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2 years ago

Three physical properties of the bowling ball.

Physics

1 answer:

ICE Princess25 [194]2 years ago

6 0

The physical properties of the bowling ball is that it is hard, smooth and very dense.

<h3>What are physical properties?</h3>

Physical properties are those properties which are easily observable and which involve only physical changes in a substance.

A bowling ball is a solid.

The physical properties of bowling ball include:

density- it has high density
texture- it is smooth
hardness - it is hard

Therefore, the physical properties of the bowling ball is that it is hard, smooth and has high density.

Learn more about physical properties at: brainly.com/question/17436856

#SPJ1

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The chief by 3 3/4 kg of Apple's 7 1/4 kilograms of peers and 10 1/8 kilograms with orange about how many kilograms of fruit did

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SOLUTION:

Total amount of fruit = Total amount of apples + Total amount of pears + Total amount of oranges

Total amount of fruit = 15 / 4 + 29 / 4 + 81 / 8

Total amount of fruit = 30 / 8 + 58 / 8 + 81 / 8

Total amount of fruit = 169 / 8

Total amount of fruit = 21.125 kg

ANSWER:

Therefore, the chief bought 21.125 kg of fruit all together.

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Jessie and Jaime complete a 5.0 km race. Each has a mass of 68 kg. Jessie runs the race at 15 km/h; Jaime walks it at 5 km/h. Ho

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The total metabolic energy used by each to complete the course is determined as 656.91 J.

<h3>Kinetic energy of Jessie and Jaime</h3>

The kinetic energy of Jessie and Jaime is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of Jaime
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15 km/h = 4.17 m/s

5 km/h = 1.39 m/s

K.E = ¹/₂(68)(4.17)² + ¹/₂(68)(1.39)²

K.E = 656.91 J

Thus, the total metabolic energy used by each to complete the course is determined as 656.91 J.

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Answer:

To make it into the pool you must run and jump at

$\displaystyle v_o=x.\sqrt{\frac{g}{2y}}$

Explanation:

Horizontal Launch

When an object is thrown with a specified initial speed in the horizontal direction, it describes a curved path that finishes when it hits the ground level after traveling certain horizontal distance x and a vertical height y from the launching point. The horizontal speed is always constant and the vertical speed increases due to the effect of gravity. It can be found that the horizontal distance reached by the object when launched at an initial speed in a given time t is

$x=v_o.t$

And the vertical distance is

$\displaystyle y=\frac{g.t^2}{2}$

If t is the total flight time, then x and y are maximum and we can find a relation between them. Solving for t in the first equation

$\displaystyle t=\frac{x}{v_o}$

Substituting in the second equation

$\displaystyle y=\frac{g}{2}\left ( \frac{x}{v_o}\right )^2$

Rearranging

$\displaystyle \left ( \frac{v_o}{x}\right )^2=\frac{g}{2y}$

Solving for $v_o$

$\displaystyle v_o=x.\sqrt{\frac{g}{2y}}$

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