Abus moves to the cast with a speed 10 ms, and a student in the bus moves with a speed 9 m/s to the cust. What is the velocity of the . relative to the bus A. 19 m/s to the cast B. 19 ms to the west c. 1 m/s to the east D. L us to the west.

Answer:

40m (40m east)

Explanation:

The fist move is 70m east, and east is the positive direction so the truck initially is at +70m.

The second move is 120m to the west, since east is the positive direction, west must be the negative direction, this means the truck now is at:

[tex]+70m -120m = -50m[/tex]

The third move is 90m to the east, again, this is the positive direction, so the new position is:

[tex]-50m + 90m = +40m.[/tex]

The truck is at + 40m, so it ended up 40m away from its initial position and this is the resultant dispacement.

Answer: Ok, so we know that them start on the same place.

First, the first one rides 1290 to the east, and 1410 to the north.

So if we put the 0 at the place where they start, north is the Y axis and east the X axis, we could describe the campground is in (1290,1410)

a) when the second cyclist turns, he traveled 1890m to the north, se the point where he is, is described by (0, 1890). We want to know how far is from the campground which we already know that is located in (1290,1410), so if we subtract we get distance = (0,1890) - (1290,1410) = (-1290, 1890 - 1410) = (1290, 480). The total distance will be D = [tex]\sqrt{1290^{2} + 480^{2}  }[/tex] = 1376.4 meters.

b) You want to know in what angle should he turn now.

So again, the cyclist is on the point (0,1890) and the campground is on the point (1290,1410) so he want to go 1290 meters to east, and -480 meters to north. If we define this two amounts as the cathetus of triangle rectangle, the route that he must follow is the hypotenuse of said triangle.

So the angle that he must turn (counting from the east, or the +x axis, counterclockwise) is defined by Tg(A) = -480/1290 = -0.372.

A = aTg(-0.372) = -20° aprox.

The magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.

The vertical displacement of the ink drop as it passes between the deflecting plates is due to the electric force acting on the charged drop. The force of gravity on the drop is negligible compared to the electric force.

Given:

Mass of the ink drop [tex](m) = 1.00 \times 10^{-11}\ kg[/tex]

Vertical displacement[tex](d) = 0.290\ mm[/tex]

Horizontal velocity [tex](v) = 23.0\ m/s[/tex]

Electric field magnitude [tex](E) = 8.20 \times 10^4\ N/C[/tex]

Plate length [tex](D_o) = 1.80\ cm[/tex]

The electric force on a charged particle in an electric field is given by Coulomb's law:

[tex]F_{electric} = q \times E[/tex]

Where:

[tex]F_{electric}[/tex] is the electric force,

q is the charge of the particle,

E is the magnitude of the electric field.

The vertical displacement (d) of the drop is caused by this electric force and is given by the equation:

[tex]d = (1/2) \times a \times t^2[/tex]

Since the drop is initially moving horizontally and is not influenced by gravity, the vertical acceleration (a) is due solely to the electric force

[tex]F_{electric} = m \times a[/tex]

Equate the two expressions for the electric force:

[tex]q \times E = m \times a\\a = (q \times E) / m\\d = (1/2) \times ((q \times E) / m) \times t^2[/tex]

The time it takes to travel through the plates using the horizontal distance [tex](D_o)[/tex] and the horizontal velocity:

[tex]t = D_o/ v[/tex]

[tex]d = (1/2) \times ((q \times E) / m) \times (D_o / v)^2\\q = (2 \times m \times d \times v^2) / (E \times D_o^2)\\q = (2 \times (1.00 \times 10^{-11}) \times (0.290 \times 10^{-3} ) \times (23.0 )^2) / ((8.20 \times 10^4) \times (1.80 \times 10^{-2})^2)\\q = 7.5 \times 10^{-14}[/tex]

Therefore, the magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.

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To find the charge necessary for an ink drop to be deflected a certain distance in an electric field, apply the equation q = Fe/E, where Fe is derived from the kinematic equations for the ink drop's motion in the field.

To calculate the magnitude of charge q that must be given to an ink drop such that it is deflected by a distance d = 0.290 mm by the time it reaches the end of the deflection plate of length D0 = 1.80 cm in a uniform electric field with magnitude E = 8.20×104 N/C, we can use the concepts of electric force and motion.

Firstly, the electric force Fe acting on the ink drop is equal to qE. This force will cause the ink drop to accelerate. Since the drops are moving horizontally at velocity v = 23.0 m/s and they need to be deflected by a certain distance d while travelling through the electric field, we can use the kinematic equations of motion to find the vertical displacement due to acceleration.

The time t it takes for the ink drop to pass through the plates is t = D0/v. From this, the vertical displacement d can be given as d = 0.5 * a * t^2 where a is the acceleration due to the electric force. Knowing d, t, and E, we can find a and subsequently use Fe = mdrop × a to solve for the charge q as q = Fe/E.

Using the given mass m = 1.00×10−11 kg for the ink drop, we can rearrange and solve the appropriate equations to find the value of the charge q necessary to cause the required deflection d.

Answer:

The momentum is not conserved.

Explanation:

When the superhero throws the piano, he exerts a force on the piano. Newton's third law says that each action has an opposite and equal reaction. In other words, this means that when the superhero throws the piano, the piano also exerts a force on the superhero, therefore, the superhero should recoil. You can visualize a gun being shot to visualize this phenomenon, when the gun is shot, it will tend to go backwards, trying to conserve momentum.

The scenario in the movie is not physically possible and violates the principles of Newton's laws of motion and conservation of momentum.

The scenario in the movie where the superhero hovers in the air and throws a piano while remaining stationary is not physically possible. It violates the principles of Newton's laws of motion and the conservation of momentum. In order for the superhero to hover and throw a piano, they would need some external force or propulsion, like jetpack or levitation powers, which are not commonly associated with superheroes.

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Explanation:

The average speed is the  displacement of a particle divided into a time in which the displacement occurs. Instead, instantaneous speed is the limit of the average speed as the time interval approaches zero.

In physics expressions

[tex]Average speed=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }[/tex]

Instantaneous speed=[tex]lim_{t \to \infty} \frac{x}{t} =\frac{dx}{dt}[/tex]

b) Speed ​​denotes distance traveled divided by time, and instantaneous speed measures how quickly and in which direction does it move

The difference between average speed and instantaneous speed is that average speed is the total distance traveled divided by the total time taken, whereas instantaneous speed is the speed of an object at a particular moment in time. Speed is a scalar quantity which means it only has magnitude and no direction, while velocity is a vector quantity which includes both magnitude and direction.

Average speed is calculated over a stretch of time and it does not consider the direction of the motion. For instance, if you drove for 100 kilometers and it took you 2 hours, your average speed would be 50 kilometers per hour. On the other hand, instantaneous speed is the speed of an object at a specific instant. It can be seen as the speedometer reading in a vehicle at a particular moment. To illustrate, while driving, your speedometer might read 60 km/h at one instant, which would be your instantaneous speed at that time.

Moreover, velocity includes direction as part of its definition and it describes the rate of change of an object's position. Average velocity takes into account the total displacement (change in position) and the total time. If the same 100 kilometers were travelled in a direct line to the north and it took 2 hours, then the average velocity would be 50 km/h north. If, however, the trip involved a return to the starting point (making the total displacement zero), the average velocity would also be zero, despite having a non-zero average speed. Therefore, while they might have the same magnitude under certain circumstances, speed and velocity are not interchangeable terms due to velocity's directional component.

Understanding the distinction between speed and velocity is crucial in physics because it determines how motion is described and analyzed.

Answer:

the magnitude of the force is 192.29 N

Solution:

As per the question:

Charges present on the corner of the triangle are same, Q = [tex]- 3.1\times 10^{- 9} C[/tex]

Since, its an equilateral triangle, distance between the charges, l = 0.50 m

Now,

The Coulomb force on a charge  due to the other is:

[tex]F_{C} = K\frac{Q^{2}}{l^{2}}[/tex]

where

K = Coulomb constant = [tex]9\times 10^{9} C^{2}/m^{2}[/tex]

[tex]F_{C} = (9\times 10^{9})\frac{(3.1\times 10^{- 9})^{2}}{0.5^{2}}[/tex]

[tex]F_{C} = 111.6 N[/tex]

The the net force on the charges in an equilateral triangle on all the charges due to each other:

[tex]F_{eq} = \sqrt{3}F_{C} = \sqrt{3}\times 111.6 = 193.29 N[/tex]

The magnitude of the force on each charge of -3.1 x 10^-9 C at the corners of an equilateral triangle 0.50 m on a side is 1.088 x 10^-7 N.

For one charge, the force due to the other two charges will be the vector sum of the forces due to each charge separately. Since the triangle is equilateral, the forces will have the same magnitude but different directions. Using Coulomb's law, we find the magnitude of the force between any two charges:

F = k × |q1 × q2| / r2

Where:

Substituting the given values:

F = (8.988 x 109 Nm2/C2) × (3.1 x 10-9 C)2 / (0.50 m)2

This results in:

F = 1.088 x 10-7 N

As all sides are equal, and charges are equal, the net force will be the same in magnitude for each charge. The direction of the resulting force on each charge will be along the perpendicular bisector of the side opposite to each charge due to the symmetry of the configuration.

Answer:

The kinetic energies are 0.46 eV and 0.78 eV.

Explanation:

Given that,

Work function = 1.5 eV

Wavelength = 632.8 nm

We need to calculate the energy for red helium neon laser

Using formula of energy

[tex]E=\dfrac{hc}{\lambda}[/tex]

Put the value into the formula

[tex]E=\dfrac{4.136\times10^{-15}\times3\times10^{8}}{632.8\times10^{-9}}[/tex]

[tex]E=1.96\ eV[/tex]

We need to calculate the kinetic energy

Using formula of K.E

[tex]K.E=E-\phi[/tex]

[tex]\phi[/tex]=work function

E = energy

Put the value into the formula

[tex]K.E=1.96-1.5[/tex]

[tex]K.E=0.46\ eV[/tex]

We need to calculate the energy for green helium neon laser

Using formula of energy

[tex]E=\dfrac{4.136\times10^{-15}\times3\times10^{8}}{543.5\times10^{-9}}[/tex]

[tex]E=2.28\ eV[/tex]

We need to calculate the kinetic energy

Using formula of K.E

[tex]K.E=2.28-1.5[/tex]

[tex]K.E=0.78\ eV[/tex]

Hence, The kinetic energies are 0.46 eV and 0.78 eV.

Answer:

The cannon ball hits the ground with a speed of 76.35 m/s.

Explanation:

We shall use the conservation of energy principle to solve the problem

The initial energy of the cannon ball is the sum of the potential and the kinetic energies.

We know that potential energy = [tex]mgh[/tex]

Applying the given values we obtain the initial potential energy as

[tex]P.E=m\times g\times 18.0[/tex]

Similarly the initial kinetic energy of the ball equals

[tex]K.E_{initial}=\frac{1}{2}mv_{o}^{2}[/tex]

Applying values we get

[tex]K.E_{initial}=\frac{1}{2}m(74.0)^{2}[/tex]

Thus the initial energy is thus [tex]E_{initial}=m\times g\times 18.0+\frac{1}{2}\times m\times (74.0)^{2}[/tex]

Now upon hitting the ground the only energy that the cannon ball will posses is kinetic energy since the potential energy of any object upon touching the surface of earth equals zero

Thus we have

[tex]Energy_{final}=\frac{1}{2}mv_{f}^{2}[/tex]

Equating initial and final energies we get

[tex]m\times g\times 18+\frac{1}{2}\times m\times (74.0)^{2}=\frac{1}{2}\times m\times v_{f}^{2}\\\\v_{f}^{2}=36g+(74.0)^{2}\\\\\therefore v_f=\sqrt{36\times 9.81+(74.0)^{2}}\\\\v_f=76.35m/s[/tex]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

Final answer:

The correct statement is that c) at the highest point of its trajectory, a rock thrown vertically has a velocity of zero and an acceleration directed downward at -9.80 m/s². The acceleration due to gravity is constant throughout the motion, whether the rock is rising or falling.

Explanation:

Among the given statements regarding a rock thrown vertically upward and neglecting air resistance, the true statement is: c) At the highest point the velocity is zero, the acceleration is directed downward. This is because at the maximum height, the rock stops for an instant (velocity = 0 m/s), but the acceleration due to gravity (acceleration due to gravity) is still acting on it, which is approximately -9.80 m/s² towards the Earth. This acceleration remains constant whether the rock is moving up or down.

It's important to note that as the rock rises, its acceleration vector always points downward, opposite to the direction of its velocity. Once the rock reaches its maximum height and begins its descent, the velocity becomes negative, which indicates the rock is falling towards the ground, yet still the acceleration is -9.80 m/s² downwards. The velocity and acceleration are in the same direction during the fall, both pointing toward the Earth.

Answer:

a) 919 mts

b) 392 mts

Explanation:

In order to solve this, we will use the formulas of acceletared motion problems:

[tex](1)Y=Yo+Vo*t+\frac{1}{2}*a*t^2\\(2)Vf^2=Vo^2+2*a*y[/tex]

We are looking to obtain the initial velocity of the canister right after it was relased, we will use formula (2):

[tex]Vf^2=(0)^2+2*(3.10)*(240)\\Vf=38.6 m/s[/tex]

we need to calculate the time the canister takes to reach the ground, we will use formula (1):

[tex]0-240m=38.6*t+\frac{1}{2}(-9.8m/s^2)*t^2\\\\-4.9*t^2+38.6*t+240=0\\t=11.9seconds[/tex]

in order to know the new height of the rocket we have to use the formula (1) again:

[tex]Y=240+38.6*(11.9)+\frac{1}{2}*(3.10)*(11.9)^2\\Y=919mts[/tex]

We can calulate the total distance the canister traveled before reach the ground by (2):

[tex](0)^2=(38.6)^2+2*(-9.8)*y\\Y=76m[/tex]

So the canister will go up another 76m, so the total distance will be:

[tex]Yc=Yup+Ydown+240m\\Yc=76+76+240\\Yc=392 mts[/tex]

Answer:

Time taken by the coin to reach the ground is 1.69 s

Given:

Initial speed, v = 11.8 m/s

Height of the building, h = 34.0 m

Solution:

Now, from the third eqn of motion:

[tex]v'^{2} = v^{2} + 2gh[/tex]

[tex]v'^{2} = 11.8^{2} + 2\times 9.8\times 34.0 = 805.64[/tex]

[tex]v' = \sqrt{805.64} = 28.38 m/s[/tex]

Now, time taken by the coin to reach the ground is given by eqn (1):

v' = v + gt

[tex]t = \frac{v' - v}{g} = \frac{28.38 - 11.8}{9.8} = 1.69 s[/tex]

Final answer:

To find the angle and tension in the wire, we can use the concept of electrostatic force and Coulomb's Law.

Explanation:

To find the angle and tension in the wire, we can use the concept of electrostatic force. The electrostatic force between the two charges can be calculated using Coulomb's Law:

FE = k * (q1 * q2) / r^2

where FE is the electrostatic force, k is the electrostatic constant (8.99 x 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

By equating this force to the tension in the wire and solving for the angle, we can find the angle:

tan(angle) = FE / T

where T is the tension in the wire.

Answer:[tex]\Delta E=2.0715\times 10^{-21} J[/tex]

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

[tex]E=\frac{3}{2}\cdot \frac{R}{N_A}T[/tex]

Where

R=Gas constant

[tex]N_A[/tex]=Avogadro's number

T=Temperature in kelvin

[tex]\frac{R}{N_A}=1.381\times 10^{-23}[/tex]

So kinetic energy increases by

[tex]\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )[/tex]

[tex]\Delta E=2.0715\times 10^{-21} J[/tex]

Answer:

c.m=(1/∑mi)*∑mi*yi

[tex]c.m.=\frac{m*d+2m*3d}{m+2m}=7d/3[/tex]

Explanation:

We solve this problem, with the equation for the center of mass:

c.m=(1/∑mi)*∑mi*yi

[tex]m_i:[/tex]   represent the different mass of the system

[tex]y_i:[/tex]    represent the position of the different mass of the system

c.m=(1/∑mi)*∑mi*yi=(1/(m+2m))*(m*d+2m*3d)=7d/3

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

[tex]R = u² sin2θ/g[/tex]

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

[tex]u² = \frac{Rg}{sin2\theta}[/tex]

[tex]u^2 = \frac{9.1 x 9.80}{sin26}[/tex]

[tex] u^2 = 113.17[/tex]

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

[tex]v^2 = (v_osin\theta)^2 -2gh[/tex]

[tex]h =\frac{(v_o^2sin\theta)^2}{2g}[/tex]

[tex]h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}[/tex]

h = 1.10 m

We have that for the takeoff speed and maximum height above ground is

From the question we are told

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo leaves the ground at a 26° angle. If this is so:

1) What is its takeoff speed

2) What is its maximum height above the ground?

Generally the equation for the Projectile motion   is mathematically given as

[tex]R=\frac{u^2sin 2\theta}{g}\\\\Therefore\\\\9.1=\frac{u^2sin 52}{9.8}[/tex]

u=10.69m/s

And

Generally the equation for the Projectile Height   is mathematically given as

[tex]Hmax=\frac{u^2sin^2 2\theta}{2g}\\\\Therefore\\Hmax=\frac{114.27*0.184}{19.6}[/tex]

Hmax=1.072m

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Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

Answer:

9.4 m

Explanation:

We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.

If the ballon is launched at 9.7 m/s at 39 degrees of elevation.

Vx0 = 9.7 * cos(39) = 7.5 m/s

Vy0 = 9.7 * sin(39) = 6.1 m/s

If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

a = -9.81 m/s^2

It will fall when Y(t) = 0

0 = 6.1 * t - 4.9 * t^2

0 = t * (6.1 - 4.9 * t)

t1 = 0 (this is when the balloon was launched)

0 = 6.1 - 4.9 * t2

4.9 * t2 = 6.1

t2 = 6.1 / 4.9 = 1.25 s

The distance from the car will be the horizonta distance it travelled in that time

X(t) = X0 + Vx0 * t

X(1.25) = 7.5 * 1.25 = 9.4 m

Answer:

a)  radio < microwaves < visible < gamma

Explanation:

Electromagnetic waves are transverse waves produced by oscillations of the electric and magnetic fields, so light speeds and must comply with the relationship of the wave speed

      v = λ f

  Name             Wavelength              Frequency

  Radio waves        Km                          10⁵

 Microwaves          cm                           10¹⁰

 Visible light           μm                          10¹⁴

 Gamma rays         pm                           10¹⁹

All waves also meet the Planck energy equation

       E = h f

h is the constant of Planck 6.6 10 -34 J s (4.136 10-15 eV-s), the unit of electron volts is very like for the waves of greater energy we substitute the frequencies and we calculate

Name            Frequency             Energy( J)    Energy (eV)

Gamma ray      10¹⁹                        10⁻¹⁵                10⁺⁴

Visible              10¹⁴                        10⁻²⁰               10⁻¹

Microwave       10¹⁰                        10⁻¹⁴                10⁻⁵

Radio waves    10⁵                         10⁻²⁹               10⁻¹⁰

You can see that frequency and energy are proportional

Answer:

In increasing order of frequency:

radio waves < microwaves< visible light < gamma rays

In increasing order of energy per photon:

radio waves < microwaves< visible light < gamma rays

Explanation:

Electromagnetic waves can transfer energy in the space as well. Electromagnetic spectrum is a range of radiation having different frequency.  The radiation which has more frequency has more energy but smaller wavelength. Gamma rays have maximum frequency where as radio waves have smaller  frequency and radiation.

Answer: 950 Kg/m^3

Explanation: We can deduce from the Archimedes principle that there is a relation between the density and the volumes displaced, as follows:

Density*Volume= Mass

So for equilibrium Density of body= Density of water *Vw/Vb

Being Vw/Vb the relation between  the displaced water and the body volume, and given the water density as 1000 Kg/m^3 we got:

Density(B)= 0.95 * 1000 Kg/m^3.

The sound would reach the dolphin in the water about 2.12 seconds earlier than the person in the boat, using given speeds of sound in air and water, and the distance provided.

The time it takes for sound to travel is calculated using the formula Time = Distance/Speed. Given, distance is 1000m in both air and sea water. We need to find the speed of sound in the air to calculate how long it will take the sound to reach the person in the boat.

The speed of sound in the air at 25°C is approximately 346.13 m/s. Using the formula: Time = Distance/Speed, the time it takes for the sound to reach the person in the boat is about 2.889 seconds (1000m/346.13 m/s).

The speed of sound in seawater is given as 1,533 m/s. Using the same formula, the time it takes for the sound to reach the dolphin in the water is about 0.652 seconds (1000m/1,533 m/s).

Therefore, the difference in time it takes for the sound to travel to the person and the dolphin is approximately 2.237 seconds (2.889 seconds - 0.652 seconds). Hence, the correct answer would be (B) 2.12. Although, it's close to the calculated difference, none of the available options are an exact match.

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Answer:

Maximum current in the inductor will be 3.824 A

Explanation:

In first case inductive reactance = 59.2 ohm

Frequency = 60 Hz

We know that inductive reactance is given by [tex]X_L=\omega L[/tex]

[tex]59.2=2\pi f\times L[/tex]

[tex]59.2=2\times 3.14\times  60\times L[/tex]

[tex]L=0.157H[/tex]

In second case frequency f = 45 Hz

Now inductive reactance [tex]X_L=\omega L =2\times 3.14\times 45\times .157=44.368ohm[/tex]

Now current [tex]i=\frac{V}{X_L}=\frac{120}{44.368}=2.70A[/tex]

Maximum current [tex]i_{max}=\sqrt{2}i=1.414\times 2.70=3.824A[/tex]

To find the maximum current when the inductor is connected to a 45Hz source, one needs to first calculate the inductance of the inductor using the provided inductive reactance at 60Hz. Then, with the inductance determined, the inductive reactance at 45Hz can be computed. Finally, Arnold's law (I = V/X₁) is used to determine the current.

In this problem, the concept of inductive reactance and Ohm's law are involved. First, we know that the inductive reactance (X₁) is directly proportional to the frequency, and it is given by the formula X₁ = 2πfL. From the information provided, we can find the inductance (L) of the inductor by using this formula at 60Hz, then looking at how X₁ changes when we use a 45Hz source. Once we determine L, we can find the new inductive reactance at 45Hz.

Next, we use Ohm's Law (I = V/X₁) to find the current at 45Hz, where V is the rms voltage across the inductor and X₁ is the inductive reactance. This would give us the maximum current in the inductor.

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Answer:

Explanation:

Conduction of heat per second Q  through a metal sheet having thermal conductivity K , thickness d , area A and temperature difference between cold and hot surface ( T₂ -T₁ ) is given by the following equation

Q = [tex]\frac{KA(T_2-T_1)}{d}[/tex]

Area A = π R² = 3.14 X 10000 X 10⁻⁶

= 3.14 X 10⁻2 m²

For aluminium plate :-

T₂ - T₁ = [tex]\frac{Q\times d}{KA}[/tex]

= [tex]\frac{600\times5\times10^{-3}}{240\times3.14\times10^{-2}}[/tex]

T₂ - T₁ = .4

T₂ = 110.4 ° C

For copper plate :

[tex][tex]\frac{600\times5\times10^{-3}}{390\times3.14\times10^{-2}}[/tex][/tex]

T₂ - T₁ = .25

T₂ = 110.25 ° C

The temperature of the surface in contact with the stove is approximately 113.0 °C for aluminum and 111.3 °C for copper, considering the thermal conductivity and dimensions of the pan.

This is a heat transfer problem involving conduction through different materials. The heat conduction through a flat wall, like the bottom of a pan, can be calculated using Fourier's law of heat conduction: q = -kA(T₂ - T₁)/d, where q is the heat transfer rate (600 W in this case), 'k' is the thermal conductivity, 'A' is the area across which heat is transferring, 'T₁' and 'T₂' represent the temperatures on either side of the wall, and 'd' represents the wall thickness.

First, convert the diameter of the pan to the radius in meters, which is (200mm/2)/1000 = 0.1 m. The area, A, is then π*(0.1 m)² = 0.0314 m². The thickness d given is already in meters: 5 mm = 0.005 m.

For aluminum, rearrange Fourier's law to find T₂, the temperature on the stove side of the pan: T₂ = T₁ + qd/(kA), where T₁ = 110°C and k for aluminum = 240 W/m K. Plugging in all values we get: T₂ = 110 + (600*0.005)/(240*0.0314) ≈ 113.0 °C.

For copper, using k for copper = 390 W/m K in the above formula will give: T₂ = 110 + (600*0.005)/(390*0.0314) ≈ 111.3 °C.

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Answer:

Rate of heat loss=7992 kJ/h

Explanation:

First, we can consider the room as a closed system, so we can use the first law of the thermodynamic:

[tex]Qnet,in-Wnet,out = Esystem[/tex]

The incoming net heat is the incoming heat less the leaving heat, the leaving net work is the leaving work less the incoming work  and the system energy is only the change of the intern energy because we have a stationary system, so:

[tex](Qin-Qout)-(Wout-Win) =U2-U1[/tex]

From the problem we consider that there is no change of the temperature thus there is no change in the intern energy(U2=U1), moreover there is no incoming heat neither exit work, all the work is made to the system as a result the leaving heat is equal to the incoming work:

[tex](0-Qout)-(0-Win) =0[/tex]

[tex]-Qout+Win =0[/tex]

[tex]Qout=Win[/tex]

Substituting the known values we can get the rate of heat loss (exit heat):

[tex]Q_{out}=Q_{loss}=W_{ref}+W_{TV}+W_{res}+W_{fan}[/tex]

[tex]Q_{loss}=250 [W]+120 [W]+1800 [W]+50 [W][/tex]

[tex]Q_{loss}=2220 [W]=2220 [\frac{J}{s} ][/tex]

Converting to Joules/hours:

[tex]Q_{loss}=2220 [\frac{J}{s} ][\frac{3600 s}{1 h} ][/tex]

[tex]Q_{loss}=7992000 [\frac{J}{h} ][/tex]

Finally the rate of heat loss is:

[tex]Q_{loss}=7992 [\frac{kJ}{h} ][/tex]

Final answer:

The rate of heat loss from a room with several electrical appliances running and constant air temperature is 7992 kJ/h, calculated by summing the power of all appliances and converting to kJ/h.

Explanation:

To find the rate of heat loss from the room, we need to consider the energy input from all the electrical appliances and the fact that the air temperature in the room remains constant. The total power consumption of the appliances is the sum of the power of the refrigerator, TV, electric resistance heater, and fan, which is 250 W + 120 W + 1.8 kW + 50 W = 2.22 kW.

Since the temperature is constant, the rate of energy input equals the rate of heat loss. This input is the combined power consumption of the appliances and is entirely converted into heat (assuming 100% efficiency for simplicity).

To express this rate in kilojoules per hour (kJ/h), multiply the power in kilowatts by the number of hours and by 1000 to convert from kW to kJ since 1 kW = 1 kJ/s:

2.22 kW = 2.22 kJ/s

So, the rate of heat loss is:

2.22 kJ/s × 3600 s/h = 7992 kJ/h

Answer:

Explanation:

Given that,

Length of wire = 9.5 cm =

Charge density = 100 NC/m

Distance = 4.50 cm

(A). We need to calculate the  the magnitude and direction of the electric field

Using formula of magnetic field

[tex]E=\dfrac{k\lambda\times l}{x\sqrt{x^2+a^2}}[/tex]

Here, x = distance from mid point of the wire

Put the value into the formula

[tex]E=\dfrac{9\times10^{9}\times100\times10^{-9}\times9.5\times10^{-2}}{4.50\times10^{-2}\sqrt{(4.50)^2+(\dfrac{9.5\times10^{-2}}{2})^2}}[/tex]

[tex]E=2.90\times10^{4}\ N/C[/tex]

The electric field is [tex]2.90\times10^{4}\ N/C[/tex].

The direction of electric field is upward.

(B). If the wire is now bent into a circle lying flat on the table,

We need to calculate the the magnitude and direction of the electric field

Using formula of electric field

[tex]E=\dfrac{k\lambda lx}{(x^2+(\dfrac{L}{\2\pi})^2)^{\dfrac{3}{2}}}[/tex]

Put the value into the formula

[tex]E=\dfrac{9\times10^{9}\times100\times10^{-9}\times9.5\times10^{-2}\times4.50\times10^{-2}}{((4.50\times10^{-2})^2+(\dfrac{9.5\times10^{-2}}{2\pi})^2)^{\dfrac{3}{2}}}[/tex]

[tex]E=3.59\times10^{4}\ N/C[/tex]

The electric field is [tex]3.59\times10^{4}\ N/C[/tex]

The direction of electric field is upward.

Hence, This is the required solution.

The electric field at a point above a charged wire depends on the configuration of the wire. For a straight wire, only the vertical components of the electric field produced by the charges along the wire add up, resulting in a field pointing upwards. If the wire is reshaped into a circle, the field still points upwards due to the circular symmetry.

The electric field E produced by a straight nonconducting plastic wire with a length of 9.5 cm carrying a uniform charge density λ of 100 nc/m can be calculated using Coulomb's law. Considering a small charge dq located at the midpoint of the wire, the electric field dE at a point P located 4.5 cm directly above this midpoint will have both a vertical (dE cos θ) and a horizontal component (dE sin θ). Given the symmetry of the problem, the horizontal components for all the dq's will cancel out, leaving only the resultant vertical component. Integrating this over the length of the wire, we can calculate the magnitude of the electric field E (directed vertically upwards).

If the wire is reshaped into a flat circle lying on the table, the situation changes. Now, because of the circular symmetry, the horizontal components of the electric field produced by the small charge elements dq around the circle do not cancel out, and instead add up to create a resultant electric field pointing vertically upwards at a point located 4.5 cm above the center of the circle.

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Answer:

mass = 61.16 kg

Explanation:

given data

distance = 50 m

energy = 500 W = 500 × 60sec = 30000 J/min

to find out

mass

solution

we will apply here energy equation  

that is

energy = m×g×h    .............1

put here all value m mass here and g = 9.81 and h is distance

energy = m×g×h

30000 = m×9.81×50

mass = 61.16 kg

Answer:

61 kg

Explanation:

Given data

distance (d): 50 m

power (P): 500 W = 500 J/s

mass (m): ?

time (t): 1 min = 60 s

The engine must do some work (w) to lift the coal. The work done in 1 minute is:

P = w / t

w = P . t = 500 J/s × 60 s = 3.0 × 10⁴ J

The work is equal to the force exerted (F) times the distance (d).

w = F × d

F = w / d = 3.0 × 10⁴ J / 50 m = 6.0 × 10² N

The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).

F = m × g

m = F / g = 6.0 × 10² N / 9.8 m/s² = 61 kg

Answer:

circumference of earth = 24000 miles

Explanation:

given data

New York and Los Angeles distance = 3000 mi

time difference = 3 h

to find out

circumference of the Earth

solution

we find speed first that is

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

speed = [tex]\frac{3000}{3}[/tex]

speed = 1000 mph

now we know speed of earth is

speed = [tex]\frac{circumference of earth}{time of rotation}[/tex]

so

circumference = speed × time of rotation

and we know earth take 24 hours to rotate at its axis

so

circumference of earth = 1000 × 24

circumference of earth = 24000 miles

Final answer:

To calculate the Earth's circumference, we establish that a 3-hour difference in time between New York and Los Angeles represents a 45° longitudinal difference. Since 3000 miles correspond to this 45°, we extend this to a full 360° rotation to find that the Earth's estimated circumference is about 24,000 miles.

Explanation:

To calculate the circumference of the Earth using the time difference between New York and Los Angeles, we use the relationship between time and longitude. Since the Earth rotates 360° in 24 hours, it rotates 15° per hour. Thus, a 3 hour time difference corresponds to a 45° difference in longitude (because 3 hours x 15° per hour = 45°).

If New York and Los Angeles are approximately 3000 miles apart, and this represents 45° of the Earth's rotation, we can find the total circumference of the Earth by setting up the proportion:

3000 miles / 45° = Circumference / 360°

Circumference = (3000 miles / 45°) x 360°

Circumference = 24,000 miles

Therefore, the estimated circumference of the Earth is approximately 24,000 miles.

Answer:

[tex]v=\frac{v_{1}v_{2}}{\left ( v_{1}+v_{2} \right )}[/tex]

Explanation:

Let the distance traveled in first half is d and then in the next half is also d.

Let the time taken in first half is t1 and the time taken  in the second half is t2.

[tex]t_{1}=\frac{d}{v_{1}}[/tex]

[tex]t_{2}=\frac{d}{v_{2}}[/tex]

Total time taken

[tex]t=t_{1}+t_{2}=\frac{d}{v_{1}}+\frac{d}{v_{2}}=\frac{d\left ( v_{1}+v_{2} \right )}{v_{1}v_{2}}[/tex]

The average speed of a body is defined as the total distance traveled by teh body to the total time taken.

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

[tex]v=\frac{2d}{t}[/tex]

[tex]v=\frac{2d}{\frac{2d\left ( v_{1}+v_{2} \right )}{v_{1}v_{2}}}[/tex]

[tex]v=\frac{v_{1}v_{2}}{\left ( v_{1}+v_{2} \right )}[/tex]

The average speed of a runner who ran at different speeds for two equal halves of a race is calculated as 2V1V2 / (V1+V2), where V1 and V2 are the speeds for the first and second halves respectively.

The question pertains to calculating average speed in a two-part race which essentially is a concept in Mathematics. We'll start by understanding the definition of average speed. It is calculated as the total distance travelled divided by the total time taken.

In this case, as the race is divided into two halves, the distance covered in the first half is equal to the distance covered in the second half. Hence, the total distance covered by the runner is 2D (where D is the distance of one half of the race).

However, the time to cover each half of the race is different. The time taken to cover the first and second halves would be D/V1 and D/V2 respectively. Therefore, the total time taken for the whole race is D/V1+D/V2.

Now we can calculate the average speed by dividing the total distance by the total time, which gives: Average Speed = 2D / (D/V1+D/V2). This can be simplified to: Average Speed = 2V1V2 / (V1+V2).

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Answer:

Their kinetic energy will increase but potential energy will decrease.

Explanation:

Given that

Initial velocities of electron and proton is zero.

We know that ,electron have negative charge and proton have positive charge it means that they will attract to each other.We know that opposite charge attract to each other and same charge repels to each other.

It means that the velocities of proton and electron will increase and that leads to increase in the kinetic energy of proton and electron.

We know that potential energy U

[tex]U\alpha -\dfrac{1}{r}[/tex]

So when r will decrease then U will increase but in negative direction it means that U will decrease.

So we can say that their kinetic energy will increase but potential energy will decrease.

Final answer:

The glass globe with a net electric charge of 5.31 × 10^-6 C has 3.32 × 10^13 fewer electrons than it did in its neutral state, because positive net charge indicates electron removal.

Explanation:

When a glittering glass globe is given a net electric charge of 5.31 × 10^-6 C, it means that electrons have been added or removed to achieve this net charge. The charge on an electron is known to be approximately -1.6 × 10^-19 C. To determine if the globe has more or fewer electrons than in its neutral state, we compare the sign of the given charge. Here, the positive charge indicates that electrons have been removed. To calculate the quantity of electrons removed, we use the formula:

Number of electrons (n) = Net charge (Q) / Elementary charge (e).

Plugging in the values:

n = 5.31 × 10^-6 C / 1.6 × 10^-19 C/e-

n = 3.32 × 10^13 electrons.

Therefore, the globe now has 3.32 × 10^13 fewer electrons than it did in its neutral state.

Answer:

An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.

Explanation:

An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.

if the +ve charge moves in the very same direction as the electrical field vector then the velocity of the particle will increase. It will slow down if it moves in the reverse direction.

if the adverse charge moves in the very same direction as the electrical field vector, the particle velocity will decelerate. If it moves in the reverse direction, it will speed up.

Answer and Explanation:

When a particle is placed in an Electric field having a charge Q and mass M, then it experiences an electric force on it. and it is given by:

[tex]F = QE[/tex]

This force is the overall force which is responsible for the acceleration of the particle.

We know, from Newton's second law:

F = Ma

and the force on a particle in electric field:

[tex]F_{E} = QE[/tex]

Therefore, from these two forces:

Ma = QE

a = [tex]\frac{QE}{M}[/tex]

In case of uniform Electric field, the particle moves with constant acceleration.

This acceleration is also capable of bending the object and to move it in a circle.