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Alexus [3.1K]
2 years ago
12

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
Physics
1 answer:
Aneli [31]2 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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romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
2 years ago
The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventuall
ozzi

Answer:

For a (1) net force will be greater than the weight of the ball

For b (2) net force will be lesser than the weight of the ball

Explanation:

For (a):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes upward, one thing is for sure, the net force is greater than the weight of the ball, because three forces are applied during upward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is also acting downward as it is creating friction between ball and air molecules, so creating hindrance in upward motion

External force to throw ball upward

So

Net Force = Upward force - Air friction - Weight

Since ball is going upward, so net force is greater than both weight and air friction which are pulling ball downward.

For (b):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes downward, one thing is for sure, the net force is lesser than the weight of the ball, because two forces are applied during downward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is acting upward as it is creating friction between ball and air molecules, so creating hindrance in downward motion

So

Net Force = Weight - Air friction

Since ball is going downward, so weight is greater than net force which is in this case is air friction which is pulling ball upward.

4 0
3 years ago
What is the resistance if the current is 15 amps and the voltage 105 volts
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3 years ago
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8 0
3 years ago
The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tis
Eduardwww [97]

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m' = 1 x 10⁻⁶ kg/s

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Given that

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Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

m'=\dfrac{m}{2}

m'=\dfrac{2\times 10^{-6}}{2}\ kg/s

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