Which of the following objects has the greatest inertia? Select one: a. A 2 kilogram object moving at 5 m/s b. A 5 kilogram object moving at 3 m/s c. A 7 kilogram object at rest d. A 3 kilogram object moving at 4 m/s

Answer:

force required to push the block = 219.714 lb

Explanation:

GIVEN DATA:

weight W of block = 250 lb

coefficient of friction = 0.2

consider equilibrium condition in x direction

[tex]P*cos(30)-W*sin(30)-\mu _{s}N = 0[/tex]

[tex]P*0.866-0.2N = 125[/tex].........................(1)

consider equilibrium condition in Y direction

[tex]N-Wcos(30)-Psin(30)= 0[/tex]

[tex]N-0.5P=216.503[/tex].....................(2)

SOLVING 1 and 2 equation we get N value

N = 326.36 lb

putting N value in either equation we get force required to push the block = 219.714 lb

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

Answer:

The marble at its highest point are hmax= 1.39 meters above the floor.

Explanation:

hi= 1m

V= 3m/s

Vx= 1.1 m/s

Vy= ?

α=cos⁻¹(1.1/3)

α= 68.48º

Vy= V * sin(α)

Vy= 2.79 m/s

Vertical Velocity in highest point are Vy(hmax) = 0

Vy(hmax)= Vy - g*t

t= Vy/g

t= 0.28 sec

t : time to get the highest point

hmax= hi + Vy*t - g*t²/2

hmax= 1.39 m

The unknown capacitance C connected parallelly with a 30-μF capacitor initially charged at 80 V is calculated to be 90 μF. After being connected, the total potential difference across both capacitors reduces to 20 V.

The question deals with capacitors and how they function in a circuit. A capacitor is an electronic component that stores electrical energy and releases it in the circuit when necessary. When a charged 30-μF capacitor is connected across an initially uncharged capacitor, the voltage or potential difference across the connected capacitors will equalize. In this case, the final potential difference is noted as 20 V.

We know that for capacitors connected in parallel, the total charge stored is the sum of the charges stored in each capacitor, i.e., Q = Q₁ + Q₂, where Q is the total charge, and Q₁ and Q₂ are the charges stored in capacitor 1 (the 30-μF capacitor) and 2 (unknown capacitance C), respectively.

By using the formula Q = CV (charge = capacitance × voltage), where C is the capacitance and V is the voltage, we understand that Q₁ (charge on the 30 uF capacitor) is initially 30 μF × 80 V = 2400 μC (micro coulombs). After connecting the uncharged capacitor, the voltage drops to 20V, thus the final charge on the 30 uF capacitor becomes 30 μF × 20 V = 600 μC. The remaining charge must then be stored in the previously uncharged capacitor.

So, the charge on the unknown capacitor C would be the difference i.e., 2400 μC - 600 μC = 1800 μC. Now using the Q = CV formula, we can calculate the unknown capacitance C. V here is 20V as capacitors connected in parallel have the same voltage. So, C = Q / V = 1800 μC / 20 V = 90 μF. Thus, the unknown capacitance is 90 μF.

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Answer:

Velocity of the particle at time t = a

        [tex]v(a)=-\frac{12}{a^3}[/tex]

Velocity of the particle at time t = 1

         [tex]v(1)=-12m/s[/tex]

Velocity of the particle at time t = 2

         [tex]v(2)=-1.5m/s[/tex]

Velocity of the particle at time t = 3

          [tex]v(3)=-0.44m/s[/tex]

Explanation:

Displacement,

          [tex]s(t)=\frac{6}{t^2}[/tex]

Velocity is given by

          [tex]v(t)=\frac{ds}{dt}=\frac{d}{dt}\left ( \frac{6}{t^2}\right )=-\frac{12}{t^3}[/tex]

Velocity of the particle at time t = a

        [tex]v(a)=-\frac{12}{a^3}[/tex]

Velocity of the particle at time t = 1

         [tex]v(1)=-\frac{12}{1^3}=-12m/s[/tex]

Velocity of the particle at time t = 2

         [tex]v(2)=-\frac{12}{2^3}=-1.5m/s[/tex]

Velocity of the particle at time t = 3

          [tex]v(3)=-\frac{12}{3^3}=-0.44m/s[/tex]

The velocity of the particle for different times (t = a, 1, 2, and 3 seconds) is found by differentiating the displacement equation s = 6/t^2 and applying the values of t to get the velocities: v(a) = -12/a^3, v(1) = -12 m/s, v(2) = -1.5 m/s, and v(3) ≈ -0.44 m/s.

The question you've asked is about finding the velocity of a particle moving in a straight line where its displacement is given by s = 6/t2, and t is the time in seconds. To find the velocity (v) at any given time, we need to take the derivative of the displacement with respect to time. So the derivative of s with respect to t gives us v = -12/t3. Let's apply this formula for t = a, 1, 2, and 3 seconds.

Note that the negative sign indicates the direction of velocity is opposite to the direction assumed as positive in the displacement equation.

The magnitude of the instantaneous acceleration of the object is 12.3 m/s².

The magnitude of the instantaneous acceleration of the object can be calculated using the formula:

a = v² / r

Where v is the speed and r is the radius of the circle. In this case, the speed is the distance traveled divided by the time taken, which is equal to the circumference of the circle divided by the time taken:

v = (2πr) / t

Substituting the values given in the question, we can calculate the acceleration:

a = [(2πr) / t]² / r = (4π²r²) / t² = (4π² * (2.5)²) / (4)² = 12.3 m/s²

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The instantaneous acceleration of a particle moving in a circular path of diameter 5.0 m at a constant speed, completing the path in 4.0 s, is approximately 6.17 m/s².

The subject of this question is the instantaneous acceleration of a particle moving in a circle, which is a topic within physics, specifically, kinematics. This is a case of uniform circular motion, which implies a constant speed, but changing direction, thereby resulting in centripetal (towards the center) acceleration.

The radius of the circle is half of its diameter: r = 5.0 m/2 = 2.5 m. As the particle completes a full circle in 4.0 s, the speed can be calculated using circumference (2πr) and time: v = 2πr/t = 2π * 2.5 m/4.0 s ≈ 3.92 m/s. The centripetal acceleration (the instantaneous acceleration for a particle in a uniform circular motion) can be calculated as: a = v²/r = (3.92 m/s)²/2.5 m ≈ 6.17 m/s².

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Answer:

3.26 x 10^9 A

Explanation:

I1 = 7.2 A, r = 18.1 cm = 0.181 m, F/l = 2.6 x 10^4 N/m

Let teh current in other wire is I2.

Use the formula of force per unit length

[tex]F / l = \frac{\mu _{0}}{4\pi }\times \frac{2 I_{1}I_{2}}{r}[/tex]

[tex]2.6 \times 10^{4} = 10^{-7}\frac{2 \times 7.2I_{2}}{0.181}[/tex]

I2 = 3.26 x 10^9 A

Final answer:

The current in the second wire is 2.0 A.

Explanation:

The current in the second wire is 2.0 A.

Given that the force per unit length between the wires is directly proportional to the product of their currents, we can set up a proportion to find the current in the second wire:

[tex](7.2 A) / (2.6 x 10^4 N/m) = (x A) / (2.6 x 10^4 N/m)[/tex]

Solving for x gives x = 2.0 A, which is the current in the second wire.

Answer:

1+5x+10x^2+10x^3+5x^4+x^5

Explanation:

You just make it (1+x)(1+x)(1+x)(1+x)(1+x) and multiply it out until it's all one big term.

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

[tex]A_{1}\times V_{1} = A_{2}\times V_{2}[/tex]

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

[tex]A_{1} = 0.070 m^{2}[/tex]

[tex]V_{1} = 3.50 m/s[/tex]

a) [tex]A_{2} = 0.105 m^{2}[/tex]

substituting the values in the continuity equation, we get

[tex]0.070\times 3.50 = 0.105\times V_{2}[/tex]

or

[tex]V_{2} = \frac{0.070\times 3.5}{0.150}m/s[/tex]

or

[tex]V_{2} = 2.33m/s[/tex]

b) [tex]A_{2} = 0.047 m^{2}[/tex]

substituting the values in the continuity equation, we get

[tex]0.070\times 3.50 = 0.047\times V_{2}[/tex]

or

[tex]V_{2} = \frac{0.070\times 3.5}{0.047}m/s[/tex]

or

[tex]V_{2} = 5.21m/s[/tex]

c) we have,

Discharge[tex]Q = Area (A)\times Velocity(V)[/tex]

thus from the given value, we get

[tex]Q = 0.070m^{2}\times 3.5m/s\[/tex]

[tex]Q = 0.245 m^{3}/s[/tex]

Also,

Discharge[tex]Q = \frac{volume}{time}[/tex]

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

[tex]0.245m^{3}/s = \frac{volume}{3600s}[/tex]

or

[tex]0.245m^{3}/s\times 3600 = Volume[/tex]

volume of flow = [tex]882 m^{3}[/tex]

Using the continuity equation for incompressible fluids, we calculated the velocity of water at points in a pipe with different cross-sectional areas and also determined the volume of water discharged from the pipe in one hour.

To answer the fluid dynamics question, we will use the principle of conservation of mass, specifically the continuity equation for incompressible fluids, which states that the product of the cross-sectional area (A) and the velocity (v) of the fluid must remain constant at all points in the flow. This can be written as A1 * v1 = A2 * v2, where A1 and v1 are the area and velocity at point 1, and A2 and v2 are the corresponding values at point 2.

To find the fluid speed at the point where the cross-sectional area is 0.105 m2, we use the provided information:

Thus, v2 = (A1 * v1) / A2 = (0.070 m2 * 3.50 m/s) / 0.105 m2 = 2.33 m/s

Next, to find the fluid speed at the point where the cross-sectional area is 0.047 m2, we have:

The velocity at this point can be calculated as v3 = (A1 * v1) / A3 = (0.070 m2 * 3.50 m/s) / 0.047 m2 = 5.21 m/s

For the volume discharged from the open end of the pipe in 1.00 hour, we use the flow rate at point 1, with A1 and v1. The flow rate Q1 = A1 * v1 = 0.070 m2 * 3.50 m/s = 0.245 m3/s. To get the volume for one hour, we need to convert seconds to hours, knowing that there are 3600 seconds in one hour:

Volume = Q1 * time = 0.245 m3/s * 3600 s = 882 m3

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Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

  m=8/32=0.25

To find the spring constant

Kx=mg    (given that c=6 in and mg=8 lb)

K(0.5)=8               (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F  

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

A steady-state response is the behavior of a circuit over a lengthy period of time when stable conditions have been achieved. The steady-state response of this system will be y= 240/901 cos 2t+ 8/901 sin 2t.

A steady-state response is the behavior of a circuit over a lengthy period of time when stable conditions have been achieved following an external stimulus.

The given data in the problem will bge;

C=0.25 lb.s/ft

Weight is defined as the product of mass and gravity.

[tex]\rm{m=\frac{W}{g} }\\\\\rm{m=\frac{8}{32}[/tex]

[tex]\rm m=0.25[/tex]

Spring constant is defined as the ratio of force per unit displaced length.

The spring force is balanced by the weight;

[tex]\rm Kx=mg\\\\ \rm x= \frac{mg}{K} \\\\ \rm x=\frac{8}{0.5} lb/ft[/tex]

The equation for the spring-mass system is given by;

[tex]\rm {my''+Cy'+Ky=F }[/tex]

[tex]\rm 0.25 y"+0.25 y'+16 y=4 cos 20 t[/tex]

Steady-state equation;

[tex]\rm y=A cos 2t+ B sin 2t[/tex]

For finding the value of A and B

[tex]\rm y'= -2A sin 2t+2B cos 2ty"=-4A cos 2t-4B sin 2t[/tex]

By putting the value we got

[tex]\rm 0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t[/tex]

The value of cofficient obtained from the equation

[tex]30 A+ B=8[/tex]

Getting the value as

[tex]A= \frac{240/901}\\\\ B=\frac{8}{901}[/tex]

The steady-state response got

[tex]y= 240/901 cos 2t+ 8/901 sin 2t[/tex]

Hence the steady-state response of this system.y= 240/901 cos 2t+ 8/901 sin 2t

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Answer:

Coefficient of static friction between the road and the car's tire is 0.81

Explanation:

It is given that,

Velocity of the car, v = 40 m/s

Radius of the curve, r = 200 m

We need to find the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding. Let it is equal to μ.

The centripetal force of the car is balanced by the force of friction as :

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

[tex]\mu=\dfrac{v^2}{rg}[/tex]

[tex]\mu=\dfrac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}[/tex]

[tex]\mu=0.81[/tex]

So, the coefficient of static friction between the road and the car's tire is 0.81 Hence, this is the required solution.

The minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816.

Given the data in the question;

For a not car not to slide of the road, the frictional force and centripetal force should balance each other.

That is; Frictional force = Centripetal force

[tex]Frictional\ Force = uF = umg \\\\Centripetal\ Force = \frac{mv^2}{r}[/tex]

So,

[tex]u_{min}mg = \frac{mv^2}{r} \\\\u_{min}g =\frac{v^2}{r}\\\\u_{min} =\frac{v^2}{gr}\\[/tex]

Where [tex]u_{min}[/tex] is the minimum coefficient friction, v is the velocity, r is the radius and g is acceleration due to gravity( [tex]g = 9.8m/s^2\\[/tex])

We substitute our values into the equation

[tex]u_{min} = \frac{(40.0m/s)^2}{9.8m/s^2\ *\ 200m}\\\\ u_{min} = \frac{1600m^2/s^2}{1960m^2/s^2} \\\\u_{min} = 0.81[/tex]

Therefore, the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816

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Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

[tex]\Delta P.E=mg(h_{2}-h_{1})[/tex]

[tex]\Delta P.E=10000\times9.8\times(10000-0)[/tex]

[tex]\Delta P.E=10000\times9.8\times10000[/tex]

[tex]\Delta P.E=980000000\ J[/tex]

[tex]\Delta P.E=980\ MJ[/tex]

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

[tex]\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)[/tex]

[tex]\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)[/tex]

[tex]\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)[/tex]

[tex]\Delta K.E=148298642\ J[/tex]

[tex]\Delta K.E=148.3\ MJ[/tex]

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Answer:

The least amount of time in which the fisherman can raise the fish to the  dock without losing it is t= 2 seconds.

Explanation:

m= 5 kg

h= 2m

Fmax= 54 N

g= 9.8 m/s²

W= m * g

W= 49 N

F= Fmax - W

F= 5 N

F=m*a

a= F/m

a= 1 m/s²

h= a * t²/2

t= √(2*h/a)

t= 2 seconds

The least amount of time in which the fisherman can raise the fish to the dock without losing it depends on the tension in the fishing line. The tension can be calculated using the equation T = mg, where T is the tension, m is the mass of the fish, and g is the acceleration due to gravity. As long as the tension in the line is below the breaking point, the fisherman can safely raise the fish to the dock.

In order to determine the least amount of time in which the fisherman can raise the fish to the dock without losing it, we need to calculate the tension in the fishing line. We know that the fishing line will break under a tension of 54 N or more. The tension in the line can be calculated using the equation T = mg, where T is the tension, m is the mass of the fish, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Given that the mass of the fish is 5.0 kg, we can calculate the tension in the line as follows:

T = (5.0 kg) (9.8 m/s²) = 49.0 N

Therefore, the least amount of time in which the fisherman can raise the fish to the dock without losing it is determined by the tension in the line. As long as the tension in the line is below 54 N, the fisherman can safely raise the fish to the dock without the line breaking.

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Answer:

3.99 m option c

Explanation:

n = 1.33

Real depth / Apparent depth = n

Real depth = n × apparent depth

Real depth = 1.33 × 3

Real depth = 3.99 m

The apparent depth of the fish is not the same as its actual depth due to refraction. The actual depth can be calculated using the equation for apparent depth and the refractive index of water. In this case, the actual depth of the fish is 2.26 m.

The apparent depth of an object submerged in water can be different from its actual depth due to refraction. In this case, the fish appears to be 3 m below the surface when viewed from above by a fisherman. To find the actual depth, we can use the equation for apparent depth, which states that the apparent depth is equal to the actual depth divided by the refractive index of the medium.

Using this equation, we can calculate the actual depth as follows:

Actual depth = Apparent depth ÷ Refractive index of water

Given that the refractive index of water is 1.33, we can substitute the values and calculate the actual depth.

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Answer:

Axial flow pumps can usually handle large flow rates. They hence have high specific speed - b.

Axial flow pumps are capable of handling large flow rates and have a high specific speed, which is a contrast to positive displacement pumps like diaphragm pumps that deliver a constant flow regardless of pressure.

Axial flow pumps are designed to handle large flow rates, which indicates their ability to move a high volume of fluid. The specific speed of a pump describes how efficient it is at handling different flow rates. In this context, axial flow pumps that can handle large flow rates usually have low specific speed (a) as they are more effective at high flow rates than pumps with higher specific speeds.

Axial flow pumps are designed to handle very large flow rates and are characterized by their ability to deliver fluid primarily in a direction parallel to the pump shaft. This capability allows them to operate efficiently under conditions where a significant volume of liquid needs to be moved across relatively short distances with lower head (pressure) requirements. The concept of specific speed is important when discussing pump performance. It is a dimensionless number that describes a pump's shape and characteristics based on its speed, flow rate, and head. Given that axial flow pumps are optimal for applications with high flow rates and lower head, they are distinguished by a high specific speed. This is in contrast to positive displacement pumps, like diaphragm pumps, which deliver a constant flow for each revolution of the pump shaft, regardless of changes in pressure, highlighting a key difference in operation and efficiency between pump types.

Answer:

Explanation:

Givens

Time taken to go down + time taken for the sound to come up = 1.17 seconds.

d = 7.92 m

a = 9.81 m/s^2

t (see below)

vi = ???

Solution to How long it takes to come back up.

v = 343 m/s

d = 7.92 meters

t = ?

t = d/v

t = 7.92 m / 343 m/s

t = 0.0231 seconds.

Solution to time taken to go down.

Time_down = 1.17 - 0.0231

time_down = 1.147 seconds

Solution to vi

d = vi*t + 1/2 a t^2

7.92 = vi*1.147 + 1/2 * 9.81 * 1.147^2

7.92 = vi*1.147 + 6.452                    Subtract 6.452 from both sides.

7.92 - 6.452 = 1.147*vi

1.468 = 1.147 * vi                              Divide by 1.147

1.468 / 1.147 = vi            

1.279 m/s = vi

Answer:

59.5 deg

Explanation:

[tex]v[/tex] = original speed at which the projectile is launched

θ = angle of launch of projectile

[tex]v_{x}[/tex] = component of speed along the horizontal direction =  [tex]v [/tex] Cosθ

At the highest position, the vertical component of velocity becomes zero and there is only horizontal component of velocity, hence

[tex]v_{highest}[/tex] = velocity at the highest point = [tex]v_{x}[/tex] = [tex]v [/tex] Cosθ

it is given that

[tex]v_{highest}[/tex] = 0.508 [tex]v[/tex]

so

[tex]v[/tex] Cosθ = 0.508 [tex]v[/tex]

Cosθ = 0.508

θ = 59.5 deg

Answer:

3 m/s

Explanation:

M1 = 87 kg, u1 = ?

M2 = 22 kg, u = 0

After grabbing, let the velocity is V.

V = 2.4 m/s

By using the conservation of momentum

Momentum before grabbing = momentum after grabbing

M1 x u1 + M2 x u2 = (M1 + M2) x V

87 x u1 + 0 = (87 + 22) x 2.4

87 u1 = 261.6

u1 = 3 m/s

Answer: 1.00

Explanation:

The index of refraction [tex]n[/tex] is a number that describes how fast light propagates through a medium or material.

Being its equation as follows:

[tex]n=\frac{c}{v}[/tex]  (1)

Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum and [tex]v[/tex] its speed in the other medium and [tex]n=1.55[/tex].

So, from (1) we can find the velocity at which the signal travels and then the time it requires to travel:

[tex]v=\frac{c}{n}[/tex]  (2)

[tex]v=\frac{3(10)^{8}m/s}{1.55}[/tex]  (3)

[tex]v=193548387.1m/s[/tex]  (4)

Now, knowing the velocity [tex]v[/tex] is the distance [tex]d=0.35m[/tex] traveled in a time [tex]t[/tex]:

[tex]v=\frac{d}{t}[/tex]  (5)

We can isolate [tex]t[/tex] from (5) and find the value of the required time:

[tex]t=\frac{d}{v}[/tex]  (6)

[tex]t=\frac{0.35m}{193548387.1m/s}[/tex]  (7)

[tex]t=0.000000001s=1(10)^{-9}s=1ns[/tex]  (8) This is the time it takes the signal to travel through the optical fiber: 1 nanosecond.

Final answer:

The time for a signal to travel 0.35 m through an optical fiber with an index of refraction of 1.55 is approximately 1.81 nanoseconds.

Explanation:

To calculate the time required for a signal to travel through an optical fiber with an index of refraction of 1.55, we use the equation that relates the speed of light in a vacuum (c) to the speed of light in a material (v) using the index of refraction (n): v = c/n. Given that the speed of light in a vacuum is approximately 3.00×108 m/s, we can find the speed of light in the optical fiber.

The speed of light in the fiber (v) is:
v = (3.00 × 108 m/s) / 1.55 = 1.935×108 m/s.

Now, we calculate the time (t) it takes for light to travel 0.35 m in the fiber, using the formula
t = distance/speed:

t = 0.35 m / (1.935 × 108 m/s).

The calculated time t is found to be approximately 1.81 nanoseconds, after converting from seconds by multiplying by 109.

Answer:

Explanation:

a) using the energy conservation  equation

mgh = 0.5mv^2 + 0.5Iω^2

I(moment of inertia) (basket ball) = (2/3)mr^2

mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)

gh = 1/2v^2 + 1/3v^2

gh = v^2( 5/6)

v =  [tex]\sqrt{\frac{6gh}{5} }[/tex]

putting the values we get

[tex]6.6 ^{2} = \frac{6\times9.8h}{5}[/tex]

solving for h( height)

h = 3.704 m apprx

b) velocity of solid cylinder

mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)

g*h = 1/2v^2 + 1/4v^2

g*h = 3/4v^2

putting the value of h and g we get

v= = 6.957 m/s apprx

Answer:

19.09 m/s

Explanation:

u = 0, h = 18.6 m

Use third equation of motion

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 18.6

v = 19.09 m/s

Answer:

The wavelength of light is [tex]4.53\times10^{-7}\ m[/tex]

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

[tex]d \sin\theta=n\lambda[/tex].....(I)

The separation of the slits

[tex]d = \dfrac{1}{N}[/tex]

[tex]d=\dfrac{1}{5000}[/tex]

[tex]d=2\times10^{-6}\ m[/tex]

Now put the value in equation (I)

[tex]2\times10^{-6}\sin13.1^{\circ}=\lambda[/tex]

Here, n = 1

[tex]\lambda=4.53\times10^{-7}\ m[/tex]

Hence, The wavelength of light is [tex]4.53\times10^{-7}\ m[/tex]

Answer:

5.14 J

Explanation:

l = 81 cm = 0.81 m

diameter = 3 mm

Radius, r = 1.5 mm = 1.5 x 10^-3 m

Change in temperature, T2 - T1 = 50 degree

t = 1.8 minutes = 1.8 x 60 = 108 s

k = 109 W/mk

Heat transfer = K A (T2 - T1) t / l

H = 109 x 3.14 x 1.5 x 10^-3 x 1.5 x 10^-3 x 50 x 108 / 0.81

H = 5.14 J

Answer:

E = 2.5 x 10⁶ N/C

Explanation:

V = Potential difference between the plates of a TV set = 25 kV = 25000 Volts

d = Distance between the plates of TV set = 1.0 cm = 0.01 m

E = Electric field in the region between the plates

Electric field between the plates is given as

[tex]E= \frac{V}{d}[/tex]

Inserting the values

[tex]E= \frac{25000}{0.01}[/tex]

E = 2.5 x 10⁶ N/C

The uniform electric field is the field in which the value of the electric field strength remain same at each point.

The magnitude of the uniform electric field in the region between the plates  [tex]2.5\times10^6 \rm N/C[/tex].

The uniform electric field is the field in which the value of the electric field strength remain same at each point.

It can be given as,

[tex]E=\dfrac{V}{d}[/tex]

Here, [tex]V[/tex] is the  potential difference between two points and [tex]d[/tex] is the distance between two points.

Given information-

The  potential difference between the accelerating plates of a TV set is about 25 k-V or 25000 V.

The distance between the plates is 1.0 cm or 0.01 meters.

Use the above formula to find the magnitude of the uniform electric field in the region between the plates as,

[tex]E=\dfrac{25000}{0.01}\\E=2.5\times10^6 \rm N/C[/tex]

Thus the magnitude of the uniform electric field in the region between the plates  [tex]2.5\times10^6 \rm N/C[/tex].

Learn more about uniform electric field here;

https://brainly.com/question/14632877

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

[tex]F=\mu mg = m \frac{v^2}{r}[/tex] (1)

where

[tex]\mu[/tex] is the coefficient of friction

m is the mass of the car

g = 9.8 m/s^2 is the acceleration due to gravity

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

[tex]\mu=0.50\\v = 20 mph = 8.9 m/s[/tex]

So the radius of the curve is

[tex]r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m[/tex]

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when [tex]\mu=1.0[/tex]. We find:

[tex]v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph[/tex]

The maximum speed at which the car can take the same curve on a sunny day is about 28 mph

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

coefficient of friction on a snowy day = μs₁ = 0.50

maximum speed of the car on a snowy day = v₁ = 20 mph

coefficient of friction on a sunny day = μs₂ = 1.0

Asked:

maximum speed of the car on a snowy day = v₂ = ?

Solution:

Firstly , we will derive the formula to calculate the maximum speed of the car:

[tex]\Sigma F = ma[/tex]

[tex]f = m \frac{v^2}{R}[/tex]

[tex]\mu N = m \frac{v^2}{R}[/tex]

[tex]\mu m g = m \frac{v^2}{R}[/tex]

[tex]\mu g = \frac{v^2}{R}[/tex]

[tex]v^2 = \mu g R[/tex]

[tex]\boxed {v = \sqrt { \mu g R } }[/tex]

[tex]\texttt{ }[/tex]

Next , we will compare the maximum speed of the car on a snowy day and on the sunny day:

[tex]v_1 : v_2 = \sqrt { \mu_1 g R } : \sqrt { \mu_2 g R }[/tex]

[tex]v_1 : v_2 = \sqrt { \mu_1 } : \sqrt { \mu_2 }[/tex]

[tex]20 : v_2 = \sqrt { 0.50 } : \sqrt { 1.0 }[/tex]

[tex]20 : v_2 = \frac{1}{2} \sqrt{2}[/tex]

[tex]v_2 = 20 \div \frac{1}{2} \sqrt{2}[/tex]

[tex]v_2 = 20 \sqrt{2} \texttt{ mph}[/tex]

[tex]\boxed{v_2 \approx 28 \texttt{ mph}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

Answer:

[tex]T = 308.6 ^0 C[/tex]

Explanation:

Here by ideal gas equation we can say

[tex]PV = nRT[/tex]

now we know that pressure is kept constant here

so we will have

[tex]V = \frac{nR}{P} T[/tex]

since we know that number of moles and pressure is constant here

so we have

[tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]

now we know that initial temperature is 17.8 degree C

and finally volume is doubled

So we have

[tex]\frac{2V}{V} = \frac{T_2}{(273 + 17.8)}[/tex]

so final temperature will be

[tex]T_2 = 581.6 k[/tex]

[tex]T_2 = 308.6 ^o C[/tex]

Answer:

[tex]v = 567.2 km/h[/tex]

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

[tex]distance = 2\pi r[/tex]

[tex]distance = \pi D[/tex]

now we will have the time to complete the rotation given as

[tex]t = 6 days[/tex]

[tex]t = 6 (24 h) = 144 h[/tex]

now the speed is given by

[tex]speed = \frac{distance}{time}[/tex]

[tex]speed = \frac{\pi D}{t}[/tex]

[tex]speed = \frac{\pi(26000 km)}{144}[/tex]

[tex]v = 567.2 km/h[/tex]

Explanation:

The electric field in an electromagnetic wave is described by:

[tex]E_y=112\ sin(1.4\times 10^7x-\omega t)[/tex]

The general equation is given by :

[tex]E=E_0\ sin(kx-\omega t)[/tex]

Here,

The amplitude in electric field is [tex]E_0=112[/tex]

Propagation constant, [tex]k=1.4\times 10^7[/tex]

[tex]\omega[/tex] is the angular frequency

(a) The amplitude of the corresponding magnetic field oscillations is given by :

[tex]B_0=\dfrac{E_0}{c}[/tex]

[tex]B_0=\dfrac{112}{3\times 10^8}[/tex]

[tex]B_0=3.73\times 10^{-7}\ T[/tex]

[tex]B_0=0.37\ \mu T[/tex]

(b) We know that the propagation constant is given by :

[tex]k=\dfrac{2\pi}{\lambda}[/tex]

[tex]\lambda=\dfrac{2\pi}{k}[/tex]

[tex]\lambda=\dfrac{2\pi}{1.4\times 10^7}[/tex]

[tex]\lambda=4.48\times 10^{-7}\ m[/tex]

[tex]\lambda=0.44\ \mu T[/tex]

(c) [tex]f=\dfrac{c}{\lambda}[/tex]

[tex]f=\dfrac{3\times 10^8\ m/s}{4.48\times 10^{-7}\ m}[/tex]

[tex]f=6.69\times 10^{14}\ Hz[/tex]

Answer:

95.47 C

Explanation:

Heat added = mass × heat capacity × rise in temperature

So, rise in temperature =

20000 / (0.05 × 4190)

Rise in temperature = 95.47 C

To calculate the final temperature of [tex]50 g[/tex]  of water heated with [tex]20000 J[/tex], you use the equation [tex]Q = mc\Delta T[/tex]. The calculated temperature change is [tex]95.5^{\circ}C[/tex], and adding this to an initial temperature of [tex]25^{\circ}C[/tex] gives [tex]120.5^{\circ}C[/tex]. Because water boils at [tex]100^{\circ}C[/tex], the final temperature would be limited to [tex]100^{\circ}C[/tex].

To calculate the final temperature of [tex]50 g[/tex] of water heated with [tex]20000 J[/tex] of energy, we can use the equation:

[tex]Q = mc\Delta T[/tex]

where:

Heat energy, [tex]Q = (20000 J)[/tex]

Mass of water. [tex]m = (50 g = 0.05 kg)[/tex]

Specific heat capacity of water, [tex]c = (4190 \, \text{J/kg°C})[/tex]

Change in temperature, [tex]\Delta T = (T_{\text{final}} - T_{\text{initial}})[/tex]

Rearrange the equation to solve for[tex]\Delta T[/tex]:

[tex]\Delta T = \frac{Q}{mc}[/tex]

Substitute the values:

[tex]\Delta T = \frac{20000 \, \text{J}}{0.05 \, \text{kg} \times 4190 \, \text{J/kgC}}[/tex]

[tex]\approx 95.5^{\circ}C[/tex]

If the initial temperature of the water was[tex]25^{\circ}C[/tex] (for example), then the final temperature would be:

[tex]T_{\text{final}} = T_{\text{initial}} + \Delta T\\T_{\text{final}} = 25^\circ \text{C} + 95.5^\circ \text{C} = 120.5^\circ \text{C}[/tex]

However, remember that water boils at 100°C under standard pressure, so the actual final temperature would be limited to .

Solution:

To calculate the pressure ratio for a sound for measuring sound, we use the following logarithmic formula of Sound Pressure Level  (SPL):

SPL = [tex]20\log_{10}\frac{p}{p_{ref}} dB[/tex]

where,

p = pressure to be measured

[tex]p_{ref} [/tex] = reference pressure

[tex]\frac{90}{20}[/tex] =  [tex]\log_{10}\frac{p}{p_{ref}} dB[/tex]

[tex]10^{4.5}[/tex] =  [tex]\frac{p}{p_{ref}}[/tex]

pressure ratio is :

[tex]\frac{p}{p_{ref}}[/tex] = [tex]3.16\times 10^{5}[/tex]

Final answer:

A 90 dB sound has a pressure ratio [tex]10^{4.5[/tex] times greater than the threshold of hearing (0 dB). Every 10 dB increase corresponds to doubling the sound pressure level, so a 90 dB sound is exponentially more intense than the reference level.

Explanation:

To compute the pressure ratio for a sound that is 90 dB SPL (Sound Pressure Level), we must first understand the relationship between decibels and pressure ratios. The reference pressure for 0 dB is 20 micropascals, which is equivalent to 10⁻⁹ atm. A change of 20 dB represents a tenfold increase in the pressure amplitude.

Since every 10 dB increase corresponds to the sound pressure level doubling, a 90 dB sound is 9 factors of 10 or 10⁹ times more intense than the threshold of hearing (0 dB). Therefore, the pressure ratio of a 90 dB sound compared to the reference sound (0 dB) is 10⁹ times greater.

To confirm, every 20 dB increase multiplies the pressure ratio by 10, so:
0 dB = 1 (reference level)
20 dB = 10
40 dB = 10²
60 dB = 10³
80 dB = 10
90 dB =[tex]10^{4.5[/tex]
Thus, a 90 dB sound signal is 1[tex]10^{4.5[/tex] times the pressure of the reference signal.