Because of the internal resistance of a real battery, we know that the potential difference between the terminals:___________

a. decreases as more current is drawn.
b. increases as more current is drawn.
c. is greater than the emf of the battery
d. increases as the internal resistance increases.

Answer:

(a) The area vector is 0.00503 m² at 30⁰ from the magnetic field

(b) The induced emf is 12.58 V

Explanation:

Given;

angle between the magnetic field and the plane of the circular coil, = 30⁰

number of turns of the coil, N = 1000

radius of the coil, r = 4 cm = 0.04 m

change in the magnetic field with time, dB/dt = 5 T/s

(a) The area vector is calculated as;

A = πr²

A = π x (0.04)²

A = 0.00503 m²

The area vector is 0.00503 m² at 30⁰ from the magnetic field.

(b) The induced emf is calculated as;

[tex]emf = N\frac{\Delta \phi}{\Delta t} \\\\where;\\\\\phi = BAcos \theta\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\where;\\\\\theta \ is \ the \ angle \ between \ a \ perpendicular \ vector \ to \ the \ area\\ and\ the \ magnetic\ field\\\\\theta = 90 - 30 = 60^0\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\emf = (1000) \times (5 )\times (0.00503) \times cos (60)\\\\emf = 12.58 \ V[/tex]

Answer: [tex]377.3\ kPa[/tex]

Explanation:

Given

Water column height [tex]h=25\ cm[/tex]

After oil is poured, the total height becomes [tex]h'=40\ cm[/tex]

Pressure at the bottom will be the sum due to the water and oil column

Suppose the density of the oil is [tex]\rho=900\ kg/m^3[/tex]

Pressure at the bottom

[tex]\Rightarrow P=10^3\times g\times 25+900\times g\times 15\\\Rightarrow P=100g[250+135]\\\Rightarrow P=3773\times 100\ Pa\\\Rightarrow P=377.3\ kPa[/tex]

This question is incomplete, the complete question is;

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s. a) θ = 30° V b) θ = 60° V c) θ = 90° V

Answer:

the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

Explanation:

Given the data in the question;

number of turns N = 70

Diameter of coil D = 11 cm

Radius r = D/2 = 11/2 = 5.5 cm = 0.055 m

magnitude of magnetic ΔB = 0.7T

Δt = 0.2 seconds

Now,

a)

For θ = 30°,

Angle of with area of vector θ' = 90° - 30° = 60°

so

emf  e = N( Δ∅ / Δt ) = N( ΔBAcosθ  / Δt )

hence

e =  NAcosθ'(ΔB / Δt )

where A is area ( πr² )

so we substitute

e =  70 × πr² × cos(60°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(60°) × ( 0.7 / 0.2 )

e = 1.16 V

b)

For θ = 60°,

Angle of with area of vector θ' = 90° - 60° = 30°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(30°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.01 V

c)

For θ = 90°,

Angle of with area of vector θ' = 90° - 90° = 0°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(0°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.33 V

Therefore, the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

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:

4

Explanation:

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

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:

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:

C

Explanation:

Similarity

Answer:

b

Explanation:

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:

Explanation:

The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .

v² = u² + 2 g H .

v² = 0  + 2 x 9.8 x 4.7 m

= 92.12

v = 9.6 m /s

The fish's final velocity will have two components

vertical component = 9.6 m/s downwards

Horizontal component = 2.6 m /s  .

Resultant velocity = √ ( 9.6² + 2.6² )

= √ ( 92.16 + 6.76 )

= 9.9 m /s

Answer:

The speed of fish at the time of hitting the surface is 9.95 m/s.

Explanation:

Horizontal speed, u = 2.6 m/s

height, h = 4.7 m

Let the vertical velocity at the time of hitting to water is v.

Use third equation of motion

[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]

The net velocity with which the fish strikes to the water is

[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]

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:

c). It wouldn't change.

Explanation:

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

Answer:

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

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

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

= 261.54 Hz

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:

 [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

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:

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:

https://brainly.com/question/25559554

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

Answer:

pha của dao động là hàm bậc nhất của thời gian.

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]

Explanation:

Given that,

Two resistors 4.5 Ω and 2.3 Ω .

Potential difference = 30 V

When they are in series, the current through each resistor remains the same. First find the equivalent resistance.

R' = 4.5 + 2.3

= 6.8 Ω

Current,

[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]

So, the current through both lightbulb is the same i.e. 4.41 A.

When they are in parallel, the current divides.

Current flowing in 4.5 resistor,

[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]

Current flowing in 2.3 ohm resistor,

[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]

In parallel combination, are brighter than bulbs in series.

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:

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:

no parallel is the correct answer

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

Calculate force of each car:

P1 = 900 kg x 12m/s = 10,800

P2= 600 x 24 = 14,400

Degree of travel = arctan(14,300/10800)

Degree of travel = 53.1 N of E

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.