Check the correctness of the following equation , a=v^2/r, where a=acceleration, v=velocity, r= radius

The resistors are connected to a constant voltage of magnitude V the RA is 2.409Ω. The equivalent resistance RB of the new resistor network when the switch is open RB is 13.5 Ω. The equivalent resistance Rc of the resistor network described in Part B when the switch is closed RC is 8.6 Ω.

A)

The entire resistance of two resistors in parallel is given by means of:

1/RA = 1/(R1+R2) + 1/(R3+R4)

Substituting the expressions we derived above, we get:

1/RA = 1/[2.00Ω + (5.00Ω/12)V] + 1/[1.00Ω + (7.00Ω /22)V]

Solving for RA, we get:

RA = 2.409Ω

B)

Equivalent resistance, RB = (R1 + R6) R2/(R1 + R6 + R2) + R3 + R4 + R5

RB = (2 + 3) x 5/(2 + 3 + 5) + 1 + 7 + 3

RB = 13.5 Ω

C)

Equivalent resistance, RC = (R1 + R6) R2/(R1 + R6 + R2) + R3 + R7 x R4/(R7 + R4) + R5

RC = (2 + 3) x 5/(2 + 3 + 5) + 1 + (3 x 7/(3+7)) + 3

RC = 8.6 Ω

A resistor is an electrical component that restricts the flow of electrical current in a circuit. It is typically made of a material that has a high resistance to the flow of electricity, such as carbon, metal, or ceramics. They can be used to control the amount of current flowing through a circuit, to limit voltage, to adjust the gain of an amplifier, or to provide a load in a circuit. They are also used in electronic filters, timing circuits, and signal processing applications.

Resistors come in a variety of shapes and sizes, including through-hole, surface mount, and wirewound resistors. They are often color-coded to indicate their resistance value and tolerance, and their wattage rating determines the amount of power they can safely handle without overheating.

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To determine the time Pete arrives at work, we can start by calculating the total time he spends on his commute and gym routine:

Time spent driving to the gym = 12.5 miles ÷ average speed

We don't know Pete's average speed, so we cannot calculate this.

Time spent in the gym = 45 minutes

Time spent driving from the gym to work = 9 miles ÷ average speed

Again, we don't know Pete's average speed, so we cannot calculate this.

Total time spent on commute and gym routine = time spent driving to gym + time spent in gym + time spent driving from gym to work

= Unknown + 45 minutes + Unknown

Next, we can convert the total time to hours and minutes:

Total time = (Unknown + 45 minutes + Unknown) ÷ 60

= (Unknown + Unknown) ÷ 60 + 45/60

= (2Unknown) ÷ 60 + 0.75

= (Unknown) ÷ 30 + 0.75

We know that Pete needs to arrive at work by 9.00am, so we can set up an equation:

Arrival time = 7.30am + Total time

9.00am = 7.30am + (Unknown/30) + 0.75

Solving for Unknown:

1.5 hours = Unknown/30

Unknown = 45 minutes

Therefore, Pete will arrive at work at 8.15am.

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Answers :-

The blanks for question 7 is as follows:-

1st row : Charge of the ion = +3

2nd row : Chemical symbol = I-

3rd row : Name of ion = Magnesium ion and type of ion = cation .

4th row : Chemical symbol = K+ , charge of the ion = +1 and type of ion = cation .

5th row : Name of ion = Nitride ion , charge of the ion = -3 , type of ion = anion .

and we are done!

The velocity, period, and total energy of a hydrogen-like atom with atomic number z can be calculated with the following formulas.

Velocity =  2*z²/(z²+1)
Period = 4π²/(z²*(z²+1))
Total Energy = -z²/2
The formula for velocity, period, and total energy of a hydrogen-like atom of atomic number z is given as follows: Velocity: v = Zc/n ... (1)Period: T = 2πa/v ... (2)

Total energy: E = -me^4z^2/8εo^2h^2n^2 ... (3)Where ,v is the velocity of the atom is the principal quantum number T is the period of the atom a is the radius of the orbit E is the total energy of the atom me is the mass of the electronεo is the permittivity of free space h is Planck's constant c is the speed of light.

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Answer:

the power output of the car is 29.43 kW (rounded to two decimal places).

Explanation:

To find the power output of the car, we need to use the formula:

power = work / time

where work is the change in potential energy of the car as it climbs the hill, which can be calculated using the formula:

work = force x distance

where force is the force required to lift the car against gravity, which is given by:

force = mass x gravity

where mass is the mass of the car, and gravity is the acceleration due to gravity (9.81 m/s^2).

So, the force required to lift the car against gravity is:

force = 1000 kg x 9.81 m/s^2 = 9810 N

The distance the car travels up the hill is 30 m.

Therefore, the work done by the car is:

work = force x distance = 9810 N x 30 m = 294300 J

The time taken by the car to climb the hill is 10 s.

Therefore, the power output of the car is:

power = work / time = 294300 J / 10 s = 29430 W

The blocks accelerate at the same charge for the reason that they're linked. The acceleration is a value between zero and g.

Acceleration is a physical concept that refers to the rate of change of an object's velocity with respect to time. When an object's velocity changes, either by speeding up or slowing down, it is said to be accelerating.

Acceleration plays an important role in many aspects of physics, from the motion of celestial bodies to the behavior of particles in a particle accelerator.  The magnitude of acceleration is the rate at which an object's velocity changes, and it is measured in units of meters per second squared (m/s^2) in the International System of Units (SI). There are several factors that can cause an object to accelerate, such as a force acting on it, a change in its direction of motion, or a combination of both.

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Answer:

10m

Explanation:

r=√Gm1m2Fg

r=√[(6.7×10−11 N m2 kg−2)(5000 kg)(5000 kg) / 1.6×10−5 N}

False. E=hf, where h is Planck's constant, c is the speed of light, f is the frequency, and is the wavelength; and E=hc/, where E is directly proportional to frequency and inversely proportional to wavelength.

With respect to the wavelength of the radiation, photon energy is inversely proportional.

Two formulas can be used to determine a photon's energy: E = h f is a formula that can be used if the photon's frequency is known. This equation, sometimes known as Planck's equation, was created by Max Planck.

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Explanation:

hope its help thank you

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Explanation:

The size of a launch window is determined by a variety of factors, including the position of the launch site, the desired orbit, the position of the destination, and the characteristics of the spacecraft being launched.

One of the most important factors is the position of the launch site relative to the desired orbit. The launch site must be positioned in such a way that the rocket can achieve the required velocity and trajectory to reach the desired orbit. The angle and speed at which the rocket is launched are also crucial, as they affect the amount of fuel required and the trajectory of the rocket.

The position of the destination is another factor that affects the size of the launch window. For example, if the spacecraft is bound for a planet that is moving in its orbit, the launch window must be adjusted to account for the changing position of the planet.

In addition, the characteristics of the spacecraft being launched, such as its size, weight, and propulsion system, can also affect the size of the launch window. A larger spacecraft may require more fuel and a longer burn time, which may limit the available launch window.

Overall, the size of a launch window is determined by a complex set of factors, including the position of the launch site, the desired orbit, the position of the destination, and the characteristics of the spacecraft being launched. Launch planners use sophisticated computer models and simulations to calculate the optimal launch window based on these factors.

The potential V(x, y, z) everywhere inside the box. Formulas give V=0 at the center of this cube. Is E=0 there[tex](A_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{16V_{0}}{nm\pi^{2}}\: \: \: n,m =odd[/tex]

Laplace equation in cartesian co-ordinates is

[tex]\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=0[/tex]

Multiply both side by [tex]sin\left ( \frac{n'\pi x}{a} \right )sin\left ( \frac{m'\pi z}{a} \right )[/tex]   and integrate over x and z from 0 to a

[tex]\int_{0}^{a}\int_{0}^{a}V_{0}sin\left ( \frac{n\pi x}{a} \right )sin\left ( \frac{m\pi z}{a} \right )dxdz=\frac{a^{2}}{4}(A_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})[/tex]

[tex](A_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{4V_{0}}{a^{2}}\int_{0}^{a}sin\left ( \frac{n\pi x}{a} \right )dx \int_{0}^{a}sin\left ( \frac{m\pi z}{a} \right )dz[/tex]

[tex](A_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{16V_{0}}{nm\pi^{2}}\: \: \: n,m =odd[/tex]

Now apply the final boundary condition V(x, y=a/2, z) = V0

Solving we get

[tex](A_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{16V_{0}}{nm\pi^{2}}\: \: \: n,m =odd[/tex]

The Laplace equation is a partial differential equation that describes the behavior of a scalar field in space. In its simplest form, it states that the sum of the second partial derivatives of the scalar field with respect to each of the spatial dimensions is equal to zero. This means that the scalar field has no sources or sinks, and its value is determined only by the boundary conditions.

The Laplace equation has many applications in physics, engineering, and mathematics. For example, it can be used to model the behavior of electric and gravitational fields, fluid flow, and heat transfer. It is also used in solving problems involving potential functions, which arise in many areas of physics and engineering.

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Complete Question: -

You have a cubical box (sides all of length a) made of 6 metal plates which are insulated from each other. The left wall is located at y=-a/2, the right wall is at y=+a/2. Both left and right walls are held at constant potential V=V0. All four other walls are grounded. Find the potential V(x, y, z) everywhere inside the box. Do your formulas give V=0 at the center of this cube? Is E=0 there? (Should they be??)

The time that was taken for the movement of the item is observed as 3 seconds.

The equations of motion describe the motion of objects in terms of their position, velocity, acceleration, and time.

For the equation;

v = u + at

This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and time (t). If three of these variables are known, the equation can be rearranged to solve for the unknown variable.

We know that;

v = u - gt

We know that the object would come to rest after being thrown.

0 = 30 - 10t

-30 = - 10t

t = 3 seconds

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a) We can use Coulomb's law to calculate the electric field vector at the point (8,16) m due to the point charge placed on the x-y plane.

The electric field vector is given by E = F/q, where F is the force exerted on the point charge and q is the magnitude of the charge. The force exerted on the charge is (21 + 4j) N. The magnitude of the charge is given by q = F/E, where E is the electric field at the point (8,16) m. Therefore, we have:

E = F/q = (21 + 4j) N / (-15 nC) = (-1.4 - 0.267j) x 10⁶ N/C

So, the electric field vector at the point (8,16) m is (-1.4 - 0.267j) x 10⁶N/C.

b) To determine the magnitude and sign of the point charge that produces the electric field calculated in part (a), we can use the formula for the electric field of a point charge. The electric field at a point P due to a point charge q located at the origin is given by:

E = kq/r²

q is the charge of the point charge, and r is the distance between the point charge and point P. We can rearrange this equation to solve for q:

q = Er²/k

for E and r (r = sqrt(8² + 16²) = 17.89 m) we get:

q = (-1.4 - 0.267j) x 10^6 N/C x (17.89 m)² / (8.99 x 10⁹ N m²/C²) = -5.37 nC

So, the magnitude of the point charge is 5.37 nC and its sign is negative, indicating that it is an additional negative charge placed at the origin that produces the electric field calculated in part (a).

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The electric field vector at the point (8, 16) m is (-5.53i - 11.07j) N/C. and

the magnitude of the point charge is 2.11 nC and the sign is negative, indicating that it is the same as the original point charge placed on the x-y plane.

The steps are as following to calculate the given question :-

a. To calculate the electric field vector at the point (8, 16) m due to the -15 nC point charge, we can use Coulomb's law:

The distance between the two points is given by:

r = sqrt[(8-0)^2 + (16-0)^2] = 17.8885 m

The electric field vector is given by:

E = k*q/r^2 * r_hat

where k is the Coulomb constant (k = 9x10^9 N*m^2/C^2), q is the charge of the point charge, r_hat is the unit vector pointing from the point charge to the point of interest.

Since the point charge is negative, the electric field vector points towards the point charge. Therefore, r_hat = -icosθ - jsinθ, where θ is the angle between the vector pointing from the point charge to the point of interest and the x-axis.

θ = atan2(16, 8) = 63.43 degrees

So, r_hat = -0.4472i - 0.8944j

Plugging in the values, we get:

E = (9x10^9 Nm^2/C^2)(-15x10^-9 C)/(17.8885m)^2 * (-0.4472i - 0.8944j)

E = -5.53i - 11.07j N/C

Therefore, the electric field vector at the point (8, 16) m is (-5.53i - 11.07j) N/C.

b. To find the magnitude and sign of the point charge that produces this electric field, we can use the formula:

E = k*q/r^2

where E is the magnitude of the electric field, k is the Coulomb constant, q is the charge of the point charge, and r is the distance between the point charge and the point of interest.

Plugging in the values, we get:

E = (9x10^9 N*m^2/C^2)*q/(17.8885m)^2

-11.07 N/C = (9x10^9 N*m^2/C^2)*q/(17.8885m)^2

Solving for q, we get:

q = -2.11x10^-9 C

Therefore, the magnitude of the point charge is 2.11 nC and the sign is negative, indicating that it is the same as the original point charge placed on the x-y plane.

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Option C, Because of the wavelength of radio waves and other forms of electromagnetic radiation, diffraction is most likely to occur as the wave travels through a classroom doorway.

Diffraction is the bending of waves around barriers or through apertures that are equivalent to or smaller than the wavelength of the wave.

Because radio waves have longer wavelengths than visible light and X-rays, they are more likely to diffract while passing through a similar-sized aperture, such as a classroom doorway.

Because X-rays have considerably shorter wavelengths and visible light has wavelengths in between, diffraction is less likely to occur in this scenario for these forms of electromagnetic energy. As a result, option C is the right answer.

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The y-component of the initial velocity of C be if all three objects are to end up moving at 0.50 m/s in the y-direction after the collision with the velocity -0.44 m/s.

Part A,

the x-component of the initial velocity of C must be 0.26 m/s. To answer Part B, the y-component of the initial velocity of C must be -0.44 m/s.

To solve this problem, we can use the law of conservation of momentum. This states that the total momentum before the collision is equal to the total momentum after the collision.

We can use this to calculate the velocity of C in each direction.

We know that A and B have an initial velocity in the x-direction of 0.50 m/s and 1.50 m/s respectively, and the velocity in the y-direction is 0 m/s for both. We also know that the total mass is 0.100 kg. So the total initial momentum in the x-direction is:

[tex]Momentum_x = (mass_A x velocity_A_x) + (mass_B x velocity_B_x)[/tex]

= (0.020 kg x 0.50 m/s) + (0.030 kg x 1.50 m/s) = 0.080 kg m/s

We also know that the final velocity of the three objects is 0.50 m/s in the x-direction and the total mass is 0.100 kg. So the total final momentum in the x-direction is:

[tex]Momentum_x = (mass_total x velocity_final_x)[/tex] = (0.100 kg x 0.50 m/s) = 0.050 kg m/s

Using the law of conservation of momentum, we can solve for the velocity of C in the x-direction:

0.080 kg m/s = [tex](mass_C x velocity_C_x) + 0.050 kg m/s velocity_C_x[/tex] = (0.080 kg m/s - 0.050 kg m/s) / 0.050 kg = 0.26 m/s

Part B,

we can do the same process in the y-direction. We know that the initial velocities of A and B are 0 m/s in the y-direction, and the total mass is 0.100 kg.

So the total initial momentum in the y-direction is:

[tex]Momentum_y = (mass_A x velocity_A_y) + (mass_B x velocity_B_y)[/tex]

= (0.020 kg x 0 m/s) + (0.030 kg x 0 m/s) = 0 kg m/s

We also know that the final velocity of the three objects is 0.50 m/s in the y-direction and the total mass is 0.100 kg.

So the total final momentum in the y-direction is:

[tex]Momentum_y = (mass_total x velocity_final_y)[/tex] = (0.100 kg x 0.50 m/s) = -0.050 kg m/s

Using the law of conservation of momentum, we can solve for the velocity of C in the y-direction:

0 kg m/s =[tex](mass_C x velocity_C_y)[/tex] + (-0.050 kg m/s)

[tex]velocity_C_y[/tex] = (-0.050 kg m/s) / 0.050 kg = -0.44 m/s

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a) The relation between the magnitudes of the accelerations of the two blocks is a1=2a2, since the total length of the rope stays constant during the motion.

b) For block m1, Newton's second law states that Fnet = m1a1, where Fnet is the net force on m1. Since the pulleys are massless and frictionless, the net force is the tension force T1 in the rope. Therefore, T1 = m1a1.
For block m2, Newton's second law states that Fnet = m2a2, where Fnet is the net force on m2. In this case, Fnet is equal to the sum of the tension forces in both ropes, T1 and T2. Therefore, T1 + T2 = m2a2.
Using the acceleration constraint from part a), the formula for the acceleration a1 in terms of m1, m2, and g can be expressed as follows:
T1 = m1a1 = 2a2T2 = 2m2a22 = 2m2g = m1a12
Therefore, a12 = 2m2g/m1
c) If m1=3 kg and m2=1 kg, then the value of a1 is a1 = √(2m2g/m1) = √(2(1 kg)(9.8 m/s2)/(3 kg)) = 1.5 m/s2.

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Answer: 1,050W is the power rating of immersion heater

Answer:

Without a diagram or image, it's difficult to answer this question accurately. However, I can provide a general answer based on the information given.

a. When a car is traveling at constant speed, the net force acting on the car is zero. Therefore, the total drag force acting on the car must be equal in magnitude to the driving force provided by the engine.

b. i. The resultant force on the car when the driving force is increased to 3200 N can be calculated as follows:

Resultant force = Driving force - Drag force

Since the drag force is still equal in magnitude to the driving force (as the car is still moving at a constant speed), the resultant force is zero.

Resultant force = 3200 N - 3200 N = 0 N

ii. The equation connecting mass, force, and acceleration is:

Force = mass x acceleration

This can be rearranged to find acceleration:

Acceleration = Force / mass

iii. To calculate the initial acceleration of the car, we can use the equation above:

Acceleration = 3200 N / 800 kg = 4 m/s²

c. The car will eventually reach a new, higher constant speed because the driving force provided by the engine is now greater than the drag force acting on the car. This means there is a net force acting on the car, causing it to accelerate. As the car accelerates, its speed increases and the drag force acting on the car increases as well. Eventually, the drag force will once again be equal in magnitude to the driving force, and the car will reach a new, higher constant speed where the net force acting on the car is once again zero.

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Explanation:

Ff = mg ( .25)

KE = 1/2 m (2.5)^2

Work of friction    mg (.25) X (d)      must equal the KE for the disc to stop

  mg(.25)(d) = 1/2 m ( 2.5)^2   <=== Divide both sides by 'm'

    g (.25)(d) = 1/2 ( 2.5)^2

    9.81 (.25) d = 1/2 ( 2.5)^2

      d = 1.27 m

Two parallel-plate capacitors with the same charge Q but different areas (A and 2A) can be compared by looking at the capacitance. The capacitance of the second capacitor is double that of the first due to the increase in area.

Two parallel-plate capacitors with the same charge Q but different areas (A and 2A) can be compared by looking at the capacitance, which is defined as the ratio of the charge stored on the capacitor to the voltage applied across the plates. The capacitance C of a capacitor is given by the equation C=Q/V. Therefore, the capacitance of the first capacitor, C1, is C1=Q/V, and the capacitance of the second capacitor, C2, is C2=(2Q)/V. It is seen that the capacitance of the second capacitor is double that of the first. This is because the area of the second capacitor is double that of the first. Therefore, the same charge Q stored on the first capacitor is distributed over twice the area in the second capacitor, resulting in the capacitance being double. This can be mathematically expressed as C2 = 2C1. Thus, two parallel-plate capacitors with the same charge Q but different areas (A and 2A) can be compared by looking at the capacitance. The capacitance of the second capacitor is double that of the first due to the increase in area.

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The above-listed characteristics apply to moving water as an energy source. Thus, moving water can be used as an energy source.

The following are the characteristics of moving water as an energy source:

Water is an excellent resource for producing electricity since it is clean, renewable, and is available in many different forms. When water moves, it has the potential to generate energy, which can be harnessed in several ways to produce electricity. As a result, moving water is an excellent source of renewable energy, as it is available in many different forms and can be used in a variety of ways.The above-listed characteristics apply to moving water as an energy source. Thus, moving water can be used as an energy source.

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The physical processes by which atoms rearrange during phase transformations in the solid state involve changes in the arrangement of the atoms in the lattice, which can be caused by changes in temperature, pressure, or both.

The physical processes by which atoms rearrange during phase transformations in the solid state involve changes in the arrangement of the atoms in the lattice.

This is typically done by changing the number of nearest neighbours of each atom or by introducing new lattice points in the solid structure. In some cases, atoms may even have to move from one position to another.

Common examples of phase transformations in the solid state include melting, recrystallization, and solidification.

Melting occurs when the thermal energy of the solid is increased and the atoms become mobile enough to break the bonds between them. This causes the solid to transition into a liquid phase.

Recrystallization occurs when the thermal energy of the solid is decreased, causing the atoms to return to their original positions and form a new, more ordered lattice.

Lastly, solidification is the reverse process of melting, where thermal energy is removed and the atoms return to their original positions in the lattice.

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Everything but blue & orange would now be present in the spectrum of light observed.

Spectrum refers to a range of different wavelengths of electromagnetic radiation. Electromagnetic radiation is a form of energy that travels through space and includes different types such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Each type of electromagnetic radiation has a different wavelength and frequency, and together they make up the electromagnetic spectrum.

The concept of spectrum is used in a variety of fields, including physics, astronomy, and telecommunications. The spectrum of electromagnetic radiation is essential for many technologies, such as radios and televisions, cell phones, and medical imaging devices, as they all rely on the transmission and reception of specific wavelengths of electromagnetic radiation.

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Complete Question: -

Imagine that the blue light and orange light from the source were blocked. What color(s) would now be present in the spectrum of light observed?

Answer:B. Microwaves

Explanation:

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The speed of the ball when it leaves the gun is approximately 17.66 m/s.

To determine the speed of the ball when it leaves the gun, we can use the principle of conservation of energy. At the moment when the trigger is pulled, the spring has a potential energy equal to:

U = (1/2) k x²

where k is the spring constant and x is the amount by which the spring is compressed from its unstretched length. Using the given values, we have:

U = (1/2) (20 N/m) (0.0125 m)² = 1.5625 mJ

When the spring is released, this potential energy is converted into kinetic energy of the ball, as well as work done against friction. The work done against friction is equal to:

W = f_friction x

where f_friction is the frictional force between the ball and the barrel, and x is the distance that the ball moves along the barrel. Using the given values, we have:

W = (0.02 kg) (9.81 m/s²) (0.20) (0.05 m) = 0.01962 J

The kinetic energy of the ball is equal to the difference between the potential energy of the spring and the work done against friction:

K = U - W = 1.5625 mJ - 0.01962 J = 1.5429 J

The speed of the ball when it leaves the gun is equal to the:

v = sqrt(2K/m)

where m is the mass of the ball. Using the given value of 20 g, we have:

m = 0.02 kg

Substituting the values and evaluating the expression, we get:

v = sqrt(2(1.5429 J)/(0.02 kg)) = 17.66 m/s

Therefore, the speed of the ball when it leaves the gun is approximately 17.66 m/s.

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Answer:

Explanation:

We can calculate the change in the man's center of gravity by considering the initial and final positions of the center of gravity of his arms.

Assuming the man's arms are initially hanging down by his sides, the center of gravity of his arms is located at the midpoint of the cylinder, which is at a distance of L/2 = 79/2 = 39.5 cm from the shoulder joint.

When the man raises his arms straight up, the center of gravity of his arms is located at the top of the cylinder, which is at a distance of L = 79 cm from the shoulder joint.

The change in the man's center of gravity is therefore:

Δh = h_final - h_initial

= L - L/2

= 79 cm - 39.5 cm

= 39.5 cm

Therefore, the man raises his center of gravity by 39.5 cm when he raises both his arms from hanging down to straight up.

To find:-

Answer:-

We are here given that two aeroplanes are travelling in opposite directions to each other that is North and South .

Speed to pilot traveling towards north is ,

[tex]\implies v_N = 400\ km/hr \\[/tex]

Speed of pilot traveling towards south is ,

[tex]\implies v_S = 300\ km/hr \\[/tex]

Since they are traveling opposite directions the relative speed will be the sum of their speeds that is ,

[tex]\implies v_{relative} = 300km/hr + 400 km/hr \\[/tex]

[tex]\implies \underline{\underline{ v_{relative}= 700km/hr}} \\[/tex]

Hence the relative speed is 700km/hr .

and we are done!

The bearing stress in the given screw is 3.32 N/mm².
The bearing stress for a screw with a mean diameter of 6 mm, a pitch of 1 mm, and 3 engaged threads carrying a load of 275 n is calculated using the formula:

Bearing Stress (σ) = (Load / (π * Mean Diameter * No. of Engaged Threads))
σ = (275 N / (π * 6 mm * 3))
σ = 23.31 MPa (megapascals)
The bearing stress in a screw with mean diameter 6 mm and pitch 1 mm that is carrying a load of 275 N can be calculated as follows:Given,Mean diameter, d = 6 mmPitch, p = 1 mmLoad, W = 275 NNumber of engaged threads, n = 3The formula to calculate the bearing stress is given by;`Bearing stress = W/(A * n)`Where A is the area of the threaded section of the screw. It is given by;`A = (π/4) * (d - 0.9382p)²

`Now, substitute the given values in the formula for A.`A = (π/4) * (6 - 0.9382 × 1)²`Solving the above equation, we get`A = 26.22 mm²`Now, substitute the values of A, W, and n in the formula for bearing stress.`Bearing stress = W/(A * n)``= 275/(26.22 × 3)``= 3.32 N/mm²`

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Answer:

The linear velocity of blood in the aorta can be calculated using the equation:

v = Q / A

where v is the linear velocity, Q is the volume flow rate, and A is the cross-sectional area of the vessel.

The volume flow rate Q can be calculated using the equation:

Q = SV / t

where SV is the stroke volume and t is the duration of systole.

The cross-sectional area of the aorta can be calculated using the equation:

A = πr^2

where r is the radius of the aorta.

Given that the radius of the aorta is 1.5 cm, the stroke volume is 60 ml, and the duration of systole is 0.25 s, we can calculate the volume flow rate Q:

Q = SV / t = 60 ml / 0.25 s = 240 ml/s

Converting the units of Q to cm^3/s:

Q = 240 ml/s × 1 cm^3/1 ml = 240 cm^3/s

We can then calculate the cross-sectional area of the aorta:

A = πr^2 = π × (1.5 cm)^2 = 7.07 cm^2

Finally, we can calculate the linear velocity of blood in the aorta:

v = Q / A = 240 cm^3/s / 7.07 cm^2 = 33.9 cm/s

Therefore, the linear velocity of blood in the aorta is 33.9 cm/s.