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Novay_Z [31]
3 years ago
5

35. What would a ship's position be if that ship started at

Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

The final position of the ship after the given time period is  42 km West of B.

Explanation:

Given;

average velocity of the ship, v = 35 km/h

time taken for the ship to reach point D, t = 1.2 hours

The position of the ship after the given time period is calculated as follows;

x = v x t

x = (35 km/h) x 1.2 h = 42 km

x  = 42 km West of B.

Therefore, the final position of the ship after the given time period is  42 km West of B.

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A racecar is traveling at a speed of 80.0 m/s on a circular
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A :-) F = mv^2 by r
Given - m = 500 kg
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explain why the EMF of a dry cell drops if a large current is drawn for a short time and then recovers if allowed to rest​
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Manganese (iv )oxide is a slow depolarizer and polarization occurs with a large current, on resting, the depolarization returns the p.d of the dry cell.

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3 years ago
15.6 grams of lead(II) nitrate is in a test tube. A 4.8-g sample of potassium iodide is an Erlenmeyer flask. The substance in th
miv72 [106K]

<u>Answer:</u>

 Mass of combined products = 20.4 grams

 Mass of combined product and Erlenmeyer flask = 136 grams

<u>Explanation:</u>

  By law of mass conservation, mass cannot be destroyed or created. So mass before reaction is mass after reaction.

  Mass of  lead(II) nitrate  = 15.6 grams

  Mass of potassium iodide = 4.8 grams

  Mass of the Erlenmeyer flask = 115.6 grams

  Mass of reactants = 15.6 + 4.8 = 20.4 grams

  So mass of products = 20.4 grams

  Mass of combined product and Erlenmeyer flask = 115.6 + 20.4 = 136 grams

6 0
3 years ago
chapter 2 linear motion problems a student launches an arrow upward with an unknown initial velocity. the arrow takes 2.3 second
anzhelika [568]

Answer:

V_{0}= 22.5\frac{m}{s} and Ymax=25.8m

Explanation:

Velocity at any time is given by V=V_{0}sin\theta -gt. but when the arrow is on the top its velocity is zero and if it is launched upward the angle is 90°, so.

0=V_{0} sin90-gt

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At the maximun height, position is given by Ymax=V_{0}sin\theta.t-\frac{1}{2}gt^{2}, replacing Ymax=22.5\frac{m}{s}x2.3s-\frac{1}{2}x9.8x(2.3)^{2}=25.8\frac{m}{s}

5 0
3 years ago
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