The velocity of an object is given by the expression v (t) = 3.00 m / s + (2.00 m / s ^ 3) t ^ 2. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.

Answer:

A= The type of plant

B= How tall the plant is

Explanation:

Answer:

a) t = 27.00 h

b) B = 6.84 MeV/nucleon

Explanation:

a) The time can be calculated using the following equation:

[tex] R = R_{0}e^{-\lambda*t} [/tex]

Where:

R: is the radiation measured at time t

R₀: is the initial radiation

λ: is the decay constant

t: is the time

The decay constant can be calculated as follows:

[tex] t_{1/2} = \frac{ln(2)}{\lambda} [/tex]

Where:

t(1/2): is the half life = 4.5 h

[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 h} = 0.154 h^{-1} [/tex]

We have that the radiation measured is 64 times the maximum permissible level, thus R₀ = 64R:  

[tex] \frac{R}{64R} = e^{-\lambda*t} [/tex]                      

[tex] t = -\frac{ln(1/64)}{\lambda} = -\frac{ln(1/64)}{0.154 h^{-1}} = 27.00 h [/tex]            

b) The binding energy (B) can be calculated using the following equation:

[tex]B = \frac{(Z*m_{p} + N*m_{n} - M_{A})}{A}*931.49 MeV/u[/tex]

Where:

Z: is the number of protons = 2 (for [tex]^{4}_{2}He[/tex])

[tex]m_{p}[/tex]: is the proton mass = 1.00730 u

N: is the number of neutrons = 2 (for [tex]^{4}_{2}He[/tex])

[tex]m_{n}[/tex]: is the neutron mass = 1.00869 u  

[tex]M_{A}[/tex]: is the mass of the He atom = 4.002602 u

A =  N + Z = 2 + 2 = 4    

The binding energy of [tex]^{4}_{2}He[/tex] is:

[tex]B = \frac{(2*1.00730 + 2*1.00869 - 4.002602)}{4}*931.49 MeV/u = 7.35\cdot 10^{-3} u*931.49 MeV/u = 6.84 MeV/nucleon[/tex]

Hence, the binding energy per nucleon is 6.84 MeV.

I hope it helps you!

_____________________________

Solution,

Work=50 Joule

Time=10 seconds

Power=?

Now,

Power=Work/time

= 50/10

= 5 Watt.

So the right answer is 5 W

Hope it helps..

Good luck on your assignment

__________________________

Answer:

A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B.  Lebron James hits at an angle of 17.48° North -East.

The direction of Stephen is   = 20° due to East of North

Explanation:

Let [tex]r ^ {\to[/tex] represent the position vector of the hole;

Also; using  the origin as starting point. Let the east direction be along the positive x axis and the North direction  be + y axis

Thus:

[tex]r ^ {\to[/tex]  = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]

[tex]r ^ {\to[/tex]  = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]

Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot

So;

[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]

[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]

Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot

[tex]r_2 ^ \to[/tex]  [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]

[tex]r_2 ^ \to[/tex]  = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]

However;

[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]

Also;

[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]

Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B .

For Lebron James ;

The angle can be determine using the trigonometric function:

[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]

Thus  Lebron James hits at an angle of 17.48° North -East.

For Stephen Curry;

[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]

The direction of Stephen is  = 90° - 70° = 20° due to East of North

Answer:

5.06*10^6 m/s

Explanation:

Given that

Initial velocity, u = 4*10^6 m/s

Acceleration, a = 6*10^12 m/s²

Distance traveled, s = 80 cm

Final velocity, v = ?

We can find the final velocity by using one of the equations of motion.

v² = u² + 2as

On substituting the values, we have

v² = (4*10^6)² + 2 * 6*10^12 * 0.8

v² = 2.56*10^13

v = √2.56*10^13

v = 5.06*10^6 m/s

Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s

The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].

The given parameters;

The final velocity of the proton over the given distance is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]

Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]

Learn more here:https://brainly.com/question/13613973

Answer:

Explanation:

purchase price = 400 tepizes / m²

1 tepiz = .625 dollar

purchase price in terms of dollar = 400 x .625 dollar / m²

= 250 dollar / m²

.9144 m = 1 yard

1 m = 1.0936 yard

1m² = 1.196 yard²

price in terms of dollar / yards²

= 250 / 1.196 dollar / yard²

= 209 dollar / yard²

Price of cloth in New York = 120 dollar / yard²

loss = 209 - 120 = 89 dollar / yard²

500 m² = 500 x 1.196 yard²

= 598 yard²

net loss in purchasing 500 m² cloth

= 598 x 89

= 53222 dollar .

Answer:

a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g,  y = 24.25 m

e) R = 138.46 m

Explanation:

This is a projectile launch exercise

a) let's use trigonometry to find the components of the initial velocity

  cos θ = v₀ₓ / v₀

  v₀ₓ = v₀ cos θ

v₀ₓ = 38 cos 35

v₀ₓ = 31.13 m / s

b) sin θ = [tex]v_{oy}[/tex] / v₀

    v_{oy} = v₀ sin θ

    v_{oy} = 38 sint 35

    v_{oy} = 21 80 m / s

c, d) to find the maximum height, the vertical speed is zero

     v_{y}² = v_{oy}² - 2 g y

     0 = [tex]v_{oy}[/tex]² - 2 gy

     y = v_{oy}² / 2g

let's calculate

     y = 21.80 2 / (2 9.8)

     y = 24.25 m

e) They ask to find the horizontal distance

    for this we can use the expression of reaches

       R = v₀² sin 2θ / g

let's calculate

      R = 38² sin (2 35) / 9.8

       R = 138.46 m

Answer:

4.8 × 10^15 N

Explanation:

Electric Field is defined as Force per unit Charge.

This is expressed mathematically as;

E= F/Q

Where E- Electric Field

F- Force

Q- charge

From the expression above by change of subject of formula for F, we have;

F=E×Q

= 1.2 * 10^6 ×4.0 * 10^9

= 4.8 × 10^15 N

Answer:

D. Presynaptic terminal button

explanation:

Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron

hope this helped!

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

Answer:

9.05 m/s ,   -14.72°  (respect to x axis)

Explanation:

To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:

[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]

[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]

m1: mass of the bowling ball = 5.50 kg

m2: mass of the bowling pin = 0.850 kg

v1xi: initial velocity of the bowling ball = 9.0 m/s

v2xi: initial velocity of bowling pin = 0m/s

v1: final velocity of bowling ball = ?

v2: final velocity of bowling pin = 15.0 m/s

θ: angle of the scattered bowling pin = ?

Φ: angle of the scattered bowling ball = 85.0°

Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.

First you solve for v1cosθ in the equation for the x component of the momentum:

[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]

and also you solve for v1sinθ in the equation for the y component of the momentum:

[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]

Next, you divide v1cosθ and v1sinθ:

[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]

the direction of the bawling ball is -14.72° respect to the x axis

The final velocity of the bawling ball is:

[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]

hence, the final velocity of the bawling ball is 9.05 m/s

Answer:

2025m

Explanation:

Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.

Volume of Sphere= volume of cylinder

4/3 ×π×R^3= π× r2× L

4/3 ×R^3= r^2×L

Hence

L = 3/4 × R^3/ r^2

But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}

r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}

Substituting R and r into the expression for L, we have :

L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm

202500/100= 2025m{ we divide by 100 because 100cm=1m}

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image

Answer:

Explanation:

capacitance of air capacitor

C = ε₀ A /  d

ε₀ is permittivity of medium , A is plate area , d is distance between plate .

C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²

= 170.19 x 10⁻¹⁴ F .

charge on the capacitor when it is charged to  potential of 255 V

= CV , C is capacitance and V is potential

= 170.19 x 10⁻¹⁴  x 255

= 4.34 x 10⁻¹⁰ C .

After it is disconnected from the source , and it is immersed in water , charge on it remains the same .

So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.

b )

When it is immersed in water its capacity increases  k times where k is dielectric constant of water which is 80 .

capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴  F

= 13615.2  x 10⁻¹⁴ F .

= 1.36 x 10⁻¹⁰ F

potential difference = charge / capacitance

= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰

= 3.2 V

c )

Energy of capacitor = 1/2 C V²

Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴

=  55.33 x 10⁻⁹ J

Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²

= .7  x 10⁻⁹ J .

decrease of energy = 54.63 x 10⁻⁹ J .

Answer:

Distance = 16.9 m

Explanation:

We are given;

Power; P = 70 W

Intensity; I = 0.0195 W/m²

Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;

I = Power/Unit area = P/(4πr²)

where;

P is the sound power

r is the distance.

Thus;

Making r the subject, we have;

r² = P/4πI

r = √(P/4πI)

r = √(70/(4π*0.0195))

r = √285.6627

r = 16.9 m

Answer:

16.9 m

Explanation:

Answer:

The apparent weight of the person as she pass the highest point is  [tex]N = 458.8 \ N[/tex]

Explanation:

From the question we are told that

   The radius of the Ferris wheel is [tex]r = 5.0 \ m[/tex]

    The period of revolution is [tex]T = 8.0 \ s[/tex]

     The weight of the person is  [tex]W = 670 \ N[/tex]

Generally the speed of the wheel is mathematically represented as

      [tex]v = \frac{2 \pi r}{T }[/tex]

substituting values

      [tex]v = \frac{2 * 3.142 * 5}{8 }[/tex]

       [tex]v = 3.9 3 \ m/s[/tex]

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

          [tex]N = mg - \frac{mv^2}{r}[/tex]

Where m is the mass of the person which is mathematically evaluated as

     [tex]m = \frac{W}{g}[/tex]

substituting values

    [tex]m = \frac{670}{9.8}[/tex]

    [tex]m = 68.37 \ kg[/tex]

So

    [tex]N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}[/tex]

    [tex]N = 458.8 \ N[/tex]

Answer:

Explanation:

From Newton's second Law of Motion,

F = ma

Ff + F = ma

Where F is the applied force, Ff is the frictional force, a is the acceleration and m is the mass of the object or box.

Magnitude of the acceleration:

a = Ff+F/m

This must act in the direction of F or the box would slide or accelerate off the negative side of the board (taking the direction of F to be positive

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

Answer:

1.25 m/s^2

Explanation:

F = m*a ...... force = mass * acceleration

force = 100 N, mass = 80 kg

100 = 80 * a

100/80 = a = 1.25 m/s^2

Answer:

The acceleration is 1.25m/s².

Explanation:

You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :

[tex]f = m \times a[/tex]

Let F = 100,

Let m = 80,

[tex]100 = 80 \times a[/tex]

[tex]100 = 80a[/tex]

[tex]a = 100 \div 80[/tex]

[tex]a = 10 \div 8[/tex]

[tex]a = 1.25[/tex]

Answer:

Work is done by friction = -165 J

Explanation:

Given:

Mass of block (m) = 15 kg

Ramp inclined = 28°

Friction force (f) = 30 N

Distance (d) = 5.5 m

Find:

Work is done by friction.

Computation:

Work is done by friction = -Fd

Work is done by friction = -(30)(5.5)

Work is done by friction = -165 J

Answer:D

Explanation:

Given

Same force is applied to each ball such that all have different masses

and Force is given by the product of mass and acceleration

[tex]F=m\times a[/tex]

[tex]a=\frac{F}{m}[/tex]

So acceleration of ball A

[tex]a_A=\frac{F}{0.5}=2F[/tex]

acceleration of ball B

[tex]a_B=\frac{F}{0.75}=\frac{4F}{3}=1.33F[/tex]

acceleration of ball C

[tex]a_C=\frac{F}{1}=F[/tex]

acceleration of ball D

[tex]a_D=\frac{F}{7.3}=\frac{F}{7.3}[/tex]

It is clear that acceleration of ball D is least.

Answer:

100 J

Explanation:

Work = change in energy

W = ΔKE

W = ½ mv² − ½ mv₀²

W = ½ m (v² − v₀²)

W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)

W = 100 J

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

Explanation:

Momentum = mass × velocity

p = mv

p = (1000 kg) (14 m/s)

p = 14000 kg m/s

The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

Given the data in the question

Momentum is the product of the mass of a particle and its velocity.

Momentum = Mass × Velocity

[tex]P = m \ * \ v[/tex]

We substitute our given values into the equation

[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]

Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

Learn more; https://brainly.com/question/265061

Answer:

rt

Explanation:

Answer:

a bale

Explanation:

a bale is a group of turtles

Answer:

A bale or nest

Explanation:

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

[tex]I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4[/tex]

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

[tex]\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa[/tex]

Calculate the first moment of area below point C

[tex]Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm[/tex]

Now calculate shear stress at point C

[tex]=\frac{FQ}{It}[/tex]

[tex]=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa[/tex]

Calculate the principal stress at point C

[tex]\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa[/tex]

Calculate the maximum shear stress at piont C

[tex]\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa[/tex]

Answer:

Plutonium–238

Explanation:

The stability of isotopes is largely dependent on their half-life.

Half life of an isotope is the time taken for the initial mass of the isotope to be halfed or we can say that the half-life of an isotope is the time taken for the mass of the isotope to become half the initial mass.

From the above definition, we discovered that if the time taken for the mass of the isotope to become half its initial mass is long, then the isotope must be very stable. On the other hand, if the time taken to become half its initial mass is short, then the isotope is unstable because.

We can thus say that, the longer the half-life the more stable the isotope and the shorter the half-life, the less stable the isotope will be.

Considering the table given in the question above and with the ideas we have obtained from the explanation above, we can see that Plutonium–238 has the longest half-life. Therefore Plutonium–238 will be more stable.

Answer:

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat

Explanation:

Fourier's Law of heat conduction, gives the following formula:

Q = - KAΔT/t

where,

Q = Rate of Heat Conduction out of window = ?

K = Thermal Conductivity of Glass = 0.84 W/m.°C

A =Surface Area of window = 2.82 m²

ΔT = Difference in Temperature of both sides of surface

ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)

ΔT = 15°C

t = thickness of window = 0.675 cm = 0.00675 m

Therefore,

Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat.

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.

Since

number of turns, N = 200 turns

cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

the magnitude of magnetic field strength, B = 30 x 10⁻³ T

the maximum value of the induced emf, E = 8 V

Now

Maximum induced emf should be

E = NBAω

here,

ω is angular velocity (ω = 2πf)

Now

E = NBA2πf

here,

f is the minimum frequency

So,

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz.

Learn more about frequency here: https://brainly.com/question/24470698