In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 28.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe

Answers

Answer 1
Answer:

Answer:

The volume required is  V  = 11.91 \ liters

Explanation:

From the question we are told that

  The cars mileage is  v =  28.0 mi/gal

   The  distance is  d  =  142 km

Converting the distance from km to  miles

        d =  142 *  0.6214 = 88.24 \ miles

Generally the volume of gasoline needed is mathematically represented as  

       V =  (d)/( v)

=>     V =  (88.24)/( 28.0 )

=>     V = 3.151 \ gal

Converting to  Liters

    =>   V = 3.151  *   3.78

     =>   V  = 11.91 \ liters


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Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 162 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.

Answers

Answer:

0.00124 V

Explanation:

Parameters given:

Initial circumference = 162 cm

Rate of decrease of circumference = 14 cm/s

Magnetic field, B = 0.5 T

Time, t = 8 secs

The magnitude of the EMF induced in the loop is given as:

V = (-NBA) / t

Where N = number of turns = 1

B = magnetic field

A = area of loop

t = time taken

First, we need to find the area of the loop.

To do this, we will find the radius after the loop circumference has decreased for 8 secs.

The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:

14 * 8 = 112 cm

The new circumference is:

162 - 112 = 50 cm = 0.5 m

To get radius:

C = 2 * pi * r

r = C / (2 * pi)

r = 0.5 / (2 * 3.142)

r = 0.0796 m

The area is:

A = pi * r²

A = 3.142 * 0.0796²

A = 0.0199 m²

Therefore, the EMF induced is:

V = (-1 * 0.5 * 0.0199) / 8

V = -0.00124V

This is the EMF induced in the coil.

The magnitude is |-0.00124| V = 0.00124 V.

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.

Answers

Answer:

force = 11.33 kg-m/s^(2)

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

P_(1)=mU

P_(1)= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion

F=(\Delta P)/(\Delta t)

F = (P_(1)-P_(2))/(T)

Kgm/s^2

F = (69.7-0)/(6.15)= 11.33[tex]kg-m/s^(2)[/tex]

Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One of the uncharged balls can become charged by touching it to the charged ball and then separating the two. This process of touching one ball to another and then separating the two balls can be repeated over and over again, with the result that the three balls can take on a variety of charges. Which one of the following distribution of charges could not possibly be achieved in this fashion, even if the process were repeated an infinite number of times?Why the answer is qA = 1/2q, qB=3/8q, qC=1/4q. Explain please.

Answers

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

         qA = ½ q

         qB = ½ q

         qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

         qA = 1/2 (q / 2) = ¼ q

         qC = ¼ q

         qB = ½ q

At this point all spheres have a charge,

      qA = ¼ q

      qb = ½ q

      qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

       Q = ½ q + ¼ q = ¾ q

When splitting the charge

        qB = ½ ¾ q = 3/8 q

        qC = ½ ¾ q = 3/8 q

        qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

       Q = qA + QB + QC

       Q = ½ q + 3/8 q + ¼ q

        Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Final answer:

The principle of charge distribution among conductive materials is violated in qA = 1/2q, qB=3/8q, qC=1/4q, as the sum of charges on B and C doesn't equate to the charge on A and 3/8q isn’t a multiple of halving the original charge.

Explanation:

The answer lies in the fact that balls made of conducting materials when in contact, distribute charges evenly among them. This is due to the free movement of electrons within the conductive material that seeks to equalize potential difference, a property exploited in charge distribution problems of this sort.

Given the scenario, every time a charged ball, A, touches an uncharged ball (B or C) the net charge is evenly split between them. Hence, each subsequent distribution halves the charge of the originating ball (A) and gives the complementary half to the ball it's being touched to (B or C).

In the distribution, qA = 1/2q, qB=3/8q, qC=1/4q, the sum of charges on B and C does not equate to A, which is a violation of the charge conservation principle. Moreover, 3/8q isn’t a multiple of halving the original charge q, which negates the manner in which the charge is distributed (i.e., by halving).

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The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.

Answers

Answer:

σ  = 0.8 N/m

Explanation:

Given that

L = 12 cm

We know that 1 m = 100 cm

L = 0.12 m

The force ,F= 0.096 N

Lets take surface tension = σ

We know that surface tension is given as

\sigma =(F)/(L)\n\sigma =(F)/(L)\nNow\ by\ putting\ the\ values\n\sigma =(0.096)/(0.12)\ N/m\n\sigma=0.8\ N/m

Therefore the surface tension σ  will be 0.8 N/m .

σ  = 0.8 N/m

Final answer:

The surface tension of the liquid in air is 0.8 N/m.

Explanation:

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

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The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

V = V_(0)(1-e^{(-t)/(RC) }  )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) })  \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )

Taking natural log on both sides,

e^{(-t)/(RC) } = 1- (V)/(V_(0) )  \n(-t)/(RC) = ln(1-(V)/(V_(0) ) )\nt = -RCln(1 - (V)/(V_(0) ))

t = -RC ln (1-(V)/(V_(0) ))        (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

  = ((8.85*10^(-12))) (20*10^(-4)) )/(1*10^(-3) )

  = 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

t = -RC ln (1-(V)/(V_(0) ))

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

Why is it important to select a coordinate system when studying motion?

Answers

Motion is detected when an object changes its position with respect to a reference point. Coordinate system is basically used to represent motion. A coordinate system uses numbers or coordinates which represent position of the reference points on a two-dimensional or three-dimensional space. The trajectory of a point or line can be studied on a coordinate system which describes various aspects of motion like velocity, acceleration, distance, displacement etc. Coordinate system is important because it helps to choose a starting point and the direction (which will be positive).