suppose the bipolar transsitor exhibits the following hypothetical charactersitic: and no early effect. computer the voltage gain for a bias current of 1 ma.

The student will have to move the loop out of the magnetic field in a time interval of approximately 0.36 seconds in order to induce an emf of 1.3 V.

To calculate the time interval required to induce the desired emf, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf (ε) in a wire loop is equal to the rate of change of magnetic flux through the loop. Mathematically, it is given by ε = -dΦ/dt, where ε is the induced emf, dΦ is the change in magnetic flux, and dt is the time interval.

In this case, the area of the loop (A) is given as 4.40 × 10^(-3) m², and the magnetic field (B) is 4.7 × 10^(-2) T. The magnetic flux (Φ) through the loop is given by Φ = B * A.

We need to rearrange the equation ε = -dΦ/dt to solve for dt. Rearranging, we have dt = -dΦ / ε.

Substituting the given values, we have dt = -(B * dA) / ε, where dA is the change in the area of the loop. Since the loop is moved perpendicular to the magnetic field lines, the change in area (dA) is equal to the area of the loop (A).

Therefore, dt = -(B * A) / ε.

Substituting the values, we have dt = -(4.7 × 10^(-2) T * 4.40 × 10^(-3) m²) / 1.3 V.

Evaluating this expression, we find that the time interval required to induce an emf of 1.3 V is approximately 0.36 seconds.

In summary, the student will have to move the loop out of the magnetic field in a time interval of approximately 0.36 seconds to induce an emf of 1.3 V.

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A key prerequisite for adaptive radiations that is shared among just about all examples of adaptive radiations is the availability of ecological opportunities or vacant ecological niches.

Adaptive radiation refers to the rapid diversification of a single ancestral lineage into a variety of species that occupy different ecological niches. It occurs when a group of organisms encounters new and vacant ecological opportunities, allowing them to exploit different resources or habitats.

The availability of ecological opportunities is crucial for adaptive radiations because it provides the necessary conditions for evolutionary divergence and speciation.

When new ecological niches become available, organisms that possess adaptations enabling them to exploit these niches can undergo rapid diversification and give rise to multiple new species.

This prerequisite of ecological opportunities is observed in various examples of adaptive radiations, such as Darwin's finches in the Galápagos Islands, Hawaiian honeycreepers, cichlid fish in African lakes, and many others.

In each case, the colonization of new habitats or the opening of new ecological niches has facilitated the adaptive radiation and subsequent diversification of species.

The key prerequisite for adaptive radiations that is shared among just about all examples is the availability of ecological opportunities or vacant ecological niches. This provides the necessary conditions for organisms to diversify and occupy different niches, leading to the rapid evolution of multiple new species.

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Given that in the first case,

the volume of the gas is

The pressure of the gas is

The temperature of the gas is

In the second case,

the pressure of the gas is

The temperature of the gas is

We have to find the volume, V2.

The volume v2 can be calculated as

Answer:

3

Explanation:

You will do 60÷20=30

This is the answer and please follow me

Using the free body diagram, let's determine the scenario the diagram illustrates.

From the free body diagram, we can see that the weight is acting on the object while the tension will be the force which supports the object.

The tension in this case, helps hang the object while the weight is the product of the mass of the object and the force of gravity acting on the object.

Tension can be defined as the pulling force transmitted through a rope or string.

Therefore, the best scenario which is represented by the free body diagram is a sign board supported by two strings.

ANSWER:

A signboard supported by

Answer:

The speed of the car is in from as the required speed limit on the road.

Explanation:

The speed limit on the road is 30 miles per hour.

Given that: 1 mile = 1609 m

Also,

60 second = 1 minutes

60 minutes = 1 hour

Then,

30 miles per  hour = \(\frac{30 *1609}{60*60}\)

                              = \(\frac{48270}{3600}\)

                              = 13.4083

30 miles per  hour  = 13.41 m/s

The speed measured by the detector is 15 m/s which is not accurate. Therefore, the car was moving at the required speed limit. The speed of the car is lesser than that measured by the detector. Thus no offence has been committed by the driver.

(a) Leak before break is a concept of designing the system or structure to minimize the risk of sudden catastrophic failure due to cracks.

The concept is to create a design that allows the development of cracks, and before the crack gets to a critical size, the system or structure should be able to detect the crack or leak. There are three stages in a leak-before-break condition:

First stage - Crack initiation: In this stage, cracks start to develop, but there is no effect on the system or structure's performance.

Second stage - Slow Crack Growth: In this stage, the crack's growth is slow, and the crack's effects on the system or structure are minimal.

Third stage - Rapid Crack Propagation: In this stage, the crack grows rapidly, and it could cause a catastrophic failure of the system or structure.  

(b) Longitudinal stress, σl = (pi x di^2)/4tσl = (28.75 x 10^6 x (0.08^2))/4 x 0.01σl = 58000000 Pa Hoop stress, σH = (pi x di x p)/2tσH = (28.75 x 10^6 x 0.08)/2 x 0.01σH = 11500000 Pa

(c) Critical crack size can be calculated as; ac = ((KIc^2 x Y)/(σpi))^(1/2)ac = ((23^2 x 0.7)/(28.75 x 10^6 x pi))^(1/2)ac = 1.6 x 10^-3 m

The crack is stable since the size of the crack is less than the critical crack size.

(d) Yes, the condition corresponds to a 'leak before break since the hemispherical crack's growth is slow, and the cylinder could detect it and initiate repairs before the cylinder fails catastrophically.

(e) No, the design will not be 'leak before break since the cheaper aluminum alloy's critical crack size is smaller than the crack's size. Therefore the cylinder will fail catastrophically.

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The inverted pendulum model is the most basic mechanical system that can replicate the motion of the centre of gravity during gait.

In this model, the hip joint serves as the pivot point for the leg's inverted pendulum. The inverted pendulum's swinging motion can be used to model how the centre of mass moves during gait. The centre of mass moves in an arc as the leg swings back and forth, imitating the natural stride. This mechanical device, which has been sped up, depicts the basic dynamics of walking and offers insights into the coordination and control of human mobility.

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An electroscope is used to detect the presence of an electric charge. It consists of two metal plates or leaves that are suspended from an insulating support.

When an electric charge is brought near the electroscope, the leaves will diverge due to the electrostatic force of repulsion between them.

The extent of the divergence will depend on the magnitude of the charge. However, it is not possible to determine the sign (positive or negative) of the charge from an electroscope alone.

An electroscope is a device used to detect the presence of an electric charge. To tell if an electroscope has an electric charge, you can observe the behavior of the metal leaves on the electroscope.

When an electroscope has an electric charge, the metal leaves on the electroscope will spread apart. This is because the same charge on the metal leaves repels each other, causing them to move away.

It is not possible to tell from an electroscope alone what kind of charge it has. To determine the type of charge, you will need to use another device to induce a charge on the electroscope and observe the behavior of the leaves. If the leaves move away from each other, the electroscope has a positive charge. If the leaves move towards each other, the electroscope has a negative charge.

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MAIN ANSWER in 30 WORDS:
A) α = 8.8 rad/s^2; B) at = 118.8 m/s^2; C) ac = 413.52 m/s^2.

EXPLANATION PART IN 120 WORDS:
A) Angular acceleration α can be calculated using the formula α = (ω - ω0)/Δθ. Substituting values, we get α = (8.8 - 4.4)/5.5 = 0.8 rad/s^2.

B) Tangential acceleration at can be calculated using the formula at = Rα, where R is the radius of the platform. Substituting values, we get at = 34 × 0.8 = 27.2 m/s^2.

C) Final centripetal acceleration ac can be calculated using the formula ac = Rω^2, where ω is the final angular velocity. Substituting values, we get ac = 34 × 8.8^2 = 413.52 m/s^2.

Therefore, the angular acceleration is 0.8 rad/s^2, the tangential acceleration at the outer edge of the platform is 27.2 m/s^2, and the final centripetal acceleration at the outer edge of the platform is 413.52 m/s^2.

The angular acceleration of the platform is 0.8 rad/s^2, the tangential acceleration at the outer edge of the platform is 27.2 m/s^2, and the final centripetal acceleration at the outer edge of the platform is 2617.6 m/s^2.

A) The angular acceleration α of the platform can be calculated using the formula: α = (ω^2 - ω0^2) / (2Δθ). Plugging in the values, we get α = (8.8^2 - 4.4^2) / (2 * 5.5) = 44/11 rad/s².

B) The tangential acceleration at of a point on the outer edge can be calculated using the formula: at = α * R. Converting R to meters, we get R = 0.34 m. Thus, at = (44/11) * 0.34 = 1.36 m/s².

C) The final centripetal acceleration ac can be calculated using the formula: ac = ω^2 * R. Plugging in the values, we get ac = 8.8^2 * 0.34 = 26.424 m/s².

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Answer:

2500 N

Explanation:

Tension force may be defined as the force which is transmitted along a cable, rope, or through a string which is stretchable and is puled tight by the forces acting from the opposite sides. It is directed along the length of the string or the rope.

In the context, a pulley string holds a 2500 N piano that is used to lift it to second story balcony. When the piano is held stationary by the mover, the tension force acting on the rope is the weight force or the gravitational force that acts on the piano which is 2500 N. The tension force balances the the weight force of the piano when it is not moving.

Answer:

3. 1.8A

Explanation:

Given the following data;

Quantity of charge, Q = 650C

Time = 6 minutes to seconds = 6 * 60 = 360 seconds.

To find the current l;

Quantity of charge = current * time

Substituting into the equation, we have;

650 = current * 360

Current = 650/360

Current = 1.8 Amperes

If you pull yourself to the center of the merry-go-round,  the rotation of the merry go round will decrease.

This is the radial acceleration experienced by a person moving in a circular path as the merry-go-round rotates about its mean position.

a = v²/r

where;

From the formula given above, as the radius of the merry go round increases, the centripetal acceleration of the merry go round decreases.

v = ωr

where;

Thus, when the radius of the circular path decreases, the speed of the merry go round will decrease as well.

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Answer:

Explanation:

The time taken by the motorcycle to reach the city can be found by using the formula

\(t = \frac{d}{s} \\ \)

d is the distance

s is the speed

From the question

d = 200 miles

s = 70 miles / hour

We have

\(t = \frac{200}{7} \\ = 28.571428 \\ \)

We have the final answer as

Hope this helps you

The force applied to bring the body to rest is 400N, and the distance covered by the body is 90 meters.

To calculate the force applied to bring a body of mass 100kg to rest, we need to use Newton's second law of motion. According to this law, the force (F) required to bring an object to rest is equal to the mass (m) of the object multiplied by its acceleration (a).

Given that the mass of the body is 100kg, we can calculate the acceleration using the formula a = (change in velocity) / (time taken). Here, the change in velocity is from 20m/s to 0m/s, and the time taken is 5 seconds.

Using the formula, a = (0 - 20) / 5 = -4m/s^2 (negative because the body is slowing down)

Now, we can calculate the force using the formula F = m * a, where m is the mass and a is the acceleration.

F = 100kg * -4m/s^2 = -400N

So, the force applied to the body to bring it to rest is 400N.

To calculate the distance covered, we can use the equation of motion, s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, t is the time taken, and a is the acceleration.

In this case, the initial velocity is 20m/s, the time taken is 5 seconds, and the acceleration is -4m/s^2.

Plugging these values into the equation, we get:

s = 20 * 5 + (1/2) * (-4) * (5^2)
s = 100 + (-10)
s = 90m

Therefore, the distance covered by the body is 90 meters.

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Based on what we know, we can confirm that this effect is often referred to as the Parallax effect.

This effect explains the differences that we see when performing this experiment with the pencil. It attempts to explain the difference in the apparent position that we see when viewing the pencil from differing lines of sight. This has to do with the change in the angle we are viewing from when we switch eyes.

Therefore, we can confirm that this effect is often referred to as the Parallax effect.

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The exposure to a carcinogen in a fish population leads to a small subset of mutated individuals with increased susceptibility to diseases, resulting in their decline while non-mutated fish continue to thrive.

When a population of fish is exposed to a carcinogen in their habitat, it can have varying effects on different individuals within the population. In this case, most fish are not significantly affected by the chemical and continue to live and reproduce as usual. However, a small subset of individuals experience mutations as a result of the carcinogen exposure. These mutations can cause significant morphological changes in these individuals, making them more susceptible to diseases or other health issues.

Over time, the mutated individuals face difficulties in reproducing due to their health challenges or genetic abnormalities. As a result, their population starts to decline as they are unable to pass on their mutated genes to future generations.

On the other hand, the non-mutated fish, being unaffected by the carcinogen or having a lower susceptibility to its effects, continue to thrive in the population. They reproduce successfully and maintain the overall survival and stability of the fish species.

This scenario highlights the potential consequences of carcinogen exposure on a fish population, where the presence of mutations can lead to a decline in certain individuals while others remain unaffected and ensure the continuation of the species.

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The net force on particle 92 is approximately 8.10 × 10¹¹ N to the right.

The net force on particle 92 due to particles 91 and 93 can be calculated using Coulomb's law:

F = k * |q₁| * |q₂| / r² where k is the Coulomb constant (9.0 × 10⁹ N·m²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them. The direction of the force is given by the signs of the charges.

The force on particle 92 due to particle 91 is:

F₁ = k * |91| * |92| / r² = k * 66.3 * 108 / (0.550)²

≈ 4.90 × 10^11 N

Since particle 91 has a negative charge and is to the left of particle 92, the force it exerts on particle 92 is to the right (positive).

The force on particle 92 due to particle 93 is:

F₃ = k * |93| * |92| / r² = k * 43.2 * 108 / (0.550)²

≈ 3.20 × 10¹¹ N

Since particle 93 has a negative charge and is to the right of particle 92, the force it exerts on particle 92 is also to the right (positive).

The net force on particle 92 is the vector sum of F₁ and F₃:

Fnet = F₁ + F₃ ≈ 8.10 × 10¹¹ N to the right.

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There is direct relationship between harmonics and resonance frequency.

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Part A: The known quantity in this problem is the y-component of vector F. This is because the problem states that "the component of the unit's weight perpendicular to the roof cannot exceed 485 N." The y-component of vector F is the component that is perpendicular to the roof.

Part B: The target variable for question (a) is the magnitude of vector F. This is because the problem is asking for the maximum allowed weight of the unit, which is equivalent to the magnitude of the weight vector.

Part C: To answer question (b), you must know the magnitude of vector F and the x-component of vector F. The magnitude of vector F is needed to calculate the work done by the weight force on the unit during its slide, and the x-component of vector F is needed to determine the displacement of the unit along the roof.

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Electromagnetic waves are those waves that are composed of electric and magnetic fields perpendicular to each other and to the direction of the propagation. All electromagnetic waves have the same speed.

That is, 3×10⁸ m/s.

X-ray is one of the electromagnetic waves. So are light, gamma rays, and radio waves.

Thus the speed of all these waves is equal.

Thus, the correct answer is option C, "Same as speed of radio waves"

Two football players are running towards each other in a straight line (exact opposite directions). Player A is running at 3.3 m/s and has a mass of 105 kg. Player B is 126 kg.

The players collide and their net momentum after the collision is 0 Ns. How fast was Player B running before they collided? The law of conservation of momentum states that in a closed system, the total momentum remains constant. Therefore, the total momentum of both players before collision equals the total momentum of both players after collision. This means: mA * VA + mB * VB = (mA + mB) * V, where VA and VB are the initial velocities of A and B, respectively, and V is their final velocity after the collision (which is 0).

So, we can rearrange the above equation to solve for VB. VB = (mA * VA) / mB Here, mA = 105 kg and VA = 3.3 m/s, and mB = 126 kg. Substituting the values, we get: VB = (105 * 3.3) / 126= 2.75 m/s Therefore, Player B was running at 2.75 m/s before they collided.

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Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

Answer:

B- It is a constant when the light is traveling in a vacuum.

Explanation:

"Light in a vacuum always travels at the same speed." Those additional three words in a vacuum are very important. A vacuum is a region with no matter in it. So a vacuum would not contain any dust particles (unlike a vacuum cleaner, which is generally full of dust particles)

Answer:

Greater acceleration of the plates towards each other

Explanation:

If more force is added to the convection currents, there would be a greater acceleration of the plates towards each other.

 A lithospheric plate is driven by underlying asthenosphere which is ductile and readily flows. The plate lies on the asthenosphere.

The rate of movement of the asthenosphere determines the rate at which the plate above is moving.

The greater the convection below, the faster the plate moves.

Step-Step Explanation:-

As the system is at equilibrium, this means that the reading on the scales have to add to the weight of the student, which means that:-

\(F^{l} + F^{r} = W\)

or

\(F^{l} =W - F^{r} \)As we know that momentum of the system have to be zero, this means that the momentum for each scale is equal and then we have:-

\(1.5F^{l} = 0.5F^{r} \)

Plugin this in the equation for forces, we have:-

\( \frac{0.5}{1.5}F^{r} =W - F^{r}\)

\( = > F^{r} + \frac{0.5}{1.5} = W\)

\( = > \frac{1.5 + 0.5}{1.5}F^{r} = W\)

\( = > F^{r} = \frac{1.5 W}{1.5 + 0.5} \)

\( = > F^{r} = \frac{1.5 W}{2} \)

\( = > \frac{1.5}{2} \times 62 \times 9.8\)

\( = > F^{r} = 455.7\)

Now we know the reading on the right scale, we plug it on the equation for the left scale.

\(F^{l} = W - F^{r} \)

\( = > F^{l} = 62 \times 9.8 - 455.7\)

\( = > F^{l} = 151.9\)

Therefore, the reading on left scale is 151.9 N and the reading on right scale is 455.7 N.

The force of reaction at the left will be  \(F_c=152.057\ N\) and at the right side will be \(F_D=456.165\ N\)

It is given that

Mass of the body m=62kg

Now the Force of the body

\(F_{body} =mg=62\times 9.81=608.22\ N\)

Now to find the effect of this force on the scale we will suppose the left point is C and the right point will be D.

Now to find the at Point D (right side ) Take a moment about point A

So the equation will become

\(F_D\times2=F_{body}\times 1.5\)

\(F_D=\dfrac{F_{body}\times 1.5}{2}\)

\(F_D=\dfrac{ 608.22\times 1.5}{2}\)

\(F_D=456.165\ N\)

Now to find the force on the left side by the equilibrium of the forces

The algebraic sum of all the forces will be zero

\(F_D+F_C-F_{body}=0\)

\(F_C=F_{body}-F_D\)

\(F_C=608.22-456.16=152.05N\)

Thus the force of reaction at the left will be  \(F_c=152.057\ N\) and at the right side will be \(F_D=456.165\ N\)

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Centripetal force is 5.61N

string length r=85cm

mass =1.5kg

velocity=1.08

Centripetal force= mv2/r= 1.5+1.08*1.08/85

=5.61N

Centripetal force is the force exerted on an object moving curvilinearly that is pointed in the direction of the object's axis of rotation or center of curvature. Newton is the symbol for centripetal force. Whenever an object moves, the centripetal force is applied perpendicular to that direction.

Gravity is the centripetal force in Newtonian mechanics that drives astronomical orbits. The scenario in which a body travels at a constant speed along a circular path is a typical illustration of centripetal force.

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​

The speed of the boat relative to the water is 9.08 m/s, when the boat takes 14 minutes to cross the river.

We can decompose the velocity of the boat into two components: one parallel to the river, and one perpendicular to it. Let's call the parallel component \(v_p_a_r\) and the perpendicular component \(v_p_e_r_p\):-

\(v_p_a_r = v_b_o_a_t cos\theta\)

\(v_p_e_r_p = v_b_o_a_t sin\theta\)

The boat will travel a distance of 102 m across the river, which will take 14 minutes or 840 seconds. Therefore, the distance traveled parallel to the river will be:

\(d_p_a_r = v_p_a_r * t = v_b_o_a_t cos\theta * 840 s\)

The distance traveled perpendicular to the river will be:

\(d_p_e_r_p = v_p_e_r_p * t = v_b_o_a_t sin\theta * 840 s\)

The boat will end up at a point directly north of its departure point on the south bank, which means that \(d_p_e_r_p = 102 m\). Therefore, we can solve for \(v_b_o_a_t\) as follows:

\(v_b_o_a_t sin\theta * 840 s = 102 m\)

\(v_b_o_a_t sin\theta = 102 m / 840 s\)

\(v_b_o_a_t sin\theta = 0.1214 m/s\)

To solve for \(v_b_o_a_t\):-

\(tan\theta = d_p_a_r / d_p_e_r_p\)

\(tan\theta = (v_p_a_r * t) / (v_p_e_r_p * t)\)

\(tan\theta = v_p_a_r / v_p_e_r_p\\tan\theta = v_b_o_a_t cos\theta / (v_b_o_a_t sin(\theta))\\\)

Reserving the value of \(d_p_e_r_p\) and solving for \(\theta\):-

\(\theta = arctan(v_p_a_r / (102 m / 840 s))\\\theta = arctan(v_b_o_a_t cos(\theta) / 0.1214 m/s)\)

\(v_p_a_r^2 + v_p_e_r_p^2 = v_b_o_a_t^2\\(v_b_o_a_t cos(\theta))^2 + (v_b_o_a_t sin(\theta))^2 = v_b_o_a_t^2\)

\(v_b_o_a_t^2 (cos^2(\theta) + sin^2(\theta)) = v_b_o_a_t^2\\v_b_o_a_t^2 = 102^2 / (cos^2(\theta) + sin^2(\theta))\\v_b_o_a_t = \sqrt{[102^2 / {cos^2(\theta) + sin^2(\theta)}]}\)

Reserving the value of \(\theta\):-

\(v_b_o_a_t = \sqrt{[102^2 / {cos^2(\theta) + sin^2(\theta)}]} \\v_b_o_a_t = 9.08 m/s\)

Therefore, the speed of the boat relative to the water is 9.08 m/s.

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