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3 years ago

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The drag force due to air resistance on a soccer ball traveling through the air at 20 m/s is 2.3 N. What is the drag force on th

at same soccer ball when it is moving through the air at 35.4 m/s

1 answer:

Aneli [31]3 years ago

3 0

The drag force on that same soccer ball when it is moving through the air at $35.4 m/s$ is 7.2 Newton.

Given the data in the question;

Initial speed; $v_1 = 20m/s$

Initial drag force; $F_1 = 2.3N$

Final speed; $v_2 = 35.4m/s$

Final drag force; $F_2 =\ ?$

From Newton's laws, Drag force is directly proportional to velocity squared:

$F\ \alpha \ v^2 \\\\F = Kv^2$

Where k is the constant of proportionality.

Hence;

$\frac{F_1}{F_2} = \frac{V_1^2}{V_2^2}$

We make " $F_2$ " the subject of the formula

$F_2 = \frac{F_1 * v_2^2}{v_1^2}$

We substitute our values into the equation;

$F_2 = \frac{2.3N * (35.4m/s)^2}{(20m/s)^2}\\\\F_2 = \frac{2.3N * 1253.16}{400} \\\\F_2 = \frac{2882.268N}{400}\\\\F_2 = 7.2N$

Therefore, the drag force on that same soccer ball when it is moving through the air at $35.4 m/s$ is 7.2 Newton.

Learn more: brainly.com/question/2867185

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Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.

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3 years ago

car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

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