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algol [13]
3 years ago
15

A car can travel 30.0 mi on one gallon of gas. How many km/L is this?

Physics
1 answer:
Leona [35]3 years ago
8 0

Answer

30 mi/gal * (1.61 km / mi ) /  (3.78 L/gal)   = 30 * .426 km/L = 12.8 km/L

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A girl rides her scooter to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the
Whitepunk [10]

Answer:

6.93 km/h

Explanation:

To calculate her average speed, we need the "speed" formula, which is:

average speed = distance / time

You plug in your numbers and it will give you the answer.

Speed = 4.5km/0.65hr

           = 6.923 km/h

8 0
3 years ago
Read 2 more answers
The figure shows a 1 cm cube that has a mass of 1 gram. Calculate the density of the cube in kg/m³.
dolphi86 [110]
Density = 1 g/cm^3 = 1000 Kg/m^3
3 0
3 years ago
The midpoint M of a guitar string is pulled a distance d = 1.7 mm from equilibrium and released. Point M is observed to undergo
nlexa [21]

Answer:

ω = 380π rad/s

Explanation:

The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π

ω = 2πf

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ω = 2π(190)

ω = 380π rad/s

8 0
3 years ago
A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell
Lisa [10]

Answer: 7.53 μC

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∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0
4 years ago
The total surface area of the human body is 1.20 m2 and the surface temperature is 30∘C=303∘K. If the surroundings are at a temp
zalisa [80]

To solve this problem we must consider the expressions of Stefan Boltzmann's law for which the rate of change of the radiation of energy H from a surface must be

H = Ae\sigma T^4

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e = Emissivity that characterizes the emitting properties of the surface

\sigma = Universal constant called the Stefan-Boltzmann constant (5.67*10^{-8}m^{-2}K^{-4})

T = Absolute temperature

The total heat loss would be then

Q = H_2 -H_1

Q =Ae\sigma T_2^4-Ae\sigma T_1^4

Q = Ae\sigma (T_2^4-T_1^4)

Q = (1.2)(1)(5.67*10^{-8})(303^4-280^4)

Q = 155.29J

Therefore the net rate of heat loss from the body by radiation is 155.29J

8 0
3 years ago
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