A rectangular in-ground pool is 45 feet long, 15.5 feet wide and 5.5 feet deep. How many cubic feet of water does the pool hold?

Answers

Answer 1

Answer:

Answer: 3836.25 cubic ft.

Explanation: Simply find the volume of the pool. In this case it's a rectangular prism so we use the volume equation for a rectangular prism:

1) Volume = length x width x height

2) Volume = 45 ft x 15.5 ft x 5.5 ft

3) 3836.25 cubic ft.

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Amusement parks are a great place to see Newton's laws of motion in action. Choose your favorite ride and describe the parts of the ride that illustrate each of Newton's laws of motion. You are only required to give a written description; however, if you would like, you may draw and label a picture in addition to the paragraph explanation. Please be sure to look at the rubric to see exactly what you are expected to include!

One example of a ______________ is the U-turn.

Answers

One example of a turnabout is the U turn

The acceleration due to gravity on the moon is about one-sixth its value on earth. If a baseball reaches a height of 67 m when wurth (jaw6696) – Ch 2 LT 3/4 – kreft – (61920) 2 thrown upward by someone on the earth, what height would it reach when thrown in the same way on the surface of the moon? Answer in units of m.

Answers

Answer:

402m

Explanation:

We can solve this easily by using the free fall formula:

$v_f^2=v_0^2+2ad$

On both cases the final velocity will be 0m/s (maximum height), so we have:

$v_0^2=-2ad$

We know that the velocity the person imparts to the ball will be the same either on Earth or on the Moon, so we write for both cases:

$v_0^2=-2a_Ed_E$

$v_0^2=-2a_Md_M$

Which means

$2a_Ed_E=2a_Md_M$

So we get (taking into account that gravity on earth is 6 times gravity on the Moon):

$d_M=(a_Ed_E)/(a_M)=(6a_Md_E)/(a_M)=6d_E=6(67m)=402m$

a missile is moving 1350 m/s at a 25.0 deg angle. it needs to hit a target 23,500 m away in 55.0 deg direction in 10.20 s. what is the magnitude of its final velocity?

Answers

In this question we have given

velocity of missile=1350m/s

angle at which missile is moving=25degree

distance between missile and targets=23500m

angle between target and missile=55degree

time=10.2s

To find the final velocity of missile we will first find the acceleration required

Let x be the horizontal component of distance

x - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity

horizontally: x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

similarly vertically:

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

Therefore

V = √(V'x² + V'y²) = 3504 m/s

therefore magnitude of final velocity of missile=3504m/s

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Answers

It's to determine if there is a charge on an object/body or not and what kind of charge as well. Hope this can help!

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Answers

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Answers

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