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Alex_Xolod [135]
2 years ago
8

The gas tank of Dave’s car has a capacity of 12 gallons. The tank was 38 full before Dave filled it to capacity. It cost him $2.

50 per gallon of gas. How much did Dave spend, in dollars, to fill the tank to capacity?
Physics
2 answers:
oee [108]2 years ago
7 0

Answer:

$ 30

Explanation:

Capacity of tank = 12 gallon

Cost of one gallon = $ 2.50

The cost of 12 gallon = 12 x 2.50 = $ 30

Thus, the cost of 12 gallon fuel is $ 30.

m_a_m_a [10]2 years ago
5 0

Answer:

$ 18.75            

Explanation:

given,                                              

capacity of the Dave’s car = 12 gallons

Assuming that the tank is 3/8 full before

Cost of gas per gallon =  $2.50 per gallon of gas

Volume Dave have to fill                      

       =1 - \dfrac{3}{8}                          

       =\dfrac{5}{8} of full tank              

       =\dfrac{5}{8}\times 12              

volume Dave have to fill = 7.5 gallons          

total money Dave spent                  

     = 7.5 gallons x $2.50                

     = $ 18.75                                            

Dave have to spend $ 18.75 to fill tank to it capacity.

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The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0
2 years ago
A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

6 0
2 years ago
Read 2 more answers
PLS ANSWER ASAP!!
Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

  FM is the magnetic force between the poles

   M1 is the strength of the first magnetic pole

   M2 is the strength of the second pole

   K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0
2 years ago
A rope breaks when the tension reaches 205 N. What is the maximum speed at which it can swing a 0.477 kg mass in a circle of rad
jekas [21]

Answer:

Maximum velocity will be 17.651 m /sec

Explanation:

We have given a rope breaks when tension reaches 205 N

Mass m = 0.477 kg

Radius of the circle r = 0.725 m

Tension in the rope is given by

T=\frac{mv^2}{r}

205=\frac{0.477\times v^2}{0.725}

0.477v^2=148.625

v^2=311.5828

v = 17.651 m /sec

8 0
2 years ago
A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.
charle [14.2K]

Answer:

(a)F= 3.83 * 10^3 N

(b)Altitude=8.20 * 10^5 m

Explanation:

On the launchpad weight = gravitational force between earth and satellite.

W = GMm/R²

where R is the earth radius.

Re-arranging:

WR² / GM = m

m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg

The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:

Fc = mω²r

where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.

ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s

When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force  between the earth and the satellite:

Fc = GMm/r²

mω²r = GMm / r²

ω²r = GM / r²

r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²  

r³ = 3.612 * 10^20

r = 7.12 * 10^6 m

(a) F = GMm/r²  

F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²

F= 3.83 * 10^3 N

(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m

4 0
2 years ago
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