what is the wavelength of 2.6 million Hz ultrasound as it revealed through human tissue ?

The energy stored in the circuit at any time t is given by \(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)The units are in seconds.

The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.

Initially, the capacitor is uncharged, so the second term is zero.

Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.

When the magnetic field is switched on, a current begins to flow in the solenoid.

This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.

Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.

The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.

Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.

This is a first-order linear differential equation, which can be solved to give \(Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))\), where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.

The current in the solenoid is given by I(t) = \(dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).\)

The energy stored in the circuit at any time t is given by\(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)

The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, \((1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.\)

This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.

The solution is t = 1.74 ms.

The units are in seconds.

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Answer:

Because chemical energy is being converted into thermal energy.

Explanation:

Answer:

Assumption: there is no object between this point charge and the observer.

Explanation:

The electric field of a point charge is inversely proportional to the square of the distance from that point charge.

Let \(k\) denote Coulomb's constant (\(k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}\).) Let the magnitude of that point charge be \(q\). At a distance of \(r\) from this charge, the electric field due to this charge would be:

\(\displaystyle E = \frac{k \cdot q}{r^{2}}\).

Convert the magnitude of the point charge in this question to standard units:

\(q = 1\; \rm nC = 10^{-9}\; \rm C\).

Apply that equation to find the magnitude of the electric field due to this point charge:

\(r = 1\; \rm m\):

\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}\).

\(r = 2\; \rm m\):

\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}\).

\(r = 3\; \rm m\):

\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}\).

The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.

Because the \((+1\; \rm nC)\) point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this \((+1\; \rm nC)\!\) point charge at any point in the field would point away from this charge.

When a car is stopped, facing upwards on a hill, the friction acts in the opposite direction to the motion that the car would naturally take if it were not stopped.

In this case, the car would roll backwards down the hill due to the force of gravity. The friction between the tires and the road surface acts in the opposite direction to this motion, providing a force that opposes the car's tendency to roll backwards. Therefore, the friction acts in the forward direction, up the hill, to prevent the car from rolling backwards.

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Answer:

θ = 720º

Explanation:

Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.

Let's reduce the magnitudes to the SI system

         θ = 2 rev (2π rad / 1 rev) = 4π rad

we assume that the pulley has a radius r

linear and angular coordinates are related

         y = θ r

let's calculate

         y = 4π r

to convert the distance in angles to radians

        θ = y / r

        θ = 4π

let's change this angle to degrees

        π rad = 180º

         θ = 4π rad (180º /π rad)

         θ = 720º

Answer:

920000

Explanation:

Each kg contains 1,000 grams

The net force acting on the rocket = 1.6 x 10⁸ N

The rocket starts from rest

The initial speed, u = 0 m/s

Final speed, v = 9.6 km/s

v = 9.6 x 1000 m/s

v = 9.6 x 10³ m/s

The time taken, t = 8 min

t = 8 x 60

t = 480 s

The mass of the rocket, m = 8.0 x 10⁶ kg

The net force is calculated as shown below

The net force acting on the rocket = 1.6 x 10⁸ N

Answer:

net force

Explanation:

net force is the total amount of force exerted on an object.

Answer:

C. 15.0 N.s

Explanation:

What is its change in momentum?15.0 N.s12.0 N.s180 N.s1.30 N.s

The change in momentum of the object is 180 N.s, which is option B.

It is an attribute of a moving body that exists due to its mass and motion and is analogous to the product of the body's mass and velocity.

The change in momentum of an object is equal to the product of the force acting on the object and the time interval during which the force is applied.

Using the given values:

Force = 15.0 N

Time = 12.0 s

Change in momentum = Force x Time

Change in momentum = 15.0 N x 12.0 s

Change in momentum = 180 N.s

Therefore, the change in momentum of the object is 180 N.s.

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Answer:

2.2

Explanation:

input force is 35 N and the output force is 77 N.

Mechanical advantage = output force / input force

Mechanical advantage = 77 N / 35 N

Mechanical advantage = 2.2

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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The index of refraction for the material is 1.612.

The index of refraction (n) of a material can be calculated using the formula:

n = c / v

where c is the speed of light in vacuum and v is the speed of light in the material.

Given that the speed of light in the material is 1.862x\(10^8\) m/s, we can substitute the values into the formula:

n = (3.00x\(10^8 m/s) / (1.862x10^8 m/s\))

Simplifying the expression:

n = 1.612

Therefore, the index of refraction for the material is approximately 1.612.

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Answer:

4161 J/kg·°C

Explanation:

We can use the principle of conservation of energy to solve this problem, which states that the total heat energy in a closed system is constant. The heat lost by the tea is equal to the heat gained by the milk.

Let's first calculate the heat lost by the tea:

Q(tea) = mcΔT

Q(tea) = (0.27 kg)(3910 J/kg·°C)(90.0°C - 85.0°C)

Q(tea) = 6555 J

where m is the mass of tea, c is the specific heat of tea, and ΔT is the change in temperature.

Next, let's calculate the heat gained by the milk:

Q(milk) = mcΔT

Q(milk) = (0.02 kg)(c)(85.0°C - 6.0°C)

Now we can equate the two expressions:

Q(tea) = Q(milk)

6555 J = (0.02 kg)(c)(79.0°C)

Solving for c, we get:

c = 4161 J/kg·°C

Therefore, the specific heat of milk is approximately 4161 J/kg·°C.

The boy throws the ball horizontally. The initial velocity of the ball as it left the boy's hand was approximately 3.72 m/s.

Initial velocity, often represented as v0, is the velocity of an object at the beginning of a time interval or at the start of a motion.

Use the following kinematic equations to arrive at the answer:

Horizontal velocity (Vx) = Distance / Time

Vertical displacement (y) = V0y*t + (1/2)gt²

Vertical velocity (Vy) = V0y + g*t

Tan(theta) = Vy / Vx

where V0y is the initial vertical velocity, g is acceleration due to gravity (9.8 m/s²), and theta is the angle of inclination.

First, let's find the time it takes for the ball to hit the ground. We can use the vertical displacement equation and set y = 0:

0 = V0y*t + (1/2)gt²

Simplifying and solving for t, we get:

t = sqrt((2y) / g)

= sqrt((21.10 m) / 9.8 m/s²)

= 0.472 s

Now, we can use the horizontal velocity equation to find Vx. Since the ball was thrown horizontally, Vx is the same as the initial velocity (V0):

Vx = Distance / Time

= (horizontal distance travelled by ball) / t

We don't know the horizontal distance travelled by the ball, but we can find it using the vertical displacement equation. At the instant the ball hits the ground, its vertical displacement (y) is:

y = V0y*t + (1/2)gt²

= 0 + (1/2)gt²

= (1/2)*9.8 m/s² * (0.472 s)²

= 1.10 m

This means the ball travelled a total distance of:

distance = horizontal distance + vertical distance

= x + 1.10 m

where x is the horizontal distance travelled by the ball. We can find x using the angle of inclination and the vertical displacement:

Tan(theta) = Vy / Vx

Vy = V0y + g*t

Solving for V0y, we get:

V0y = Vy - g*t

Plugging in the numbers, we get:

V0y = Tan(theta) * Vx - g*t

= Tan(30 deg) * Vx - 9.8 m/s² * 0.472 s

= 0.577 * Vx - 4.62 m/s

Now, we can use the vertical displacement equation again to find x:

y = V0yt + (1/2)gt²

= (0.577Vx - 4.62 m/s) * 0.472 s + (1/2)*9.8 m/s² * (0.472 s)²

= 1.10 m

Simplifying and solving for Vx, we get:

Vx = (2y - 0.577V0t) / t

= (21.10 m - 0.577*(0.577*Vx - 4.62 m/s)*0.472 s) / 0.472 s

= 3.72 m/s

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Answer:

The Answer is C. the student changed too many variables.

Explanation:

For the quiz in K12

The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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Answer:

Iron(III) nitrate

Explanation:

The speed of the car just after the gravel is loaded is 2.8m/s.

An isolated system experiences a change in momentum to zero when the starting and final velocities are equal. The reactions between the particles are separated from the surroundings. Momentum is conserved, so we use the rule of conservation of momentum, which is expressed by the equation

   \(P_{f}\)=\(P_{i}\)(1)

Railway is frictionless due to isolated mechanism. The thing travels at a momentum p-based speed while having mass. m

An object's bulk and velocity v are combined to form a vector.Equation of the form gives the velocity.

\(p=mv\)

Using expression p into equation (1)

\(m_{f}v_{f}\)=\(m_{i}v_{i}\)(2)

The final mass of a car is its original mass plus the mass of the gravel added. A car has an initial mass of.10000Kg  preliminary pace is

\(m_{f}\)=10000Kg+4000Kg=14000Kg

initial speed 4m/s

the final speed should be found after solving the equation(2) for \(v_{f}\)

\(v_{f}\)=\(\frac{m_{i}v_{i} }{m_f} }\)=\(\frac{10000Kg(4m/s)}{1400Kg}\)=2.8m/s

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Centripetal force is used for making an object move in a circular path.

Centripetal force is the force on a rotating or orbiting body in the direction of the centre of rotation.

In other words, centripetal force is the force that makes an object move in a circular pattern.

Centripetal force is denoted by "F" and can be calculated by dividing the product of the mass and velocity squared by radius as follows:

F = mv²/r

Where;

Therefore, it can be said that Centripetal force is used for making an object move in a circular path.

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The dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

Start by converting the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Since 1 revolution is equal to 2π radians, we can use the conversion factor:

Angular speed (ω) = (1100 rpm) × (2π rad/1 min) × (1 min/60 s)

ω ≈ 115.28 rad/s

The moment of inertia (I) is given as 2.0 × 10^-7 kg⋅m².

Use the formula for rotational kinetic energy:

Rotational Kinetic Energy (KE_rot) = (1/2) I ω²

Substituting the given values:

KE_rot = (1/2) × (2.0 × 10^-7 kg⋅m²) × (115.28 rad/s)²

Calculate the value inside the parentheses:

KE_rot ≈ (1/2) × (2.0 × 10^-7 kg⋅m²) × (13274.28 rad²/s²)

KE_rot ≈ 1.331 × 10^-3 J

Round the result to the proper number of significant figures, which in this case is three, as indicated by the given moment of inertia.

KE_rot ≈ 4.8 × 10^-4 J

Therefore, the dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

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The maximum number of chips is 33,333

What is a chip?

A chip is a tiny but complex modules that store computer memory or provide logic circuitry for microprocessors.

We need to find the number of wafers that can be cut from a single crystal.

The crystal is 25cm and each wafer is 0.3mm. We perform division to get the number of wafers per crystal after getting everything into the same units. 25cm=250mm.

250mm/0.3mm=833.3333 wafers.

Each wafer yields 400 chips so we multiply 400 chips per wafer by 833.33 wafers to get 33,333 chips.

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Answer:

he fish would travel a horizontal distance of 1.78 meters during the 2 seconds of tail motion

Explanation:

The initial horizontal velocity of the bluefish is 0.45 m/s. When it begins tail motion, it experiences an additional force due to the thrust produced by the tail. The thrust produced by the tail is 0.65 N. We can use Newton's second law to find the acceleration produced by this force:

F = ma

0.65 N = 1.7 kg * a

a = 0.38 m/s^2

This acceleration will cause the velocity of the bluefish to increase over time. The distance the fish travels during this time can be calculated using the kinematic equation:

d = vit + 1/2 at^2

where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time. Since the fish is initially coasting horizontally, its initial vertical velocity is 0 m/s. Therefore, vi = 0.45 m/s. The time interval for which the fish is tail-motoring is not given, so let's assume it is 2 seconds:

d = (0.45 m/s)(2 s) + 1/2 (0.38 m/s^2)(2 s)^2

d = 1.78 meters

Therefore, the fish would travel a horizontal distance of 1.78 meters during the 2 seconds of tail motion.

Answer:

Explanation:

Loss of potential energy = mgh.

h = d sin 20

= 400 sin20 = 136.8 m

Loss of potential energy = m x 9.8 x  136.8

= 1340.64 m

negative work done by friction = μ mg cosθ x d

= .2 x m x 9.8 x cos 20 x 400

= 736.72 m

Net loss of potential energy = 1340.64 m - 736.72 m

= 603.92 m

= gain of kinetic energy = 1/2 m v²

1/2 m v² = 603.92 m

v² = 1207.84

v = 34.75 m /s .

The statement" The direction of the force on the charge placed in the electric field is upward" is false because the direction of the force on a negative charge (-6 C) placed in an upward-directed uniform electric field of magnitude 24 N/C would be downward.

The direction of the force on a charged particle placed in an electric field is determined by the charge of the particle and the direction of the electric field. In this case, a charge of -6 C is placed in an electric field directed upward with a magnitude of 24 N/C.

The force on a charged particle in an electric field can be calculated using the formula:

F = q * E

Where F is the force, q is the charge of the particle, and E is the electric field.

Since the charge q in this case is negative (-6 C) and the electric field E is directed upward, we can substitute the values into the formula:

F = (-6 C) * (24 N/C)

F = -144 N

The negative sign in the force value indicates that the force is in the opposite direction to the electric field. Therefore, the force on the charge placed in the electric field is downward, not upward.

The force on a negative charge is always opposite to the direction of the electric field. This is because negative charges experience an attractive force towards positive charges, and electric fields are directed from positive charges to negative charges.

Therefore, the statement "The direction of the force on the charge placed in the electric field is upward." is false.

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Answer:

Explanation:

Let the initial rotational inertia be I and final rotational inertia be I / 6 .

Let the initial angular velocity be ω₁ and final angular velocity be ω₂.

Applying conservation of angular momentum law

I x ω₁ = I / 6 x ω₂

6 ω₁  = ω₂

initial rotational kinetic energy = 1/2 I x ω₁ ²

Final  rotational kinetic energy = 1/2 ( I / 6 ) x ω₂ ²

Final  rotational kinetic energy / initial rotational kinetic energy

= ( 1 / 6 ) x ω₂ ² / ω₁ ²

=  ω₂ ² / 6ω₁ ²

= 36 ω₁ ² / 6ω₁ ²

= 6 .

The ratio will be "6". A further solution is below.

Let,

The initial rotational inertia,

The final rotational inertia,

By applying the conservation of angular momentum law, we get

→ \(1\times \omega_1 = \frac{1}{6}\times \omega_2\)

→    \(6 \ \omega_1 = \omega_2\)

Now,

Initial rotational K.E = \(\frac{1}{2}1\times \omega_1^2\)

Final rotational K.E = \(\frac{1}{2} (\frac{1}{6} )\times \omega_2^2\)

hence,

→ \(\frac{Final \ rotational \ K.E}{Initial \ rotational \ K.E} = \frac{(\frac{1}{6} )\times \omega_2^2}{\omega_1^2}\)

                                  \(= \frac{\omega_2^2}{6 \omega_1^2}\)

                                  \(= 6\)

Thus the answer above is right.

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Explanation:

Current flowing through the circuit, I = 2A Applying Kirchoff’s Ist law at junction P, I = I1 + I2 I2 = I – I1 …(1) Applying Kirchoff’s Ind law at junction PQSP 5 I1 + 10 Ig – 15 I2 = 0 5 I1 + 10 Ig -15(I – I1) = 0 20 I1 + 10 Ig = 15 I 20 I1 + 10 Ig = 15 x 2 ÷ by 10 21 I1 + Ig = 3 … (2) Applying Kirchoff’s II law at junction QRSQ 10(I1 – Ig ) – 20(I – I1 – Ig ) – 10 Ig = 0 10 I1 – 10g – 20(I – I1 – Ig ) – 10 Ig = 0 10 I1 – 10 – 20 I + 20 I1 -20Ig – 10 Ig = 0 30 I1 – 40 Ig = 20 I ÷ by 10 ⇒ 3 I1 – 4 Ig = 20 I 3 I1 – 4 Ig = 2 I 3 I1 – 4 Ig = 2 x 2

hope this helps :)

Answer:

The brain is a part of the body that is particularly sensitive to high temperature. Hence some ungulates, like the Thomson's gazelle, use a counter-current heat exchanging structure known as the carotid rete to keep the brain cooler than the body.The cooled arterial blood then continues toward the brain.

The acceleration of the aircraft is determined as -8.2 m/s².

The acceleration of the aircraft is calculated as follows;

a = Δv/t

where;

a = (5 - 250)/(30)

a = -8.2 m/s²

Thus, the acceleration of the aircraft is determined as -8.2 m/s².

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Explanation:

PE=mgh

5=m(9.8)(2)

m=5/19.6

m=0.2251 kg

m=225.1 grams