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GenaCL600 [577]
4 years ago
9

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Physics
1 answer:
azamat4 years ago
7 0

Answer:

1.After 11 hours, the cruise ship will be  _233.75__nautical miles from the lighthouse.

2.At the start of the journey, the cruise ship was _10.5__ nautical miles from the lighthouse.

3.The cruise ship is traveling at a speed of_21.25_  nautical miles per hour.

Explanation:

Distance covered in 2 hours= 95.5-53NM

=42.5NM

therefore,

Speed= Distance/time................(1)

put value of distance and time in equation (1)

therefore,

speed= 42.5/2hour

speed=21.25NM/hour

Therefore distance covered in 11 hours= speed* time

Distance=21.25*11

=233.75NM

2. Since distance covered in 2 hours is 42.5

therefore, at starting of the journey the cruise distance from light house is

= 53NM-42.5NM

=10.5NM

3. Distance covered in 2 hours= 95.5-53NM

=42.5NM

therefore,

Speed= Distance/time................(1)

put value of distance and time in equation (1)

therefore,

speed= 42.5/2hour

speed=21.25NM/hour

Therefore Cruise ship is traveling at speed of 21.25NM/hour

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3 years ago
When an object's final velocity is less than its initial velocity, however, it has ________________ acceleration.
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The body has negative acceleration PR a deceleration.

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

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Assume you are given an int variable named nElements and a two-dimensional array that has been created and assigned to a2d. Writ
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Explanation:

public int dimension(int [][]a2d,int nElements)

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3 years ago
Define an astronomical unit. Choose all that apply. a. 8.3 minutes Average distance from Earth to the Sun. b. Distance that ligh
Naddik [55]

Answer:

a. 8.3 minutes average distance from earth to the sun

d. 93 miles or 150 million km

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6 0
3 years ago
To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

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t = 2.35 s

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Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

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The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
4 years ago
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