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2 years ago

describe the time of day that an early explorer might have planned to enter a harbor and when he might have planned to leave

1 answer:

4 0

He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.

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A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0

2 years ago

The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

max2010maxim [7]

Answer:

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = $2\pi \sqrt{\frac{L}{g} }$

where g is the acceleration due to gravity

substituting values, we have

0.55 = $2\pi \sqrt{\frac{1.44}{g} }$

0.0875 = $\sqrt{\frac{1.44}{g} }$

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = <em>18808.7 m/s^2</em>

3 0

2 years ago

Summary about down hill on a bicycle story

Ne4ueva [31]

Answer:

The poem, Going Down the Hill on a Bicycle, written by Henry Charles Beeching describes the thrilling ride of a boy going downhill. ... The poet mentions how he lifts his feet from the pedals and keeps his hands still so that he would not lose his balance and fall off the bicycle, while it is dashing down the hill.

6 0

2 years ago

The ____ of an object depends on the diameter of the object and the distance to the object

inn [45]

Apparent diameter is the answer

5 0

2 years ago

1. Asimov the robot needs to travel from his charging station to school. Asimov travels 5 km south, 3 km west, 2 km north and 3

aivan3 [116]

1. Asimov displacement is 3km south
2. Peters distance traveled is 13 km

6 0

2 years ago