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Illusion [34]
2 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]2 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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An empty parallel plate capacitor is connected between the terminals of a 16.7-v battery and charges up. the capacitor is then d
katen-ka-za [31]
16.7V * 2 = 33.4V
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4 0
3 years ago
A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a
klemol [59]

Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

7 0
3 years ago
Read 2 more answers
A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One
Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\

To get the required work done, we will divide the mass by the speed of one knot to have:

w=\frac{7230}{0.51}\\w= 14,176.47Joules

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0
2 years ago
A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
Sounds are produced by objects that are______
ahrayia [7]

Answer: Vibrating

Explanation:

Sound is produced when an object vibrates. The sound vibrations cause waves of pressure that travel through a medium, such as air, water, wood or metal. Sound is a form of mechanical energy.

7 0
3 years ago
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