If Mars were the same size as Mercury (instead of its actual size), which surface features would it have?

Complete question:

In the circuit shown in the figure below (See image attached), suppose that the value of R1 is [tex] 500\,k\Omega [/tex]. To obtain a multimeter reading of 1 V between points B and C in the circuit, the value of R2 would have to be.

Answer:

[tex]R2=0.1\Omega [/tex]

Explanation:

First, we are going to the find current trough the circuit, because the resistors are on series the current is the same on each resistor so I=I1=I2. The Ohm's law for the circuit is:

[tex] V=R_{T}*I [/tex] (1) , with V the voltage of the battery (6V), I the current trough the circuit and [tex]R_{T} [/tex] the total resistance of the circuit, but for resistors on series the total resistance is the sum of the individual resistance so [tex] R_{T} = R1+R2[/tex] (2).

Using (2) on (1) and solving for I:

[tex]I=\frac{V}{R1+R2} [/tex] (3)

Ohm's law is true for the individual resistors too so we're going to apply that on R2:

[tex]V2=R2*I2 [/tex], but remember I2=I

[tex]V2=R2*I [/tex] (4), using (3) on (4)

[tex]V2=R2* \frac{V}{R1+R2} [/tex], solving for R2:

[tex] R1*V2+R2*V2=R2V[/tex]

[tex] R1*V2=R2(V-V2)[/tex]

[tex] R2=\frac{R1V2}{V-V2}=\frac{500\times10^{3}\Omega*1V}{6V-1V}[/tex], V2= 1V because we want that reading on the multimeter.  

[tex]R2=100\,k\Omega [/tex]

Answer:

10k

Explanation:

Answer:

The answer is: The increased voltage causes an increase in power usage, and the device will over-heat.

Explanation:

First, we must consider the variables of the electrical system that will allow us to respond. In this case, power, current and voltage, which are related by

[tex]P=VI[/tex]

Where P=Power, V=Voltage, I=Current.

In the equation it can be observed that power is directly proportional to the system voltage. Thus, if the voltage increases as in this case, the power will also increase, which overheats the device and can cause damage to it.

Answer: Astronauts only float around in the shuttle when they are outside the gravitational pull of the earth

Explanation: when astronauts takes off from the earth, they get to a point (space) where the earth's gravity can no longer pull them. At this state, they experience weightlessness because there is no gravity. Since there is no gravity to pull them down, hence they start floating.

Astronauts and objects float in the shuttle due to the microgravity state from continual free-fall around the Earth, creating the feeling of weightlessness.

Astronauts and objects like cans of soda float in the shuttle because of the lack of gravity in space, a state known as microgravity. When the shuttle is orbiting the earth, it's actually falling towards the earth but also moving forward. This forward motion allows the shuttle and everything in it to keep missing the Earth, so they keep falling towards it but never hitting it. This continual state of free-fall creates the feeling of weightlessness and is why astronauts and objects inside the shuttle appear to float.

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Final answer:

The question is related to determining the critical value of x, or xcritical, that ensures a bar with a heavy block remains stable by balancing torques around the pivot point.

Explanation:

The question asks about stability conditions in a physical system involving a block on a bar and requires the application of concepts such as the center of mass, force equilibrium, and tension. When the mass of the block is too large and positioned too close to the left end of the horizontal bar, the system may become unstable, potentially causing the bar to tilt. To determine the smallest value of x, denoted as xcritical, we must set up equations considering the torques about the pivot point, ensuring that the sum of torques must equal zero for stability. This involves calculating the torques due to the weight of the bar, the weight of the block, and any other forces acting on the system. The position of the center of mass of the bar is crucial as it influences the bar's propensity to rotate and the distribution of weight. Hence, xcritical is that specific distance at which the torques balance each other out, and the system remains in a stable, horizontal state.

Answer:

409.87803 m/s

Explanation:

v = Velocity of bullet

L = Latent heat of fusion = 6 cal/g

c = Specific heat of lead = 0.03 cal/g°C

[tex]\Delta T[/tex] = Change in temperature = (327-27)

m = Mass of bullet

[tex]1\ J=4.2\ J/cal[/tex]

The heat will be given by the kinetic energy of the bullet

[tex]Q=\dfrac{1}{2}mv^2[/tex]

According to the question

[tex]Q=0.75\dfrac{1}{2}mv^2[/tex]

This heat will balance the heat going into the obstacle

[tex]Q=mc\Delta T+mL\\\Rightarrow 0.75\dfrac{1}{2}mv^2=m(c\Delta T+L)\\\Rightarrow v^2=\dfrac{2}{0.75}\times (0.03\times (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40\times 4.2\times 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=\sqrt{168000}\\\Rightarrow v=409.87803\ m/s[/tex]

The speed of the bullet is 409.87803 m/s

Explanation:

when a main-sequence star exhausts its core hydrogen fuel supply the core starts to shrink ( lack of fusion reactions) and the rest of the star starts to expands. The fusion reaction leaves the main sequence and begin to fuse helium in a shell outside the core. This mass stars become red supergiant  and then evolve to become blue super giant.

Answer:  Their temperature decreases dramatically, but their luminosity increases only slightly.

Explanation:  This is exact from Plato

Answer:

Induced EMF,[tex]\epsilon=6.28\ volts[/tex]

Explanation:

Given that,

Radius of the circular loop, r = 1 m

Time, t = 2 s

Initial magnetic field, [tex]B_i=2\ T[/tex]

Final magnetic field, [tex]B_f=6\ T[/tex]

The expression for the induced emf within the circular loop is given by :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

[tex]\phi[/tex] = magnetic flux

[tex]\epsilon=\dfrac{d(BA\ cos\theta)}{dt}[/tex]

Here, [tex]\theta=90\ degrees[/tex]

[tex]\epsilon=A\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=A\dfrac{B_f-B_i}{t}[/tex]

[tex]\epsilon=\pi (1)^2\times \dfrac{6-2}{2}[/tex]

[tex]\epsilon=6.28\ volts[/tex]

So, the induced emf in the loop is 6.28 volts. Hence, this is the required solution.  

Answer:

a) [tex]V_s=4989.7895\ in^3[/tex]

b) [tex]mass=0.1803\ lb[/tex]

c) [tex]V_i=39.93\ gallons[/tex]

Explanation:

Therefore,

a)

External volume of the structure:

[tex]V=B.H.W[/tex]

[tex]V=84\times47\times59\div 2.54^3[/tex]

[tex]V=14214.3828\ in^3[/tex]

Internal volume of the structure:

[tex]V_i=B_i.H_i.W_i[/tex]

[tex]V_i=76\times 39\times 51\div 2.54^3[/tex]

[tex]V_i=9224.5932\ in^3[/tex]

∴Volume of Styrofoam used:

[tex]V_s=V-V_i[/tex]

[tex]V_s=14214.3828-9224.5932[/tex]

[tex]V_s=4989.7895\ in^3[/tex]

b)

given that density of Styrofoam, [tex]\rho=1\ kg.m^{-3}=3.613\times 10^{-5}\ lb.in^{-3}[/tex]

we know,

[tex]\rm mass= density \times volume[/tex]

[tex]mass=4989.7895\times 3.613\times 10^{-5}[/tex]

[tex]mass=0.1803\ lb[/tex]

c)

Volume of liquid it can hold [tex]=V_i[/tex]

[tex]V_i=39.93\ gallons[/tex]

The volume of the Styrofoam used is 14214.37 in³. The mass is 9.6 lbm. The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

To find the volume of the Styrofoam used in cubic inches, we need to convert the outside dimensions from centimeters to inches, subtract the volume of the cooler, and divide by the thickness of each wall. The volume of the Styrofoam used is 14214.37 in³.

To find the mass in lbm, we need to convert the density from kg/m³ to lbm/in³ and multiply by the volume of the Styrofoam used. The mass is 9.6 lbm.

To find the number of gallons of liquid that could be stored in the cooler, we need to convert the volume of the cooler from cm³ to gallons. The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

Answer:

12 A

Explanation:

Voltage, V = 120 V

Resistance, R  10 ohm

By using ohm's law

V = i x R

where, i is the current

i = V / R

i = 120 / 10

i = 12 A

thus, the current is 12 A.

When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials can give up electrons more readily than others.

Answer: Option C

Explanation:

We know that charges can neither be created nor be destroyed by law of conservation of charges. So when we rub two objects, it is natural to have a transfer of charges. But the charges which get transferred may be negligible in most of the cases leading to no significant observations.

But for some materials, like when we rubbed a balloon with human hair, we observed clouding of excess static charge on the balloon surface. This indicates that hair can easily give up electrons to balloon leading to clouding of excess static charge on it.

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

[tex]F_{1}=\frac{Kq_{1}q}{a^{2}}[/tex]

[tex]F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}[/tex]

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

[tex]F_{2}=\frac{Kq_{2}q}{a^{2}}[/tex]

[tex]F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}[/tex]

F2 = 0.26 N

Write the forces in the vector form

[tex]\overrightarrow{F_{1}}=0.35\widehat{i}[/tex]

[tex]\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )[/tex]

[tex]\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}[/tex]

Net force

[tex]\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}[/tex]

[tex]\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}[/tex]

Magnitude of the force

[tex]F=\sqrt{0.48^{2}+0.23^{2}}[/tex]

F = 0.53 N

Direction of force with x axis

[tex]tan\theta =\frac{0.23}{0.48}[/tex]

θ = 25.6°

To calculate the net force on charge 1 due to the other two charges, we need to find the individual forces between charge 1 and the other charges and then combine them vectorially using Coulomb's law.

To calculate the net force on charge 1 due to the other two charges, we need to find the individual forces between charge 1 and the other charges and then combine them vectorially. The magnitude of the force between two charges can be calculated using Coulomb's law:

F = k * (|q1| * |q2|) / (r^2)

Where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.

Let's calculate the individual forces:

Next, we can calculate the net force on charge 1 by adding the forces vectorially:

Net Force on charge 1 = F12 + F13

Answer:

-5.985*10^7Ns=momentum

ds=31.578m

Explanation:

The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?

from newton's second law of motion , which states that the rate of change in momentum is directly proportional to the force applied

f=kdv/dt

k=1

ft=dv/dt

dp=dv/dt

f=negative because its moving in the opposite direction

2.85× 10^6*21=

-5.985*10^7Ns=momentum

-dp=mv2-mv1

-dp+mv1=mv2

-5.985*10^7+2.0× 10^7 kg*3.0 m/s=2.0× 10^7kgV2

v2=150000/(2.0× 10^7)

v2=0.0075m/s to the right

ds=1/2(v2+v1)dt

ds=0.5*(3.0075)*21

ds=31.578m

the barge will travel 31.578m before it is brought to rest.

Answer:

∆p = –5.985×10⁷ kg×(m/s)

∆x = 31.57875 m

The boat's change in momentum is –5.985×10⁷ kg×(m/s), and it will travel 31.57875 meters before it stops.

Explanation:

First off, let's take a quick look at the formula for momentum: momentum equals the force times the time the force is applied for. This means:

p = FΔt

Why is momentum represented by the letter p? That's actually a good question, to be honest—I guess they were running out of letters because the letter m is already used for "mass". But I digress.

Let's go back to the equation. We have both those values—the force applied and the time it is applied for—and so we can substitute those into the equation. But be careful: here, the force is applied in the opposite direction as the actual motion. This means the force must be negative. We get:

p = (–2.85×10⁶ N)(21 s)

p = –5.985×10⁷ kg·(m/s)

That's our value for initial momentum, not the final momentum. Why? Momentum and impulse are equivalent and their units are the same. If we calculated this as the impulse (J = mΔv), we'd be using the initial velocity, not the final velocity (the velocity after the force is applied).

But this problem asked us to solve for two things: the change in momentum (Δp) and the stopping distance (Δx). We can't find either value if we don't know the velocity of the block after the force is applied. This means we need to solve for it.

This is where knowing impulse and momentum are equal comes in handy: if the two values are equal, their formulas should be equal, too. It's like solving a system of equations with the substitution method. With the impulse and momentum formulas, we make them equivalent and get:

mΔv = FΔt

The change in something is like the difference in subtraction: it's the final value minus the initial value. We can then rewrite the equation as:

m(v₂ – v₁) = FΔt

Here, I used (v₂) for the final (second) velocity and (v₁) for the initial (first) velocity. Now, since we need to find the final velocity, (v₂), we need to isolate it by solving the equation for this value.

m(v₂ – v₁) = FΔt

[m(v₂ – v₁)] ÷ m = (FΔt) ÷ m

v₂ – v₁ = (FΔt) ÷ m

v₂ – v₁ + v₁ = [(FΔt) ÷ m] + v₁

v₂ = [(FΔt) ÷ m] + v₁

That's our formula to find the final velocity. The problem already gave us all the values we need to solve this equation: the force applied, the time the force is applied for, the mass of the object being stopped, and the object's initial velocity. When we substitute those into the equation, we get:

v₂ = [(FΔt) ÷ m] + v₁

v₂ = [(–2.85×10⁶ N)(21 s) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = [(–5.985×10⁷ kg·(m/s)) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = 3.0 (m/s) – 2.9925 (m/s)

v₂ = 0.0075 (m/s)

That gives us the final velocity that we need to find (Δp) and (Δx). Now, using the equations for each value, we can substitute in and finish this up!

Δp = m(Δv) = m(v₂ – v₁)

Δp = (2.0×10⁷ kg)[0.0075 (m/s) – 3.0 (m/s)]

Δp = (2.0×10⁷ kg)[–2.9925 (m/s)]

Δp = –5.985×10⁷ kg·(m/s)

Δx = ¹/₂(v₂ + v₁)(Δt)

Δx = ¹/₂[0.0075 (m/s) + 3.0 (m/s)](21 s)

Δx = ¹/₂[3.0075 (m/s)](21 s)

Δx = [3.0075 (m/s)](10.5 s)

Δx = 3.157875 m

And there we go! Problem solved! I hope this helps you! Have a great day!

The speed of the wave is 208 m/s

Explanation:

The wave equation states that ; speed of wave= wavelength *frequency

Mathematically, v=λ*f

Given;

λ=52 cm= 52/100 =0.52 m

f= 400 Hz

v=?

v=0.52*400 =208 m/s

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Answer:

The Potential difference across the 8 Ω resistor = 9.6 V

Explanation:

In a series combination of resistor, The total resistance

Rt = R₁ + R₂............... Equation 1

Where Rt = total resistance, R₁ = Resistance of the first resistor, R₂ = resistance of the second resistor.

Given: R₁ = 12Ω, R₂ = 8Ω.

Substituting these vales into equation 1

Rt = 12 + 8

Rt = 20 Ω.

Since both resistors are connected in series, The same current flows through both resistors.

from ohms law,

V = IR..................... Equation 3

I = V/R....................... Equation 2

where I = current flowing through both resistors, V= Emf, R = total resistance of the circuit

Also given : V = 24 V, R = 20 Ω,

Substituting these values into equation 3,

I = 24/20

I = 1.2 A.

Note: Since both resistors are connected in series, The same current flows through both resistors.

The Potential difference across the 8 Ω resistor

V = 1.2×8

V = 9.6 V

The Potential difference across the 8 Ω resistor = 9.6 V

Explanation:

Mass, m = 899 kg

Initial velocity, u = 0 m/s

Final velocity, v = 26.3 m/s

Time, t = 0.59 s

a) We have equation of motion v = u + at

   Substituting

                     v = u + at

                     26.3 = 0 + a x 0.59

                       a = 44.58 m/s²

     Acceleration of dragster = 44.58 m/s²

b) Force = Mass x Acceleration

   F = ma

   F = 899 x 44.58 = 40074.07 N

   F = 40.07 kN

  Average net force on the dragster during this time is 40.07 kN

Answer:

The answer to your question is 1.35 Watts

Explanation:

Data

Work = W = 5 J

time = t = 3.7 s

Power = P = ?

Formula

Power is a rate in which work is done or energy is transferred over time

P = [tex]\frac{W}{t}[/tex]

Substitution

[tex]P = \frac{5}{3.7}[/tex]

Result

P = 1.35 W

Answer:

0

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

d = Vertical height from the ground

F = Force = Weight = mg

Net work done would be

[tex]W_n=W_{up}+W_{down}\\\Rightarrow W_n=Fdcos180+Fdcos0\\\Rightarrow W_n=-mgd+mgd\\\Rightarrow W_n=0[/tex]

Hence, the work done on the person by the gravitational force is 0

The electric potential at a point within a uniformly charged solid sphere can be calculated using the given formula. By substituting the given values of charge density, radius of the sphere, and distance from the center of the sphere, the electric potential can be computed.

The potential Vr within a uniformly charged solid sphere of charge density ρ, radius R, at a distance r from the center is given by the formula:

Vr = ρr²/6ε₀ + (ρR²/2ε₀) [ 1 - 3r²/R² ] for r < R

Where ρ is the charge density (4.40*10⁻⁷ C/m³), r is the distance from the center of the sphere (0.160 m), R is the radius of the sphere (0.370 m) and ε₀ is the permittivity of free space (8.854 * 10⁻¹² C²/N.m²).

Substituting the given values into the formula, we can calculate the electric potential at a point located at r = 0.160 m from the center of the sphere.

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The potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]

The potential  at a point located at a distance \(r\) from the center of a uniformly charged solid sphere can be calculated using the formula:

[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]

 where \(k\) is the Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(Q_{enc}[/tex]\) is the charge enclosed within the radius \(r\), and \(r\) is the distance from the center of the sphere to the point where the potential is being calculated.

 For a uniformly charged sphere, the charge enclosed within a radius \(r\) is given by:

[tex]\[ Q_{enc} = \frac{4}{3}\pi r^3 \rho \][/tex]

where [tex]\(\rho\)[/tex] is the charge density of the sphere.

 Given that [tex]\(\rho = 4.40 \times 10^{-7} \, \text{C/m}^3\)[/tex] and [tex]\(r = 0.160 \, \text{m}\),[/tex] we can calculate \(Q_{enc}\) as follows:

[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.160 \, \text{m})^3 (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]

[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.004096 \, \text{m}^3) (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]

[tex]\[ Q_{enc} = \frac{4}{3}\pi (1.8192 \times 10^{-9} \, \text{C}) \][/tex]

[tex]\[ Q_{enc} = 2.42528 \times 10^{-9} \, \text{C} \][/tex]

 Now, we can calculate the potential [tex]\(V_r\):[/tex]

[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]

[tex]\[ V_r = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) (2.42528 \times 10^{-9} \, \text{C})}{0.160 \, \text{m}} \][/tex]

[tex]\[ V_r = \frac{2.17836 \times 10^{-8} \, \text{Nm}^2/\text{C}}{0.160 \, \text{m}} \][/tex]

[tex]\[ V_r = 1.36148 \times 10^{-7} \, \text{V} \][/tex]

 Therefore, the potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]

Answer:

Elevation =31.85[m]

Explanation:

We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.

We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.

The point where the ball is thrown shall be taken as a reference point of potential energy.

[tex]E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J][/tex]

m = mass of the ball = 300 [gr] = 0.3 [kg]

v = initial velocity = 25 [m/s]

[tex]E_{k}=\frac{1}{2}  * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J][/tex]

[tex]93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m][/tex]

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

Answer:

F = -2.205N

Explanation:

First, we have to find the angular aceleration due to the knife following the next equation:

W = Wo + at

where W is the final angular velocity and Wo is the initial angular velocity, a the angular aceleration and t the time.

Now, we will change the angular velocity to rad/s as:

Wo = 200 rpm = 20.94 rad/s

W = 180 rpm = 18.84 rad/s

Replacing in the previus equation, we get:

18.84rad/s = 20.94rad/s + a(10s)

solving for a:

a = -0.21rad/s^2

Now, we have to find the moment of inertia of the grindstone using:

I = [tex]\frac{1}{2}MR^2[/tex]

Where M is the mass of the stone and R the radius of the stone. Replacing values:

I = [tex]\frac{1}{2}(28kg)(0.15m)^2[/tex]

I = 0.315 kg*m^2

Adittionally:

T = Ia

where T is the torque, I the moment of inertia and a the angular aceleration.

so:

[tex]U_kFd = Ia[/tex]

where [tex]U_k[/tex] is the coefficient of the kinetic friction, F is the force with which the man presses the knife and d the lever arm. So, replacing values, we get:

[tex](0.2)F(0.15m) = (0.315)(-0.21rad/s^2)[/tex]

solving for F:

F = -2.205N

it is negative because the stone is stopping due of this force.

Answer:

Radius of curvature of the path is 1063 meters

Explanation:

It is given that,

Force acting on the pilot is about seven times of his weight. Speed with which pilot moves, v = 250 m/s.

As per Newton's second law of motion, the net force acting on the pilot at the bottom is given by :

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

Where

N is the normal force

r is the radius of curvature

According to given condition,

[tex]7mg-mg=\dfrac{mv^2}{r}[/tex]    

[tex]6mg=\dfrac{mv^2}{r}[/tex]    

[tex]r=\dfrac{mv^2}{6mg}[/tex]  

[tex]r=\dfrac{mv^2}{6mg}[/tex]      

[tex]r=\dfrac{v^2}{6g}[/tex]      

[tex]r=\dfrac{(250)^2}{6\times 9.8}[/tex]  

r = 1062.92 meters

or

r = 1063 meters

So, the radius of curvature of the path is 1063 meters. Hence, this is the required solution.

Answer:

Part A

Mass = 50g

Vmax = 3.2m/s

Amplitude= 6cm

Position x from the equilibrium= 5.1cm

Part B

Kinetic energy = 0.185J

Part C

Potential energy = 0.185J

Explanation:

Kinetic energy = 1/2mv×2

Vmax = wa

w = angular velocity= 53.33rad/s

Kinetic energy = 1/2mv^2×r^2 = 0.185J

Part c

Total energy = 1/2m×Vmax^2= 0.256J

1/2KA^2= 0.256J

K= 142.22N/m (force constant)

Potential energy = 1/2kx^2

=1/2×142.22×0.051^2

= 0.185J

To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.

In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.

To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.

Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.

Substituting the known values into the equations, we can calculate the kinetic energy of the toy.

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Answer:braking

Explanation:

Large vehicles as compared to small vehicles require long braking distance, otherwise, it could topple the heavy vehicles. Heavy vehicles provide high torque thus it is used to carry heavy loads.

They run at relatively low speed as compared to the light vehicles as they are slow to accelerate and thus require long braking distance as Momentum associated with them is very high.

If sudden brakes are applied it may cause the vehicle to skid and flip over it due to the presence of large momentum.

Neither Technician A nor B is correct.

Answer: Option D

Explanation:

In AC generator, diodes are usually used to rectify the alternator current. This is the transformation into direct current from alternating current. Sinusoidal voltage is generated by generators when it rotates in both directions. The output voltage [tex]V_{i n}[/tex] is converted to DC with diode help.

This process is called rectification. The output voltage [tex]V_{i n}[/tex]  is treated as the input signal of the diode circuit. So, after rectification, the sine wave is not produced by an AC generator. And, the sine wave resulted by AC generator when the output passes via diodes and not a straight line.

The waveform produced by an AC generator after rectification is a pulsating DC waveform, not a sine wave. Technician A is correct in this case.

Technician A is correct. The waveform produced by an AC generator after rectification is not a sine wave, but a pulsating DC waveform. After the output of the generator moves through the diodes, the negative half-cycles are clipped, resulting in a waveform that resembles a series of straight lines instead of a smooth sine wave. This pulsating DC waveform still contains alternating current components, but it is not a constant DC voltage.

Answer:

[tex]R= \frac{1job}{1.429 hours}[/tex]

So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.

Explanation:

If we want to express the situation in math terms and find the number of hours that takes to complete 1 job with the 3 at the same time, we can do this.

For this case we have the following rates:

[tex]R_A =\frac{1job}{6hours}[/tex]

[tex]R_B =\frac{1 job}{5 hours}[/tex]

[tex]R_C=\frac{1job}{3 hours}[/tex]

And we know that working together the rate would be the addition of the rates like this:

[tex]R=R_A +R_B +R_C = \frac{1job}{6hours}+\frac{1job}{5hours}+\frac{1job}{3hours} =\frac{7 jobs}{10hours}[/tex]

And if we divide the numerator and denominator by 7 we got:

[tex]R= \frac{1job}{1.429 hours}[/tex]

So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train  = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train  =  4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

Answer:

Option d

Explanation:

Interference is the phenomenon in which two or more waves having the same frequency combines or cancel out each other to form a resulting wave with the amplitude equals the sum of the amplitudes of the waves that are combined.

Thus it is clear that the particle with the wave nature can produce interference pattern. Hence, electron, light, sound, all of these produces interference pattern when directed towards the slits that are suitably spaced.  

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice.

Answer: Option C

Explanation:

The closure temperature can also be termed as blocking temperature. It is mostly used in radiometric dating. As the temperature decreases, below a certain point the isotopes may get freeze in their lattice positions. And there may be slowing of diffusion.

At the closure temperature, that rate of diffusion will be zero as the isotopes will be no longer free to move out of crystal lattice. So, this is termed as closure or blocking temperature. As the isotopes loose their ability to move, their concentration will remain fixed in their position leading to measurement of radiation dating.

Final answer:

Closure temperature is the point where isotopes are trapped within a crystal lattice, marking the mineral as a closed system, which is crucial for isotopic dating.

Explanation:

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice. This is a point in the cooling process of a rock or mineral, after crystallization, at which its isotopic constituents are no longer able to escape from its crystal lattice. Essentially, it is the temperature below which a mineral becomes a closed system for certain isotopes. Until reaching this temperature, isotopic parent and daughter products can freely exchange with the environment. As the temperature drops below the closure temperature, the system locks in place, preserving the isotopic date that will be analyzed to determine the history of the rock.

The speed of the block after the bullet exits is 7.5 m/s. This velocity is found by using the conservation of momentum principle and plugging in the known values into the momentum equation to solve for the unknown variable.

To solve for the speed of the wooden block after the bullet exits, we will use the principle of conservation of momentum. This principle states that in an isolated system, the total momentum before an event is equal to the total momentum after the event. The formula to calculate the final velocity of the block is:

[tex]m_{bullet} \times u_{bullet} + m_{block} \times u_{block} = m_{bullet} \times v_{bullet} + m_{block} \times v_{block}[/tex]

Where:

[tex]m_{bullet}[/tex] = mass of the bullet

[tex]u_{bullet}[/tex] = initial velocity of the bullet

[tex]m_{block}[/tex] = mass of the block

[tex]u_{block}[/tex] = initial velocity of the block (0 m/s, since it's stationary)

[tex]v_{bullet}[/tex] = final velocity of the bullet after passing through the block

[tex]v_{block}[/tex] = final velocity of the block

Given:

[tex]m_{bullet}[/tex] = 0.050 kg (50 g)

[tex]u_{bullet}[/tex] = 500 m/s

[tex]m_{block}[/tex] = 2 kg

[tex]u_{block}[/tex] = 0 m/s

[tex]v_{bullet}[/tex] = 200 m/s

Plugging in the values:

0.050 × 500 + 2 × 0 = 0.050 × 200 + 2 × [tex]v_{block}[/tex]

Solving for [tex]v_{block}[/tex]:

[tex]v_{block}[/tex] = (0.050 × 500 - 0.050 × 200) / 2

[tex]v_{block}[/tex] = (25 - 10) / 2

[tex]v_{block}[/tex] = 15 / 2

[tex]v_{block}[/tex] = 7.5 m/s

Therefore, the velocity of the block after the bullet exits is 7.5 m/s.