A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp, measured from the horizontal, at a speed of 40.0 m/s (144 km/h). The top of the ramp is at the same height as the roofs of the buses and each bus is 20.0 m long.

Answer:

Option (A) is correct.

Explanation:

for a near sighted person, distance of object from the lens = u = ∞

distance of image from the lens, v = - 80 cm

Use lens formula

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{1}{f}=\frac{1}{-80}-\frac{1}{\infty }[/tex]

So, f = - 80 cm

Power of the lens is the reciprocal of the focal length of the lens.

P = 100/f

where, f is the focal length when it is measured in the units of cm.

P = - 100 / 80 = - 1.3 Dioptre

Thus, option (a) is correct.

The ball's velocity just after leaving the bat is -25.93 m/s.

To find the ball's velocity just after leaving the bat, we can use the principle of conservation of momentum. The impulse on the ball caused by the bat is equal to the change in momentum of the ball. Since impulse is defined as force multiplied by time, we can use the given impulse of -8.4 N s and the mass of the ball (0.145 kg) to find the change in velocity of the ball.

The formula for impulse is impulse = change in momentum = mass * change in velocity. Rearranging the formula, we can solve for the change in velocity: change in velocity = impulse/mass = -8.4 N s / 0.145 kg = -57.93 m/s.

Since the initial velocity of the ball was 32 m/s in the +x direction, the final velocity of the ball can be found by adding the change in velocity to the initial velocity: final velocity = initial velocity + change in velocity = 32 m/s + (-57.93 m/s) = -25.93 m/s.

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Answer

given,

diameter of steel = d = 26 c m

         radius = 13 cm = 0.13 m

length = L = 2.4 m

mass = 7800 Kg

g = 9.8 m/s²

a) stress = F/A

stress = [tex]\dfrac{mg}{\pi\ r^2}[/tex]

stress = [tex]\dfrac{7800 \times 9.8}{\pi\ 0.13^2}[/tex]

stress = 1.44 x 10⁶ N/m²

b) Young's modulus x  strain = stress

 Young's modulus for steel = 200 x 10⁹ N/m²

    200 x 10⁹ x strain = 1.44 x 10⁶

     strain = 7.2 x 10⁻⁶ m

c) change in length

  [tex]Strain= \dfrac{\Delta L}{L}[/tex]

  [tex]7.2 \times 10^{-6} = \dfrac{\Delta L}{2.4}[/tex]

  [tex]\Delta L= 17.28\times 10^{-6}\ m[/tex]

To solve this problem it is necessary to apply the concepts related to Kepler's second law and the conservation of angular momentum.

Kepler's second law tells us that the vector radius that unites a planet and the sun sweeps equal areas at equal times, that is, when the planet is farther from the sun, the speed at which it travels is less than when it is close to the sun.

The angular momentum is defined as

[tex]L = m*r*v[/tex]

Where,

m= mass

r = Radius

v = Velocity

For conservation of angular momentum

[tex]L_{apogee}=L_{perigee}[/tex]

[tex]mv_a*r_a = mv_p*r_p[/tex]

[tex]v_a*r_a= v_p*r_p[/tex]

[tex]r_a = \frac{v_p*r_p}{v_a}[/tex]

[tex]r_a = \frac{(4.46)(2.23*10^4)}{(3.54)}[/tex]

[tex]r_a = 2.81*10^4km[/tex]

Therefore the corresponding distance at apogee is [tex]2.81*10^4km[/tex]

When an ice cube, which floats with 90% submerged in water, is placed in a liquid that is less dense than water, less than 90% of the cube will be submerged. This is due to the interplay between the densities of the ice cube and the liquid.

The extent to which an object is submerged in a fluid depends on the density of both the object and the fluid. In the case of an ice cube in water, the ice cube is less dense than the water, causing about 90% of it to be submerged under the surface. Now if you place the same ice cube in a liquid that is less dense than water, less than 90% of the ice cube will be submerged. This is because the ice cube is denser than this new liquid, and it displaces less liquid to balance its weight.

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Answer:

 x = 259 Hz

Explanation:

given,

frequency of one tuning fork = 250 Hz

frequency of another tuning fork = 266 Hz

when a tuning fork is sounded together beat frequency heard = 9

 let x be the frequency of unknown

 x - 250 = 9 Hz..............(1)

 x = 259 Hz

when a another tuning fork is sounded together beat frequency heard = 7

266 - x  = 7 Hz..............(2)

 x = 259 Hz

now, on solving both the equation the frequency comes out to be 259 Hz.

so, The frequency of the tuning fork is equal to 259 Hz

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

[tex]KE = \frac{1}{2} mv^2[/tex]

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

[tex]m*V_i = m_1V_1+m_2V_2[/tex]

Here,

m = Total mass (8Kg at this case)

[tex]m_1=m_2 =[/tex] Mass each part

[tex]V_i =[/tex] Initial velocity

[tex]V_2 =[/tex] Final velocity particle 2

[tex]V_1 =[/tex] Final velocity particle 1

The initial kinetic energy would be given by,

[tex]KE_i=\frac{1}{2}mv^2[/tex]

[tex]KE_i = \frac{1}{2}8*5^2[/tex]

[tex]KE_i = 100J[/tex]

In the end the energy increased 100J, that is,

[tex]KE_f = KE_i KE_{increased}[/tex]

[tex]KE_f = 100+100 = 200J[/tex]

By conservation of the moment then,

[tex]m*V_i = m_1V_1+m_2V_2[/tex]

Replacing we have,

[tex](8)*5 = 4*V_1+4*V_2[/tex]

[tex]40 = 4(V_1+V_2)[/tex]

[tex]V_1+V_2 = 10[/tex]

[tex]V_2 = 10-V_1[/tex](1)

In the final point the cinematic energy of EACH particle would be given by

[tex]KE_f = \frac{1}{2}mv^2[/tex]

[tex]KE_f = \frac{1}{2}4*(V_1^2+V_2^2)[/tex]

[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex](2)

So we have a system of 2x2 equations

[tex]V_2 = 10-V_1[/tex]

[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex]

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: [tex]V_1 = 10m/s[/tex]

Then replacing in (1) we have that

PART B: [tex]V_2 = 0m/s[/tex]

Answer:

a) 2.1 × 10^-6 m b) 1.3 × 10^-4 m

Explanation:

The thermal expansion of super Invar 0.20 ×10^-6oC^-1 and the length of Super is 1.9 m

using the linear expansivity  formula

ΔL = αLoΔt where ΔL is the change in length meters, α is the linear expansivity of Super Invar in oC^-1 and Δt  is the change in temperature in oC. Substituting the value into the formula gives

ΔL = 0.2 × 10 ^-6 × 1.9 m ×5.5 = 2.1 × 10^-6 meters to two significant figure

b) the thermal expansion of steel is 1.3 × 10^-5 oC^-1 and the length of steel is 1.8 m

using the formula

ΔL of steel = 1.3 × 10^-5 × 1.8 ×  5.5 = 1.3 × 10^-4 m to two significant figure.

Answer:

Angular acceleration will be [tex]18.84rad/sec^2[/tex]

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity [tex]\omega _i=0rev/sec[/tex]

And final angular velocity [tex]\omega _f=24rev/sec[/tex]

Time is given as t = 8 sec

From equation of motion

We know that [tex]\omega _f=\omega _i+\alpha t[/tex]

[tex]24=0+\alpha \times 8[/tex]

[tex]\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2[/tex]

So angular acceleration will be [tex]18.84rad/sec^2[/tex]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

[tex]\tau = I \alpha[/tex]

Where,

I=Inertial Moment

[tex]\alpha =[/tex] Angular acceleration

Also Torque with linear equation is defined as,

[tex]\tau = F*d[/tex]

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

[tex]d*F = I\alpha[/tex]

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

[tex]d*F= \frac{mR^2a}{R}[/tex]

[tex]d*F= mRa[/tex]

[tex]a = \frac{rF}{ mR}[/tex]

[tex]a = \frac{0.04*20}{1.5*0.3}[/tex]

[tex]a=1.77 m/s^2[/tex]

Then the velocity of the wheel is

[tex]V = a *t \\V=1.77*4 \\V=7.11 m/s[/tex]

Therefore the correct answer is D.

Answer

given,

weight of the oak board = 600 N

Weight of Joe = 844 N

length of board = 4 m

Joe is standing at 1 m from left side

vertical wire is supporting at the end.

Assuming the system is in equilibrium

T₁ and T₂ be the tension at the ends of the wire

equating all the vertical force

T₁ + T₂ = 600 + 844

 T₁ + T₂ = 1444...........(1)

taking moment about T₂

 T₁ x 4 - 844 x 3 - 600 x 2 = 0

 T₁ x 4 = 3732

 T₁ = 933 N

from equation (1)

 T₂ = 1444 - 933

 T₂ = 511 N

Answer:

I will do more work against friction going around the floor.

ΔW = 5 J

Explanation:

Given

Ff₁ = 40 N

Ff₂ = 55 N

a = 3 m

b = 4 m

We have to get the distance d₁ as follows

d₁ = a + b = 3 m + 4 m = 7 m

And  d₂:

d₂ = √(a² + b²) = √((3 m)² + (4 m)²)

⇒   d₂ = 5 m

then we use the equations

W₁ = Ff₁*d₁ = (40 N)(7 m) = 280 J   (work against friction going around the floor)

W₂ = Ff₂*d₂ = (55 N)(5 m) = 275 J  (work against friction going across the rug)

since  W₂ < W₁  I will do more work against friction going around the floor.

Now, we apply the formula

ΔW = W₁ - W₂ = 280 J - 275 J = 5 J

When pushing the heavy box from one corner of the rug to the opposite corner, you will do more work against friction going across the rug than going around the floor. The extra work against friction across the rug is calculated to be 375 J.

When pushing the heavy box from one corner of the rug to the opposite corner, you will do more work against friction going across the rug. The force of friction on the rug is greater than the force of friction on the smooth concrete floor. To calculate the extra work, you can use the formula:

Extra work = (force of friction on the rug - force of friction on the floor) x distance

Extra work = (55 N - 40 N) x (diagonal distance of the rug)

Since the rug is 3 m by 4 m, the diagonal distance can be calculated using the Pythagorean theorem:

diagonal distance = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 m

Therefore, the extra work against friction going across the rug is (55 N - 40 N) x 5 m = 75 N x 5 m = 375 J.

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (in this case the -z direction) and the curling of the fingers point in the direction of the magnetic field  B→ (in this case the +x direction), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

Answer:

 ΔL = 1.43 10 -9 m

Explanation:

a) Let's start from Newton's second law, the force in a spring is elastic

    F = - k Δx

Let's divide the two sides by the area

    F / A = -k Δx / A

In general area is long by wide, the formulated pressure is

    P = F / A

    P = - k Δx / l x

    P = (-k / l) Δx/x

Call us at Young's constant module

    P = E Δx / x

Let's change x for L

   E = P / (ΔL/L)

b) we cleared

   ΔL = P L / E

Let's reduce the magnitudes to the SI system

   E = 70 GPa = 70 109 Pa

   L = 100mm (1m / 1000mm) = 0.100m

   P = 1 kN = 1 103 N

   ΔL = 1 103  0.100/ 70 109

   ΔL = 1.43 10 -9 m

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x =  position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t      v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²    

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²  

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

Answer:

4d

Explanation:

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

[tex]dQ = dU + dW[/tex]

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

[tex]W = -168J \righarrow[/tex]  Work is done ON the system

[tex]Q = -305.6J \rightarrow[/tex] Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

Answer:

a)Uo= 2 m/s

b)[tex]u_{avg}=1 \ m/s[/tex]

c)Q=7.54  x 10⁻⁴ m³/s

Explanation:

Given that

[tex]u(r)=2\left(1-\dfrac{r^2}{R^2}\right)[/tex]

Diameter ,D= 3.1 cm

Radius ,R= 1.55 cm

We know that in the pipe flow the general equation for laminar fully developed flow given as

[tex]u(r)=U_o\left(1-\dfrac{r^2}{R^2}\right)[/tex]

Uo=Maximum velocity

Therefore maximum velocity

Uo= 2 m/s

The average velocity

[tex]u_{avg}=\dfrac{U_o}{2}[/tex]

[tex]u_{avg}=\dfrac{2}{2}\ m/s[/tex]

[tex]u_{avg}=1 \ m/s[/tex]

The volume flow rate

[tex]Q=u_{avg}. A[/tex]

[tex]Q=\pi R^2\times u_{avg}\ m^3/s[/tex]

[tex]Q=\pi \times (1.55\times 10^{-2})^2\times 1\ m^3/s[/tex]

Q=0.000754 m³/s

Q=7.54  x 10⁻⁴ m³/s

Answer:

(B) You should catch the hall

Explanation:

B) You should catch the ball.

case 1 : when ball is catched.

Mass of person catching = M

mass of ball = m

velocity of the ball just before catching = v

velocity of ball just after catching = 0

[tex]\Delta P_{ball}[/tex] = change in momentum of the ball = m (0 - v) = - mv

since,

[tex]\Delta P_{person} =-\Delta P_{ball}[/tex]

[tex]\Delta P_{person} =mv[/tex]

case 2 : when the ball is deflected back

velocity of the ball just before catching = v

velocity of ball just after catching = - v

[tex]\Delta P_{ball}[/tex] = change in momentum of the ball = m (- v - v) = - 2mv

since,  

[tex]\Delta P_{person} =- \Delta P_{ball}[/tex]

[tex]\Delta P_{person} = 2mv[/tex]

clearly we can see that the change in momentum is minimum in case when the ball is catched and hence keeps the speed of skateboard minimum

Answer:

Enthalpy is 44.95 kJ/mol

Solution:

As per the question:

Temperature, T = 270.6 K

Temperature, T' = 287.5 K

Pressure, P = 324.5 mmHg

Pressure, P' = 626.9 mmHg

Now,

To calculate the enthalpy, we make use of the Clausius-Clapeyron eqn:

[tex]ln\frac{P}{P'} = \frac{\Delta H}{R}(\frac{1}{T'} - \frac{1}{T})[/tex]

where

[tex]\Delta H = Enthalpy[/tex]

R = Rydberg's constant

Substituting suitable values in the above eqn:

[tex]ln\frac{324.5}{626.9} = \frac{\Delta H}{8.31447}(\frac{1}{287.5} - \frac{1}{270.6})[/tex]

[tex]- 0.658 = \frac{\Delta H}{8.31447}(\frac{1}{287.5} - \frac{1}{270.6})[/tex]

[tex]\Delta H = 44.95\ kJ/mol[/tex]

To develop this problem it is necessary to apply the concepts related to the calculation of the Force through density and volume as well as the ideal gas law.

By definition, force can be expelled as

F = ma

Where,

m = mass

a = Acceleration

At the same time  the mass can be defined as function of density and Volume

[tex]m = \rho V[/tex]

Therefore if we do a sum in the spherical balloon we have,

[tex]\sum F = 0[/tex]

[tex]F_w +F_h-F_b=0[/tex]

Where,

[tex]F_W[/tex]= Force by weight of balloon

[tex]F_h[/tex]= Force by weight of helium gas

[tex]F_b[/tex]= Buoyant force

[tex]mg + V \rho g - V\rho_a g = 0[/tex]

Re-arrange to find [tex]\rho,[/tex]

[tex]\rho = \rho_a - \frac{m}{V}[/tex]

Our values are given as,

[tex]r= 1.55m[/tex]

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]V = \frac{4}{3} \pi (1.55)^3[/tex]

[tex]V = 15.59m^3[/tex]

Replacing the values we have,

[tex]\rho = 1.19kg/m^3 - \frac{2.7}{15.59}[/tex]

[tex]\rho = 1.0168kg/m^3[/tex]

Applying the ideal gas law we have finally that

[tex]P = \frac{\rho}{M_0} RT[/tex]

Where,

P = Pressure

[tex]\rho =[/tex] Density

M_0 Molar mass (0.004Kg/mol for helium)

R= Gas constant

T = Temperature

Substituting

[tex]P = \frac{1.0168}{0.004} *8.314*290[/tex]

[tex]P = 612891.452Pa[/tex]

[tex]P = 0.613Mpa[/tex]

Therefore the absolute pressure of the helium gas is [tex]0.613Mpa[/tex]

To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.

PART A) For the calculation of the velocity we define the area and the flow, thus

[tex]A = \pi r^2[/tex]

[tex]A = pi (2*10^{-3})^2[/tex]

[tex]A = 12.56*10^{-6}m^2[/tex]

At the same time the rate of flow would be

[tex]Q = \frac{1L}{2s}[/tex]

[tex]Q = 0.5L/s = 0.5*10^{-3}m^3/s[/tex]

By definition the discharge is expressed as

[tex]Q = NAv[/tex]

Where,

A= Area

v = velocity

N = Number of exits

Q = NAv

Re-arrange to find v,

[tex]v = \frac{Q}{NA}[/tex]

[tex]v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}[/tex]

[tex]v = 0.9047m/s[/tex]

PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe

[tex]P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2[/tex]

Replacing with our values we have that

[tex]1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)[/tex]

[tex]v_1^2 = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)[/tex]

[tex]v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}[/tex]

[tex]v_1 = 3.54097m/s[/tex]

PART C) Assuming that water is an incomprehensible fluid we have to,

[tex]Q_{pipe} = Q_{shower}[/tex]

[tex]v_{pipe}A_{pipe}=v_{shower}A_{shower}[/tex]

[tex]3.54097*A_{pipe}=0.9047*12.56*10^{-6}[/tex]

[tex]A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}[/tex]

[tex]A_{pipe = 3.209*10^{-6}m^2[/tex]

The answer provides calculations for the speed of water from the shower head, the speed in the connected pipe, and the cross-sectional area of the pipe based on the given parameters.

Answer:

the length of stretched spring  in cm is 22

Explanation:

given information:

spring length, x1 = 20 cm = 0.2 m

force, F = 100 N

the length of spring streches, x2 = 22 cm = 0.22 m

According to Hooke's law

F = - kΔx

k = F/*=(x2-x1)

  = 100/(0.22 - 0.20)

  = 5000 N/m

if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end

m = 10.2 kg

W = m g

    = 10.2 x 9.8

    = 99.96 N

F = - k Δx

Δx = F / k

     = 99.96 / 5000

     = 0.02

Δx = x2- x1

x2 = Δx + x1

    = 0.20 + 0.02

    = 0.22 m

     = 22 cm

Answer:

Velocity of the first car after the collision, [tex]v_1=8.93\ m/s[/tex]

Explanation:

It is given that,

Mass of the car, [tex]m_1 = 480\ kg[/tex]

Initial speed of the car, [tex]u_1 = 14.4\ m/s[/tex]    

Mass of another car, [tex]m_2 = 570\ kg[/tex]

Initial speed of the second car, [tex]u_2 = 13.3\ m/s[/tex]  

New speed of the second car, [tex]v_2 = 17.9\ m/s[/tex]  

Let [tex]v_1[/tex] is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]m_1u_1+m_2u_2-m_2v_2=m_1v_1[/tex]

[tex]480\times 14.4+570\times 13.3-570\times 17.9=480v_1[/tex]

[tex]v_1=8.93\ m/s[/tex]

So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.

Answer:

The induced emf for this silver circle during this period of expansion is 0.0314 V.

Explanation:

Given that,

Initial radius = 0.075 m

Magnetic field = 1.5 T

Radius = 0.125 m

Time t =1.5 s

We need to calculate the induced emf for this silver circle

Using formula of emf

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

[tex]\epsilon=\dfrac{B(A_{f}-A_{i})}{t}[/tex]

[tex]\epsilon=\dfrac{B(\pi r_{f}^2-\pi r_{i}^2)}{t}[/tex]

Put the value ino the formula

[tex]\epsilon=\dfrac{1.5(\pi\times(0.125)^2-\pi(0.075)^2)}{1.5}[/tex]

[tex]\epsilon=0.0314\ V[/tex]

Hence, The induced emf for this silver circle during this period of expansion is 0.0314 V.

The induced emf for the silver circle during this period of expansion is approximately [tex]\(\boxed{0.0314 \, \text{V}}\).[/tex]

The induced emf  in the silver circle during the expansion can be calculated using Faraday's law of induction, which states that the induced emf in a closed loop is equal to the negative rate of change of the magnetic flux through the loop. Mathematically, this is expressed as:

[tex]\[ \varepsilon = -\frac{d\Phi_B}{dt} \][/tex]

 The initial area [tex]\( A_i \)[/tex] when the radius [tex]\( r_i \[/tex]) is 0.075 m is:

[tex]\[ A_i = \pi r_i^2 = \pi (0.075 \, \text{m})^2 \][/tex]

 The final area [tex]\( A_f \)[/tex] when the radius [tex]\( r_f \)[/tex] is 0.125 m is:

[tex]\[ A_f = \pi r_f^2 = \pi (0.125 \, \text{m})^2 \][/tex]

 The change in area [tex]\( \Delta A \)[/tex] is:

[tex]\[ \Delta A = A_f - A_i = \pi (0.125 \, \text{m})^2 - \pi (0.075 \, \text{m})^2 \][/tex]

 The time taken for this change [tex]\( \Delta t \)[/tex] is 1.5 s.

 The induced emf [tex]\( \varepsilon \)[/tex] is then:

[tex]\[ \varepsilon = -\frac{d\Phi_B}{dt} = -B \frac{\Delta A}{\Delta t} \][/tex]

[tex]\[ \varepsilon = -1.5 \, \text{T} \times \frac{\pi ((0.125 \, \text{m})^2 - (0.075 \, \text{m})^2)}{1.5 \, \text{s}} \][/tex]

 Now, let's calculate the numerical value:

[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{(0.125^2 - 0.075^2)}{1.5} \][/tex]

[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{(0.015625 - 0.005625)}{1.5} \][/tex]

[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{0.01}{1.5} \][/tex]

[tex]\[ \varepsilon = -1.5 \times \pi \times 0.00666\ldots \][/tex]

[tex]\[ \varepsilon \approx -0.0314 \, \text{V} \][/tex]

The negative sign indicates the direction of the induced emf according to Lenz's law, which is opposite to the direction of the change in magnetic flux. However, the magnitude of the induced emf is approximately 0.0314 V.

Therefore, the induced emf for the silver circle during this period of expansion is approximately [tex]\(\boxed{0.0314 \, \text{V}}\).[/tex]

Answer:

The point in space where the net magnetic field is zero lies by specifying its perpendicular distance from the wire is 0.01153 m.

Explanation:

Given that,

Current = 380 A

Magnetic field [tex]B=6.59\times10^{-3}\ T[/tex]

We need to calculate the distance

Using formula of magnetic field

[tex]B = \dfrac{\mu_{0}I}{2\pi r}[/tex]

[tex]r=\dfrac{\mu_{0}I}{2\pi B}[/tex]

Where, B = magnetic field

I = current

Put the value into the formula

[tex]r=\dfrac{4\pi\times10^{-7}\times380}{6.59\times10^{-3}\times2\pi}[/tex]

[tex]r=0.01153\ m[/tex]

Hence,  The point in space where the net magnetic field is zero lies by specifying its perpendicular distance from the wire is 0.01153 m.

Answer:

0 V

Explanation:

given,                                        

Inductance  = L = 0.0345 H        

resistor of resistance = R = 30.5 Ω                          

connected in series with battery of voltage = 3.20 V

at t= 0 switch is closed                                                      

potential difference when switch is closed = ?                                      

The potential difference when the switch is just closed across the resistor will be equal to Zero.                                        

Because at t =0 the capacitor will act as a short circuit.

hence, the potential will be 0 V                                

This question asks for the rate of heat transfer from a copper fin, which requires the application of heat transfer theory, including the use of specific equations for fin efficiency and heat loss. However, a sample calculation provided deals with finding the initial temperature of a copper piece mixed with water, based on the conservation of energy and specific heat capacities.

The subject of this question involves the calculation of heat transfer from a fin made of copper. Given the material's thermal conductivity (k), the fin's dimensions, and the conditions surrounding it (base temperature, air temperature, and convection heat transfer coefficient h), we can find the rate of heat transfer. However, the actual calculation for the rate of heat transfer from this fin isn't provided directly in the context of the question. For such a problem, one would typically employ the fin equations from heat transfer theory, which might involve the use of Biot number, fin efficiency, and heat loss calculations. Instead, to answer a closely related question with given data, we can calculate the initial temperature of a 248-g piece of copper when dropped into 390 mL of water at 22.6°C, where the final temperature is measured as 39.9°C, assuming all heat transfer occurs between the copper and the water. This problem relies on the conservation of energy principle, requiring the application of the specific heat capacities of copper and water

Answer:

P₂ = 392720.38 Pa = 392.72 kPa

Explanation:

Given

D₁ = 5 cm = 0.05 m

D₂ = 10 cm = 0.10 m

v₁ = 8 m/s

P₁ = 380 kPa = 380000 Pa

α = 1.06

ρ = 1000 kg/m³

g = 9.8 m/s²

We can use the following formula

(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL

knowing that z₁ = z₂ we have

(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL   (I)

Where

V₂ can be obtained as follows

V₁*A₁ = V₂*A₂      ⇒     V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)

⇒    V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)  

⇒    V₂ = 2 m/s

and

hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)

⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)

⇒ hL = 1.9469 m

Finally we get P₂ using the equation (I)

⇒  P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)

⇒  P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)

⇒  P₂ = 392720.38 Pa = 392.72 kPa

The velocity of the water in the larger section of the pipe is 3.2 m/s and the pressure is 412 kPa. The downstream pressure P2 is 356.7 kPa and the error if Bernoulli's equation had been used is approximately 55.3 kPa.

First, we can calculate the velocity of the water in the larger section of the pipe using the continuity equation, which states that the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region. Using this equation, we can find that the velocity in the larger section is 3.2 m/s.

Next, we can use Bernoulli's equation to solve for the pressure in the larger section of the pipe. Bernoulli's equation states that the pressure in a fluid decreases as the velocity increases. Plugging in the given values and the calculated velocity of 3.2 m/s, we can find that the pressure in the larger section is 412 kPa.

Finally, we can use the pressure drop equation, P2 - P1 = RQ, to solve for the downstream pressure P2. The resistance R can be calculated using the kinetic energy correction factor and the given values. Plugging in all the values, we can find that the downstream pressure P2 is 356.7 kPa. To estimate the error that would have occurred if Bernoulli's equation had been used, we can calculate the difference between the actual downstream pressure and the pressure calculated using Bernoulli's equation. This error is approximately 55.3 kPa.

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To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

[tex]T^2 = \frac{4\pi^2}{GM}a^3[/tex]

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

[tex]T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s[/tex]

PART A) Replacing the values to find a, we have

[tex]a^3= \frac{T^2 GM}{4\pi^2}[/tex]

[tex]a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}[/tex]

[tex]a^3 = 6.46632*10^{39}[/tex]

[tex]a = 1.86303*10^{13}m[/tex]

Therefore the semimajor axis is [tex]1.86303*10^{13}m[/tex]

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

[tex]R = a(1-e)[/tex]

[tex]R = 1.86303*10^{13}(1-0.997)[/tex]

[tex]R= 5.58*10^{10}m[/tex]

Answer:

a = - 0.248 m/s²

Explanation:

Frictional drag force

F = ½ *(ρ* v² * A * α)

ρ = density of air  , ρ = 1.295 kg/m^3

α = drag coef , α = 0.250

v = 100 km/h x 1000m / 3600s

v =  27.77 m/s

A = 2.20m^2

So replacing numeric in the initial equation

F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)

F = 298.6 N

Now knowing the force can find the acceleration

a = - F / m

a = - 298.6 N / 1200 kg

a = - 0.248 m/s²

Final answer:

The question involves using aerodynamic drag force and Newton's second law to calculate the deceleration of a sports car shifted into neutral.

Explanation:

The question revolves around calculating the initial acceleration of a sports car weighing 1200 kg that has been shifted into neutral and allowed to coast, assuming the car was traveling at 100 km/hr, and given an aerodynamic drag coefficient of 0.250 and a frontal area of 2.20 m2. To calculate acceleration due to the forces acting on the car, we use the formula for aerodynamic drag force which is Fd = 0.5 × Cd × ρ × A × v2 where Cd is the drag coefficient, ρ is the air density (approximately 1.225 kg/m3 at sea level), A is the frontal area, and v is the velocity in meters per second. To find the acceleration, we can then apply Newton's second law, F = m × a, where F is the net force acting on the car, m is the mass, and a is the acceleration. In this case, the net force is opposite to the direction of the velocity vector due to drag force, thus providing the deceleration or negative acceleration of the car.