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


Answer 1


The volume required is  V  = 11.91 \ liters


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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You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle



29.5 m/s


Volumetric flowrate = (average velocity of flow) × (cross sectional area)

Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

Cross sectional Area of flow = πr²

Diameter = 0.00579 m,

Radius, r = d/2 = 0.002895 m

A = π(0.002895)² = 0.0000037629 m²

Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)

v = 0.000111/0.0000037629

v = 29.5 m/s

Given Information:  

diameter of the nozzle = d = 5.79 mm = 0.00579 m

flow rate =  0.111 liters/sec

Required Information:  

Velocity = v = ?


Velocity = 4.21 m/s


As we know flow rate is given by

Flow rate = Velocity*Area of nozzle


Area of nozzle = πr²


r = d/2

r = 0.00579/2

r = 0.002895 m

Area of nozzle = πr²

Area of nozzle = π(0.002895)²

Area of nozzle = 2.6329x10⁻⁵ m²

Velocity = Flow rate/area of nozzle

Divide the litters/s by 1000 to convert into m³/s

0.111/1000 = 1.11x10⁻⁴ m³/s

Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

Velocity = 4.21 m/s

Therefore, the water exit the nozzle at a speed of 4.21 m/s

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.



The induced emf is 0.0888 V.


Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field \Delta B=(6.683-0.997)= 5.686\ T

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

\epsilon=(NA\Delta B\cos\theta)/(\Delta T)

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula


\epsilon =0.0888\ V

Hence, The induced emf is 0.0888 V.

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.



a. 12.12°

b. 412.04 N


Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N


Magnitude - 11.83 Degree

Direction - 422.42 N


Given data:

Downward force on wire 176 N

Angle made by left section of wire 12.5 degree with horizontal

Tension force = 413 N

From figure

Applying quilibrium principle at point A

The vertical and horizontal force is 0

then we have

Tcos\theta = 413 N   ........1

176 = 413 sin 12.5 + Tsin\theta     .......2

Tsin\theta = 176 - 89.39  = 86.6.......3

divide equation 3 by 1

we get

\theta = tan^(-1) (0.2096)

theta = 11.83^o  ...........4

from equation 3 and 4

T = (86.6)/(sin 11.83)

T = 422.42 N

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.



Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³


Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

As you go up the y-axis, what happens to the number of sprouted bean seeds?A. Sprouted bean seeds decrease.

B. Sprouted bean seeds increase.

C. Sprouted bean seeds remain constant.

D. None of the above


As we go up the y-axis, the number of sproutedbean seeds increase (option B).

What is a graph?

Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.

In a graph, there are two axes as follows;

  • Y-axis or vertical axis
  • X-axis or horizontal axis

According to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.

We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.

Learn more about graphs at: