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Shkiper50 [21]
3 years ago
15

A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

way that would allow for a safe takeoff?

Physics
1 answer:
pickupchik [31]3 years ago
3 0
Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

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The sun’s energy is most useful to humans after it is converted to _____.
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The sun’s energy is most useful to humans after it is converted to chemical energy. Living organisms like humans and animals could not directly utilize the energy from the sun because of lack in organs that would allow photosynthesis. Humans and animals obtain the energy from the sun from the intake of food. These foods contain chemical energy which is obtained from the solar energy of the sun. Plants use solar energy for photosynthesis which converts the energy into chemical energy. Then, animals and humans eat the plants obtaining the chemical energy which can readily processed by the body as compared to solar energy.
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1. What is the new kinetic energy of the 1900kg ship on the right moving at 4 m/s?
neonofarm [45]

Explanation:

That`s is the answer, just check

6 0
3 years ago
A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if
Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

 F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

                                   2.5 ∝ 1.68  --------------(1)

                                   4.4 ∝ d'      --------------(2)

From (1) & (2),     4.4/2.5 = d'/1.68

                                   d' = 2.96 cm ⇒ the required extension.

7 0
3 years ago
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
2 years ago
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