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Paha777 [63]
3 years ago
8

In a posteroanterior (pa) projection of the chest being used for cardiac evaluation, the heart measures 14.7 cm between its wide

st points. if the magnification factor is known to be 1.2, what is the actual diameter of the heart?
Physics
1 answer:
Ilya [14]3 years ago
8 0

The actual diameter of the heart is 12.25 cm. Given : Heart measure = 14.7 cm Magnification factor = 1.2

14.7 / 1.2 = 12.25

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Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

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As a gaseous element condenses, the atoms become ________ and they have ________ attraction for one another.
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Answer:

blank 1: close together

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1 year ago
If each pull-up requires 300 J and Ben does a pull-up in 2 seconds, what is his power? 150 watts 300 watts 600 watts 750 watts
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Answer:

150 watts

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300/2 = 150 watts

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On a sunny day, a student poured a cup of water on the sidewalk to make a puddle. When he returned later, the puddle was gone. T
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The students conclusion is wrong.

Explanation:

The students conclusion is wrong because liquids can not soak into concrete. The water evaporated because it was a hot sunny day. The water turned into a gas called water vapor.

8 0
3 years ago
An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
Liono4ka [1.6K]

Answer:

16.25^{\circ}

Explanation:

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The angle at which the arrow is to be released is 16.25^{\circ}.

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