A traveling electromagnetic wave in a vacuum has an electric field amplitude of 69.1 V/m . Calculate the intensity ???? of this wave. Then, determine the amount of energy ???? that flows through area of 0.0247 m^2 over an interval of 13.1 s, assuming that the area is perpendicular to the direction of wave propagation.

Explanation:

To answer how far and how long the object travels, we need to know either the length of the table or the kinetic coefficient of friction between copper and glass.  We also need to know the kinetic coefficient of friction between the copper object and the steel incline.  This information wasn't provided, so for sake of illustration, I'll assume both coefficients are the same, and that μk = 0.35.

Let's divide the path into four sections to stay organized:

a) object moves up incline

b) object is in free fall

c) object slides across table

d) object falls onto books

a)

First, the object slides up the incline.  There are four forces acting on the object.  Weight pulling down, normal force perpendicular to the incline, tension up the incline, and friction down the incline.

Sum of the forces normal to the incline:

∑F = ma

N - W cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

T - F = ma

T - Nμ = ma

T - mgμ cos θ = ma

Now sum of the forces in the y direction on the hanging mass:

∑F = ma

T - W = M(-a)

T = Mg - Ma

Substituting:

Mg - Ma - mgμ cos θ = ma

Mg - mgμ cos θ = (m + M) a

a = g (M - mμ cos θ) / (m + M)

Given m = 33.1 g, M = 50.0 g, θ = 30°, μ = 0.35, and g = 9.8 m/s²:

a = 4.71 m/s²

So the velocity it reaches at the top of the incline is:

v² = v₀² + 2a(x - x₀)

v² = 0² + 2(4.71)(1.5)

v = 3.76 m/s

The time to reach the top of the incline is:

x = x₀ + v₀ t + ½ at²

1.5 = 0 + (0) t + ½ (4.71) t²

t = 0.798 s

The horizontal distance traveled is:

x = 1.5 cos 30°

x = 1.299 m

And the vertical distance traveled is:

y = 1.5 sin 30°

y = 0.750 m

b)

In the second stage, the object is in free fall.  v₀ = 3.76 m/s and θ = 30°.  The object lands on the table at the point where its speed is a minimum.  This is at the highest point of the trajectory, when the vertical velocity is 0.

v = at + v₀

0 = (-9.8) t + (3.76 sin 30°)

t = 0.192 s

The horizontal distance traveled is:

x = x₀ + v₀ t + ½ at²

x = 0 + (3.76 cos 30°) (0.192) + ½ (0) (0.192)²

x = 0.625 m

The height it reaches is:

v² = v₀² + 2a(y - y₀)

(0)² = (3.76 sin 30°)² + 2(-9.8)(y - 0)

y = 0.180 m

c)

The object is now on the glass table.  As it slides, there are three forces acting on it.  Normal force up, gravity down, and friction to the left.

In the y direction:

∑F = ma

N - W = 0

N = mg

In the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

The time it takes to reach the end of the table is:

v = at + v₀

0 = (-9.8×0.35) t + (3.76 cos 30°)

t = 0.949 s

And the distance it travels is:

v² = v₀² + 2a(x - x₀)

(0)² = (3.76 cos 30°)² + 2(-9.8×0.35) (x - 0)

x = 1.546 m

d)

Finally, the object falls onto the books.  There are 9 books, each 5 cm thick, so the height is 45 cm.  As the book falls, there are two forces acting on it: drag up and gravity down.

∑F = ma

D - W = ma

0.05 - (0.0331)(9.8) = 0.0331 a

a = -8.29 m/s²

Adding the vertical distances from parts a and b, we know the height of the table is 0.930 m.  So the time it takes the object to land is:

y = y₀ + v₀ t + ½ at²

0.45 = 0.93 + (0) t + ½ (-9.8) t²

t = 0.313 s

Adding up the times from all four parts, the total time is:

t = 0.798 + 0.192 + 0.949 + 0.313

t = 2.25 s

Adding up the horizontal distances, the total distance traveled is:

x = 1.299 + 0.625 + 1.546

x = 3.47 m

Answer:

D

Explanation:

Centripetal acceleration is the square of velocity divided by radius:

a = v² / r

Velocity is equal to angular velocity times radius, so this can also be written as:

a = ω² r

And angular velocity is 2π divided by the period, so:

a = (2π / T)² r

We can use these two equations to determine which scenario results in double the centripetal acceleration.

A) Keep the radius fixed and increase the period by a factor of two.

(2π / 2T)² r

1/4 (2π / T)² r

1/4 a

B) Keep the radius fixed and decrease the period by a factor of two.

(2π / ½T)² r

4 (2π / T)² r

4a

C) Keep the speed fixed and increase the radius by a factor of two.

v² / 2r

1/2 a

D) Keep the speed fixed and decrease the radius by a factor of two.

v² / ½r

2a

Only D doubles the centripetal acceleration.

To double the centripetal acceleration of a ball moving in a circular path at a constant velocity, halve the radius of the path while keeping the speed constant. This is because centripetal acceleration is inversely proportional to the radius.

This is possible because centripetal acceleration is inversely proportional to the radius (as described by the formula ac = v²/r). Conversely, increasing the radius or altering the period would not have the desired effect on the centripetal acceleration.

If the radius is halved and the speed remains constant, the force required to keep the object moving in a circular path increases, which is reflected in an increase in the centripetal acceleration. Hence, if you were to plot centripetal acceleration against radius, you would see it decrease as the radius increases and vice versa.

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Answer:

6300 J = 1.51 Calories

Explanation:

We have 1 Calorie = 1000 cal

1 cal = 4.184 J

That is 1 Calorie = 1000 cal = 1000 x 4.184 = 4184 J

Here we have 6300 J

That is

         [tex]1Calorie=4184J\\\\1J=\frac{1}{4184}Calorie\\\\6300J=\frac{6300}{4184}Calorie=1.51Calorie[/tex]

    6300 J = 1.51 Calories

Final answer:

The magnitude of the boat’s velocity with respect to the ground is 18.0 km/h downstream. The child’s velocity with respect to the ground is 14.0 km/h upstream.

Explanation:

To find the magnitude and direction of the boat’s velocity with respect to the ground, we can use vector addition. The boat's velocity with respect to the ground is the sum of its velocity with respect to the water and the water's velocity with respect to the ground. The magnitude of the boat's velocity with respect to the ground is the sum of its speed upstream and the speed of the water downstream: 10 km/h + 8.0 km/h = 18.0 km/h. The direction of the boat's velocity with respect to the ground is the same as the direction of the water's velocity, which is downstream.

To find the magnitude and direction of the child’s velocity with respect to the ground, we can also use vector addition. The child's velocity with respect to the ground is the sum of their velocity with respect to the boat and the boat's velocity with respect to the ground. The magnitude of the child's velocity with respect to the ground is the difference between their speed on the boat and the boat's speed with respect to the ground: 4.0 km/h - 18.0 km/h = -14.0 km/h (negative sign indicating direction opposite to the boat's velocity with respect to the ground). Therefore, the magnitude of the child's velocity with respect to the ground is 14.0 km/h. The direction is upstream, opposite to the direction of the boat's velocity with respect to the ground.

Answer:

a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m

Explanation:

This can be answered with either force analysis and kinematics, or work and energy.

a) Using force analysis, we can draw a free body diagram for the snowboarder.  There are two forces: normal force perpendicular to the slope and gravity down.

Sum of the forces parallel to the slope:

∑F = ma

mg sin θ = ma

a = g sin θ

Therefore, the velocity at the bottom is:

v² = v₀² + 2a(x - x₀)

v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)

v = 29.9 m/s

Alternatively, using energy:

PE = KE

mgh = 1/2 mv²

v = √(2gh)

v = √(2×9.8×45.7)

v = 29.9 m/s

b) Drawing a free body diagram, there are three forces on the snowboarder.  Normal force up, gravity down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N - mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Therefore, the snowboarder's final speed is:

v² = v₀² + 2a(x - x₀)

v² = (29.9)² + 2(-9.8×.102) (10 - 0)

v = 29.6 m/s

Using energy instead:

KE = KE + W

1/2 mv² = 1/2 mv² + F d

1/2 mv² = 1/2 mv² + mgμ d

1/2 v² = 1/2 v² + gμ d

1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)

v = 29.6 m/s

c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.

∑F = ma

-mg sin θ = ma

a = -g sin θ

Therefore, the final height reached is:

v² = v₀² + 2a(x - x₀)

(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)

h = 44.7 m

Using energy:

KE = PE

1/2 mv² = mgh

h = v² / (2g)

h = (29.6)² / (2×9.8)

h = 44.7 m

In the case of resistors in series and in parallel to a battery, the resistances can be determined by applying Ohm's law and using equations for total resistance in both series and parallel configurations. The calculations yield the values R1 = 1.53 ohms and R2 = 5.48 ohms.

The problem involves applying Ohm's law and equations for series and parallel resistors. When the resistors are in series, we add the resistances so R1 + R2 = 12.7 V / 1.81 A = 7.01 ohms. In a parallel connection, the total resistance R is given by 1/R = 1/R1 + 1/R2 or 12.7 V / 9.06 A = 1.40 ohms. Solving this pair of equations we get R1 = 1.53 ohms and R2 = 5.48 ohms, giving the pair of resistances asked for.

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Answer:

It's a long explanation

Explanation:

The Planck mass is approximately the mass of the Planck particle, a hypothetical minuscule black hole whose Schwarzschild radius equals the Planck length. In physics, the Planck mass, denoted by mP, is the unit of mass in the system of natural units known as Planck units. It is approximately 0.02 milligrams. Plank length as the minimum distance at which space can be thought. In physics, the Planck length, denoted ℓP, is a unit of length that is the distance light travels in one unit of Planck time. It is equal to 1.616255×10⁻³⁵ m. It is a base unit in the system of Planck units, developed by physicist Max Planck...I miss you, Landon.

Answer:

2 m/s

Explanation:

Momentum is conserved:

mv = MV

where m is mass of the boxcar, v is its initial velocity, M is the mass of all six box cars, and V is the final velocity.

v = 12 m/s, and M = 6m, so:

m (12 m/s) = 6m V

12 m/s = 6V

V = 2 m/s

Answer:

[tex]F = 1.5 \times 10^{-16} N[/tex]

this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

[tex]KE = 1 keV[/tex]

[tex]KE = 1 \times 10^3 (1.6 \times 10^{-19}) J[/tex]

[tex]KE = 1.6 \times 10^{-16} J[/tex]

now the speed of electron is given as

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

now we have

[tex]v = \sqrt{\frac{2 KE}{m}}[/tex]

[tex]v = 1.87 \times 10^7 m/s[/tex]

now the maximum force due to magnetic field is given as

[tex]F = qvB[/tex]

[tex]F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})[/tex]

[tex]F = 1.5 \times 10^{-16} N[/tex]

Now if this force is compared by the gravitational force on the electron then it is

[tex]\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}[/tex]

[tex]\frac{F}{F_g} = 1.68 \times 10^{13}[/tex]

so this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force

Final answer:

The maximum magnetic force on an electron in Earth's magnetic field is calculated using the charge of the electron, its velocity derived from kinetic energy, and the field strength. This force is much larger than the gravitational force acting on the electron due to its minuscule mass.

Explanation:

The maximum possible magnetic force on an electron moving through Earth's magnetic field can be found using the formula F = qvB sin(\theta), where q is the charge of the electron, v is its velocity, B is the magnetic field strength, and \theta is the angle between the velocity and the magnetic field. Since the maximum force occurs when the angle is 90 degrees (sin(\theta) = 1), we first need to calculate the velocity from its kinetic energy, 1 keV. Given the kinetic energy E = 1 keV = 1 × 103 eV and using the relation E = \frac{1}{2}mv2, we can find v. The electron charge is q = -1.6 × 10-19 C, and the magnetic field is B = 0.5 G = 5 × 10-5 T.

For the gravitational force, we use F = mg, where m is the mass of the electron and g is the acceleration due to gravity, approximately 9.8 N/kg on the surface of Earth. The mass of the electron is about 9.11 × 10-31 kg. Comparing these forces, the magnetic force is several orders of magnitude greater than the gravitational force acting on an electron due to its negligible mass.

Answer:

The frequency is 0.34 Hz and the wave speed is 1.05 m/s.

Explanation:

Given that,

The distance between two successive minima of a transverse wave is wavelength.

Wavelength = 3.10 m

Time = 14.7 s

(I). We need to calculate the frequency

[tex]f=\dfrac{number\ of\ wave\ propagated\ per\ second }{times}[/tex]

[tex]f=\dfrac{5}{14.7}[/tex]

[tex]f=0.34\ Hz[/tex]

(II). We need to calculate the wave speed

Formula of wave speed

[tex]v= \lambda\times f[/tex]

Where,

[tex]\lambda = wavelength[/tex]

[tex]f = frequency[/tex]

Put the value into the formula

[tex]v=3.10\times0.34[/tex]

[tex]v=1.05\ m/s[/tex]

Hence, The frequency is 0.34 Hz and the wave speed is 1.05 m/s.

The frequency of the wave is 0.34 Hz and the wave speed is 1.054 m/s.

The distance between two successive minima in a transverse wave is the same as the wavelength (λ). Given that the question provided this distance as 3.10 m, we can use this as the wavelength. On the other hand, the frequency (f) of the wave is the number of wave cycles that pass a point per unit time. The information from the question gives us five crests (complete cycles) over 14.7 seconds. Thus, we can calculate the frequency by dividing the total cycles by the total time, i.e., f = 5 cycles / 14.7 seconds = 0.34 Hz.

The speed of the wave (v) can be obtained from the equation v = λf. Substituting the values we have, v = (3.10 m) * (0.34 Hz) = 1.054 m/s. Therefore, the frequency of the wave is 0.34 Hz, and the wave speed is 1.054 m/s.

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Answer:

Escape velocity, v = 12.6 km/s

Explanation:

It is given that,

Mass of earth like planet, [tex]m=6.5\times 10^{24}\ kg[/tex]

Radius of planet, [tex]r=55\times 10^5\ m[/tex]

The escape velocity of a planet is given by the following formula as:

[tex]v=\sqrt{\dfrac{2Gm}{r}}[/tex]

G = universal gravitational constant

[tex]v=\sqrt{\dfrac{2\times 6.67\times 10^{-11}\ m^3kg^{-1}s^{-2}\times 6.5\times 10^{24}\ kg}{55\times 10^5\ m}[/tex]

v = 12556.05 m/s

or

v = 12.6 km/s

Hence, this is the required solution.

Try this option (shortly):

According to the 'law of consevation of pulse' the passenger in the smaller car receives greater energy then the one in the larger car.

Almost all the received energy is spent to injure the passenger in the smaller car. :(

Final answer:

Passengers in a smaller mass car are at greater risk in a head-on collision due to higher acceleration and force resulting from the collision. Cars with similar masses experience a more even distribution of momentum changes, potentially reducing impact severity. The duration of impact also affects the decelerating force felt by occupants, illustrating the importance of seatbelts and crumple zones.

Explanation:

A passenger in the smaller mass car is more likely to be injured in a head-on collision with a larger mass car because the smaller car will experience a larger acceleration due to the force of impact, according to Newton's second law (force equals mass times acceleration). This greater acceleration translates to a stronger force exerted on the occupants of the smaller car. Additionally, according to the principle of conservation of momentum, if both cars are moving at the same speed and collide, the car with less mass will undergo a more significant change in velocity.

When cars with more similar masses collide, the change in momentum is distributed more evenly between the two, potentially reducing the severity of the impact on both vehicles. This is due to the momentum of the two-car system remaining constant if external forces are negligible (as stated by the law of conservation of momentum). The impulse, or change in momentum, is also important to consider. If cars crumple together instead of rebounding, there's a smaller change in momentum, leading to a smaller force on the occupants, which is why crumple zones are critical safety features in cars.

The scenario with a seatbelt demonstrates the importance of the time over which a force acts. A seatbelt helps extend the time of a collision's impact, reducing the average decelerating force experienced by the occupant, compared to a sudden stop against the dashboard.

Answer:

All materials are superconducting at temperatures near absolute zero kelvin.

Explanation:

All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.

c. All materials are superconducting at temperatures near absolute zero kelvin.

A superconductor is a material that achieves superconductivity, which is a state of matter that has no electrical resistance and does not allow magnetic fields to penetrate.

An electric current in a superconductor can persist indefinitely. Superconductivity can only typically be achieved at very cold temperatures.

Superconductivity is a phenomenon observed in several metals and ceramic materials.

When these materials are cooled to temperatures ranging from near absolute zero ( 0 degrees Kelvin, -273 degrees Celsius) to liquid nitrogen temperatures ( 77 K, -196 C), their electrical resistance drops with a jump down to zero.

Therefore,

All materials are superconducting at temperatures near absolute zero kelvin.

Learn more about superconductor here:

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Answer:

Minimum stopping distance = 164.69 ft

Explanation:

Speed of car = 35.6 mi/hr = 15.82 m/s

Stopping distance = 40.8 ft = 12.44 m

We have equation of motion

             v² = u² + 2as

             0²=15.82²+ 2 x a x 12.44

              a = -10.06 m/s²

Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.

          Speed of car = 71.5 mi/h = 31.78 m/s

We have

           v² = u² + 2as

          0² = 31.78² -  2 x 10.06 x s

           s = 50.2 m = 164.69 ft

Minimum stopping distance = 164.69 ft

Answer: [tex]V=\sqrt{\frac{2GM}{R}}[/tex]

Explanation:

Taking into account what is stated in this problem and considering there is no friction during the takeoff of the rocket of the planet, the rocket will escape the gravitational attraction of the massive body when its kinetic energy [tex]K[/tex] and its potential energy [tex]P[/tex] are equal in magnitude.

Written mathematically is:

[tex]K=P[/tex]  (1)

Where:

[tex]K=\frac{1}{2}mV^{2}[/tex] (2)

Being [tex]m[/tex] the mass of the rocket

And:

[tex]P=-\frac{GMm}{R}[/tex]   (3)

Being  [tex]M[/tex] the mass of the planet,  [tex]G[/tex] the gravitational constant and  [tex]R[/tex] the radius of the planet.

Substituting (2) and (3) in (1):

[tex]\frac{1}{2}mV^{2}=-\frac{GMm}{R}[/tex] (4)

Finding [tex]V[/tex], which is the escape velocity:

[tex]V=\sqrt{\frac{2GM}{R}}[/tex]  this is the velocity the rocket must have in order to escape from the surface of the planet

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

[tex]F = ma[/tex]

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

[tex]-10000 =2000\times a[/tex]

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

[tex]a = -\dfrac{10000}{2000}[/tex]

[tex]a=-5\ m/s^2[/tex]

For the distance,

Using third equation of motion

[tex]v^2-u^2=2as[/tex]

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

[tex]0-30^2=2\times(-5)\times s[/tex]

[tex]s = 90\ m[/tex]

Hence, The car traveled the distance before stopping is 90 m.

Answer:

it will move by d = 8.00 m in x direction before it will turn back

Explanation:

Here the initial velocity of particle is along +x direction

it is given as

[tex]v_i = 4.90 m/s[/tex]

now its acceleration is given as

[tex]a_x = -1.50 m/s^2[/tex]

[tex]a_y = 3.00 m/s^2[/tex]

now when it turns back then the velocity in x direction will become zero

so we will say

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 4.90^2 = 2(-1.50) d[/tex]

[tex] d = 8.00 m[/tex]

so it will move by d = 8.00 m in x direction before it will turn back

Final answer:

The particle moves 8.01 m in the x direction before it turns around, determined by using kinematic equations with the given initial velocity and constant acceleration.

Explanation:

To determine how far the particle moves in the x direction before turning around, we need to find the point at which its velocity in the x direction is zero. Since the particle starts with an initial velocity of 4.90 m/s in the x direction and has a constant acceleration of ax = -1.50 m/s², we can use the kinematic equation v = u + at to find the time when the velocity becomes zero:

v = 0 m/s (the velocity at the turnaround point)
u = 4.90 m/s (initial velocity)
a = -1.50 m/s² (constant acceleration)

Solving for the time (t) gives:

0 = 4.90 + (-1.50)t
t = 4.90 / 1.50
t = 3.27 s

Now using the equation s = ut + (1/2)at² to find the displacement in the x direction:

s = 4.90 m/s ⋅ t + (1/2)(-1.50 m/s²) ⋅ t²
s = 4.90 ⋅ 3.27 + (0.5) ⋅ (-1.50) ⋅ (3.27)²
s = 8.01 m

The particle moves 8.01 m in the x direction before turning around.

Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

[tex]F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N[/tex]

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

v₀ = 20 m/s

θ = 35°

aᵧ = -9.8 m/s²

1) Find t when y = 0.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

0 = 0 + (20 sin 35°) t + ½ (-9.8) t²

0 = t (20 sin 35° - 4.9 t)

t = 0, t = 2.34

The ball stays in the air 2.34 seconds.

2) Find y when vᵧ = 0.

vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)

0² = (20 sin 35)² + 2(-9.8) (y - 0)

y = 6.71 m

The ball reaches a maximum height of 6.71 meters.

3) Find x when y = 0.

x = x₀ + v₀ₓ t + ½ aₓ t²

x = 0 + (20 cos 35°) (2.34) + ½ (0) (2.3)²

x = 38.4 m

The ball lands 38.4 meters from Tom.

4) Find v when y = 0.

vₓ = aₓ t + v₀ₓ

vₓ = (0) (2.34) + 20 cos 35°

vₓ = 16.4 m/s

vᵧ = aᵧ t + v₀ᵧ

vᵧ = (-9.8) (2.34) + 20 sin 35°

vᵧ = -11.5 m/s

v = √(vₓ² + vᵧ²)

v = √((16.4)² + (-11.5)²)

v = 20 m/s

The ball has a speed of 20 m/s just before it lands.

Answer:

because the troposphere has more oxygen than the stratosphere so when he felt like he was floating even though he was free falling it was due to the absence of air.

Explanation:

It is given that,

Number of turns in primary coil, [tex]N_p=60[/tex]

Number of turns in secondary coil, [tex]N_s=300[/tex]

Voltage in primary coil, [tex]V_p=120\ V[/tex]

Current in primary coil, [tex]I_p=2\ A[/tex]

We have to find the voltage and current in the secondary coli. Firstly calculating the voltage in secondary coil as :

[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]

[tex]\dfrac{60}{300}=\dfrac{120}{V_s}[/tex]

[tex]V_s=720\ Volts[/tex]

Now, calculating the current present in the secondary coil as :

[tex]\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}[/tex]

[tex]\dfrac{60}{300}=\dfrac{I_s}{2\ A}[/tex]

[tex]I_s=0.4\ A[/tex]

Hence, this is the required solution.

Final answer:

The voltage present in the secondary is 600 V and the current is 0.4 A.

Explanation:

The question involves an ideal transformer where the primary coil has 60 turns and the secondary coil has 300 turns, with 120 V at 2.0 A applied to the primary. To find the voltage and current present in the secondary, we can use the transformer equations:

Vp/Vs = Np/Ns (where V is voltage and N is the number of turns in the primary (p) and secondary (s) coils)

According to the conservation of energy in an ideal transformer, Pp = Ps (where P is power), and since P = VI, we get:

VpIp = VsIs (where I is current)

Using these equations:

Is = (Np/Ns) \\times Ip = (60/300) \\times 2.0 A = 0.4 A

Thus, the voltage present in the secondary is 600 V and the current is 0.4 A.

Answer:

Mass of water, [tex]m=1.47\times 10^{12}\ kg[/tex]

Explanation:

Given that,

Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²

Average depth, d = 49 m

The volume of the reservoir is V such that,

[tex]V=A\times d[/tex]

[tex]V=3\times 10^7\ m^2\times 49\ m[/tex]

[tex]V=1.47\times 10^9\ m^3[/tex]

We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³

Density, [tex]d=\dfrac{m}{V}[/tex]

[tex]m=d\times V[/tex]

[tex]m=1000\ kg/m^3\times 1.47\times 10^9\ m^3[/tex]

[tex]m=1.47\times 10^{12}\ kg[/tex]

Hence, this is the required solution.

Answer:

197.2 s

Explanation:

First of all, let's convert the mass of the ocean liner into kilograms:

[tex]m=40,000 Mg = 40,000,000 kg = 4\cdot 10^7 kg[/tex]

and the initial velocity into m/s:

[tex]u=4 km/h =1.11 m/s[/tex]

The force applied is

[tex]F=-225 kN = -2.25\cdot 10^5 N[/tex]

So we can find first the deceleration of the liner:

[tex]a=\frac{F}{m}=\frac{-2.25\cdot 10^5 N}{4\cdot 10^7 kg}=-5.63\cdot 10^{-3} m/s^2[/tex]

And now we can use the following equation:

[tex]a=\frac{v-u}{t}[/tex]

with v = 0 being the final velocity, to find t, the time it takes to bring the liner to rest:

[tex]t=\frac{v-u}{a}=\frac{0-1.11 m/s}{-5.63\cdot 10^{-3} m/s^2}=197.2 s[/tex]

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, [tex]\theta_c=52^o[/tex]

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

[tex]n_1sin\theta_c=n_2sin\theta_2[/tex]

Here, [tex]\theta_2=90[/tex]

[tex]sin\theta_c=\dfrac{n_2}{n_1}[/tex]

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

[tex]n_1=\dfrac{n_2}{sin\theta_c}[/tex]

[tex]n_1=\dfrac{1}{sin(52)}[/tex]

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

Answer:

8.4 ft-lb

Explanation:

Work = change in energy

W = ½ kx²

When x = 3 ft, W = 15 ft-lb:

15 ft-lb = ½ k (3 ft)²

k = 30/9 lb/ft

When x = 27 in = 2.25 ft:

W = ½ kx²

W = ½ (30/9 lb/ft) (2.25 ft)²

W = 8.4375 ft-lb

Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.

Final answer:

The work required to stretch the spring 27 in. beyond its natural length is 12.728 N-m.

Explanation:

To find the work required to stretch the spring 27 in. beyond its natural length, we can use the work done formula:

Work = (1/2)kx²

Where k is the spring constant and x is the displacement of the spring. We know that the work required to stretch the spring 3 ft beyond its natural length is 15 ft-lb. Let's convert the given measurements into consistent units:

3 ft = 36 in.

15 ft-lb = 12.728 N-m

Now, we can solve for the work required to stretch the spring 27 in. beyond its natural length:

(1/2)k(27)² = (1/2)k(729) = 12.728 N-m

We can rearrange the equation to solve for k:

k = (2 * 12.728) / 729 = 0.0349 N/m

Answer:

approximately 3.68

Explanation:

Explained in the attached picture.

(a) 4.03 s

The initial angular velocity of the wheel is

[tex]\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s[/tex]

The angular acceleration of the wheel is

[tex]\alpha = -3.40 rad/s^2[/tex]

negative since it is a deceleration.

The angular acceleration can be also written as

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_f = 0[/tex] is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

[tex]t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s[/tex]

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

[tex]\theta= \omega_i t + \frac{1}{2}\alpha t^2[/tex]

where we have

[tex]\omega_i=13.7 rad/s[/tex] is the initial angular velocity

[tex]\alpha = -3.40 rad/s^2[/tex] is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

[tex]\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad[/tex]

The grinding wheel takes 38.5 seconds to stop, covering an angular displacement of 105 radians.

(a) To calculate the time it takes for the grinding wheel to stop, we can use the equation: t = ωf / α where ωf is the final angular velocity (0), and α is the angular acceleration (-3.40 rad/s²). Solving gives t = 38.5 seconds.

(b) The total angular displacement can be found using the equation: θ = ωi*t + 0.5*α*t^2 where ωi is the initial angular velocity (1.31 × 10² rev/min converted to rad/s), t is the time found in part (a), and α is the angular acceleration. This gives θ = 105 radians.

Answer:

[tex]1.69\cdot 10^4 N/C[/tex]

Explanation:

The relationship between electric field strength and potential difference is:

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

where

E is the electric field strength

V is the potential difference

d is the distance between the plates

Here we have

V = 218 V

d = 1.29 cm = 0.0129 m

So, the electric field is

[tex]E=\frac{218 V}{0.0129 m}=16900 N/C = 1.69\cdot 10^4 N/C[/tex]

the speed of the ball before it reaches the pool of water would be 9.91 m/s.

The initial speed given to the ball is 32.35m/s.

The speed of the ball immediately before it reaches the pool of water is 28m/s

Speed is a rate of change of distance with respect to time. i.e. v=dx÷dt. Speed can also be defined as distance over time i.e. speed= distance ÷ time it is denoted by v and its SI unit is m/s. it is a scalar quantity. Speed shows how much distance can be traveled in unit time.

To find dimension for speed is, from formula

Speed = Distance ÷ Time

Dimension for distance is [L¹] ,

Dimension for Time is [T¹],

Dividing dimension of distance by dimension of time gives,

[L¹] ÷ [T¹] = [L¹T⁻¹]

Dimension for speed is [L¹T⁻¹].

In this problem,

Given,

Hight of the cliff y = 40m

speed of the sound = 343 m/s

vertical direction,

y = u(y)t + 1/2 at²  where u(y) is initial velocity of ball along y direction which is zero.

-40 = 0t + 1/2 (-9.8)*t²

80 = 9.8t²

t = 2.85s

time taken by sound to hear is = 3-2.85 = 0.15s

distance covered by the sound,

343m/s × 0.15s = 51.45m

By applying Pythagoras theorem,

AC = √(x² + y²)

51.45² = x²+40²

x = 32.35

The initial velocity of the ball is,

u = 32.35m / 2.85s = 11.35m/s

The speed of the ball immediately before it reaches the pool of water is

v² = u² + 2as

v² = 0 +2(-9.8)40m

here initial velocity of the ball along y direction is 0. and distance along it is 40m.

v² = 784

v = 28m/s ......this is velocity along verticle.

Hence initial velocity of ball is 32.35m/s.

To know more about speed :

https://brainly.com/question/7359669

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