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oksano4ka [1.4K]
3 years ago
10

A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the

pipe to eject the marshmallow from the right end. The average net force exerted on the marshmallow while it is in the pipe is 0.7 N. The speed of the marshmallow as it leaves the pipe is most nearly
Physics
2 answers:
alexdok [17]3 years ago
5 0

Answer:

15 m/s

Explanation:

Given:

F = 0.7N   m = 0.25g = 0.0025kg   x = 40cm = 0.4m v_{0} = 0

Since F = ma, a = F/m = 0.7N/0.0025kg = 280m/s^{2}

Using the Big Five, v ^{2} = v_0^{2} + 2ax, v = \sqrt{0 + 2*280m/s^2*0.4m} = \sqrt{216} m/s=  15m/s

frosja888 [35]3 years ago
3 0

Answer:

Your answer would be

A person 40 cm- blows into the left end of the pipe to eject the marshmallow from the right end. ... A strain of sound waves is propagated along an organ pipe and gets reflected from an. play · like-icon ... The velocity of sound in air is 340ms^(-1). ... The two pipes are submerged in sea water, arranged as shown in figure. Pipe.Explanation:

I belive this is the answer sorry if im wrong!

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A 16 g rifle bullet traveling 240 m/s buries itself in a 3.6 kg pendulum hanging on a 2.5 m long string, which makes the pendulu
iren [92.7K]

Answer:

x = 0.54 m

y = 0.058 m

Explanation:

m = mass of the bullet = 16 g = 0.016 kg

v = speed of bullet before collision = 240 m/s

M = mass of the pendulum = 3.6 kg

L = length of the string = 2.5 m

h = height gained by the pendulum after collision

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Using conservation of momentum

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V = 1.062 m/s

Using conservation of energy

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(m + M) g h = (0.5) (m + M) V²

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h = 0.058 m

y = vertical displacement = h = 0.058 m

x = horizontal displacement

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x = sqrt(L² - (L - h)²)

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3 years ago
Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as
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Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

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