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Illusion [34]
3 years ago
8

A 0.3 g mosquito is flying toward a girl with a speed of 4.5 mph. Just before landing on the girl, the fly is swatted straight b

ack at a speed of 12 mph. If the fly swatter and the fly were in contact for 0.2 s, what is the force that was exerted on the fly
Physics
1 answer:
luda_lava [24]3 years ago
5 0

Answer:

1.1x10^-2N

Explanation:

We have the change in momentum as

P = 0.3(4.5+12)g.mph

= 0.3x0.447x(4.5+12)x10^-3

Then the force that is exerted will be

F = p/∆t

∆t = 0.2

= 0.3x0.447x(4.5+12)x10^-3/0.2

= 0.1341x16.5x10^-3/0.2

= 1.1x10^-2

Therefore the force that was exerted is equal to 1.1x10^-2

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(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. Whe
Fittoniya [83]

Answer:

Acceleration = 1.428m/s2

Tension = 102.85N

Explanation:

The detailed solution is attached

5 0
3 years ago
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Which is not a step in the scientific method?
tensa zangetsu [6.8K]
A. Ask other people for their opinion. :)
5 0
3 years ago
Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a
Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

block = 0.5  \cdot m  \cdot v^2

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

  • The force of the weight of the block on the string, F = m \cdot g  \cdot sin(\theta)

The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

Which gives;

0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = 0.5  \cdot m  \cdot v^2

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b  is an ellipse

Therefore;

  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

6 0
2 years ago
A 0.00287 C charge is 6.52 m from
yKpoI14uk [10]

Answer:

I know dude

Explanation:

-22000

7 0
3 years ago
a red ball moves horizontally in a 30 m long tube what is the displacement of the red ball between 0s and 24s?
gayaneshka [121]

Answer:

30 metres.

Explanation:

Given that a red ball moves horizontally in a 30 m long tube.

Displacement is the distance travelled in a specific direction. It has both magnitude and direction.

Since the motion is horizontal, it moves is a certain direction.

Within the stipulation of time, the displacement will be the distance covered in the horizontal direction which is 30 metres.

Therefore, the displacement of the motion of the red ball is 30 metres.

7 0
4 years ago
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