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3 years ago

Look at paths of the projectiles. What effect does air resistance have on the projectile?

Physics

1 answer:

inna [77]3 years ago

6 0

Answer:

Therefore the resistance of the air makes the movement not parabolic but shorter in each direction

Explanation:

The projectile motion is described by the kinematics equations giving a parabolic trajectory, where on the x axis there is no acceleration and on the y axis the acceleration is the acceleration of gravity.

When the air resistance is taken into account it can be approximated as a force that opposes the movement that for low speeds is proportional to the speed of the space.

Consequently, the movement in the axis and the acceleration is less, in some cases it can be so small that the constant handle speed, in this case, is called terminal velocity.

On the x-axis the friction force creates an acceleration in the negative direction of the movement that the projectile has to brake.

Therefore the resistance of the air makes the movement not parabolic but shorter in each direction.

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3 years ago

What is the Kinetic Energy of a 120 kg object that is moving with a velocity of 15 m/s

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A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$

The frictional force on the block W₂, $F_{f2}$ ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ $F_{w2}$ = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + $\mathbf{F_{w2}}$

The frictional force from the ground, $\mathbf{F_{f1}}$ = N×μ + $\mathbf{F_{f2}}$ = P

Where;

P = The horizontal force applied to the block

P = (W₁ + $\mathbf{F_{w2}}$ ) × μ + $\mathbf{F_{f2}}$

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

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Answer:

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Explanation:

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