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nikdorinn [45]
3 years ago
12

A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?

Physics
1 answer:
ella [17]3 years ago
8 0

To calculate the mass of the fuel, we use the formula

m = V \times  \rho

Here, m is the mass of fuel, V is the volume of the fuel and its value is V =302 \ L =  302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL and  \rho is the density and its value of  0.821 g/mL.

Substituting these values in above relation, we get  

m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg

Thus, the mass of the fuel 247 .94 kg.

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A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north
Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

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Explanation:

Let east represent positive x axis and north represent positive y axis.

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1.93 km due wast

           s ₁ = 1.93 i km

1.03 km due south

           s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

           s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

          s = s ₁+  s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

  Magnitude of displacement, =\sqrt{(-0.39)^2+2.03^2}=2.07km

Time taken = 1.771 hour

Magnitude of average velocity, =\frac{2.07}{1.771}=1.17km/hr

7 0
3 years ago
Read through the and calculate the predicted change in kinetic energy of the oblect compared to 50 kg ball traveling at 10 m/s .
Sliva [168]

Answer:

A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.

A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.

A 50 kg person falling at 10 m/s would have the same kinetic energy.

Explanation:

hope this helps:)

5 0
3 years ago
Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e
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Answer:

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Explanation:

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for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

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E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

 we calculate

E_total = 9 10⁹ / (4 10⁻²)²   (6.0 10⁻⁶ +3.0 10⁻⁶)

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