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3 years ago

A pelican flying along a horizontal path drops

Physics

1 answer:

Ludmilka [50]3 years ago

8 0

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s$

The initial horizontal speed of the pelican is given by;

X = vₓt

vₓ = X / t

vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

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1. The statement about Newton's reflecting telescope contains 1 error. Find the error, write it, and

Jlenok [28]

Answer:

i. The error is the rough convex mirror.

ii. This should be replaced with a smooth convex morror.

Explanation:

Reflection is dependent on the surface involved and has two types; diffuse and specular. When the surface is rough, diffused reflection is observed. The surface causes a distortion of the incident light (the rays would be reflected at different angles to one another) after reflection. This makes some rays to interfere with one another. While specular reflection is observed with a smooth surface.

In the statement, the rough convex mirror would produce a distorted reflection which would produce diffused reflection. The effect is that few or no rays (depending on the degree of how rough the surfce is) would be reflected to the other smooth, flat diagonal mirror.

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2 years ago

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

$T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s$

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m$

From Hooke's law

$mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2$

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0

2 years ago

Perform the operation of 147.02 / 0.338, and report the result with the proper number of significant figures

Ede4ka [16]

Answer:

434.97041

Explanation:

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2 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a

timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack ( $v$ ), measured in meters per second, is equal to the product of the angular speed of the wheel ( $\omega$ ), measured in radians per second, and the distance of the tack respect to the rotation axis ( $R$ ), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second: