# 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

The volume required is

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

Generally the volume of gasoline needed is mathematically represented as

=>

=>

Converting to  Liters

=>

=>

## Related Questions

The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

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

Explanation:

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

Explanation:

As we know flow rate is given by

Flow rate = Velocity*Area of nozzle

Where

Area of nozzle = πr²

where

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.

Explanation:

Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

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

Explanation:

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

Explanation:

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

........1

.......2

.......3

divide equation 3 by 1

we get

...........4

from equation 3 and 4

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

Explanation:

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.