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nirvana33 [79]
2 years ago
15

An airplane during departure has a constant acceleration of 3 m / s².

Physics
1 answer:
Rama09 [41]2 years ago
6 0

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>

t

3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>

8

24 = v - 20

v = 44m/s

After 8s the plane would've 44m/s speed.

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pogonyaev

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

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7 0
3 years ago
A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0.2 kg, and the robo
Natali5045456 [20]

Answer:

<h3>1.43m/s²</h3>

Explanation:

According to newtons second law.

F = mass * acceleration

If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg

Force applied = 1N

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Substitute the values and get acceleration

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3 years ago
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of rho0 is placed in a container of water. Initially
ivanzaharov [21]

Answer:

a) s,f,r  b) r c) f

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For the sphere to be in equilibrium the sum of forces is zero

    B - W = 0

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Now let's use the concept of density for the body and water

Solid sphere

   ρ = m / V

  V = 4/3 π r³

   m = ρ₀ (4/3 π r³)

   W = ρ₀ (4/3 π r³) g

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   ρ = mₐ / Vₐ

   mₐ = ρ Vₐ

   B = ρ Vₐ g

Let's replace and simplify

   ρ Vₐ g = ρ₀ (4/3 π r³) g

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For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case

a) We have the same mass, but less radius, as density is mass over volume density increases

   r  <ro        V <V₀   ⇒      ρ₁> ρ₀

When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant

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- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)

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In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)

c) you have the same radius, but the mass decreases

      r = r₀     V = V₀     m <m₀        ρ₁ <ρ₀

The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)

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Answer:

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