/tex]

Answer:

Explanation:

I’ll help

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

[tex]Power = P = \frac{E}{t}[/tex]

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

[tex]P = \frac{2.927\ x\ 10^6\ J}{1200\ s}[/tex]

P = 2439.5 W = 2.439 KW

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

[tex]V_i=50 \ m/h[/tex]

Now,

[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]

    [tex]=73.333 \ ft/sec[/tex]

Now,

Final velocity,

[tex]V_f=0[/tex]

Initial velocity,

[tex]V_{initial} = 73.333 \ ft/sec[/tex]

hence,

⇒ [tex]V_f^2=V_i^2+2aD[/tex]

By putting the values, we get

      [tex]0=(73.333)^2+2\times( -22) D[/tex]

  [tex]44D=(73.333)^2[/tex]

      [tex]D=\frac{(73.333)^2}{44}[/tex]

          [tex]=122.2211[/tex]

Answer:

The correct answer is - low pitch

Explanation:

Now for the case it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

if frequency increases then pitch will be increase as well as pitch depends on frequency.

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.

As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = [tex]\frac{P_o ( 1-0.850)}{\rho \ g}[/tex]

        h = [tex]\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}[/tex]

        h = 1.55 m

Answer:

The speed at the bottom is 11.2 m/s.

Explanation:

length, s = 10 m

Angle, A = 45 degree

coefficient of friction = 0.1

let the velocity is v.

The acceleration is given by

[tex]a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2[/tex]

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s[/tex]

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical  component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

Answer:

8.3 m/s, 2196 degree from + X axis

Explanation:

m = 320 g , u = 2 m/s along X axis

m'  = 355 g, u' = 1.5 m/s along Y axis

m'' = 100 g, u'' = v

Let the speed of the third piece is v makes an angle A from the X axis.

use conservation of momentum along X axis

0 = 320 x 2 + 100 x v cos A

v cos A =  - 6.4 ..... (1)

Use conservation of momentum along Y axis

0 = 355 x 1.5 + 100 x v sin A

v sinA = - 5.3 ... (2)

Squaring and adding

[tex]v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s[/tex]

The angle is given by

[tex]tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree[/tex] from + X axis

Answer:

b

Explanation:

Answer:

A teacher is more important than a famer.

Explanation:

A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.

105 km/h ≈ 29.2 m/s

60.0 km/h ≈ 16.7 m/s

Before the collision the test car has an acceleration a of

a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²

During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is

a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²

i.e. with magnitude about 19.7 m/s².

Overall, the car has an average acceleration of

a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²

Answer:

u are the very good person I know if u will do itself u will becam3 a rising star

Answer:

B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction