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LuckyWell [14K]

3 years ago

14

Doctors can release medical information without a signed medical release for if the information is regarding organ or tissue don

ation. True or false

1 answer:

qaws [65]3 years ago

4 0

Answer:False

Explanation:

False

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91.0x26x504 ...............

Hoochie [10]

Answer:

1192464

Explanation:

91.0 times 26 = 2366

2366 times 504 = 1192464

7 0

3 years ago

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The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?

Bas_tet [7]

Answer:

God is omnipresent.

Explanation:

This means God is everywhere and He works where ever we are in the world

3 0

3 years ago

In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.

nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

$p=mv$

$p=25kg*3m/s= 75kg*m/s$

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

$75kg*m/s = 5.0kg*v$

$v=\dfrac{75}{5}$

$\boxed{v=15m/s}$

8 0

3 years ago

Two individuals start the same training program at the same time, but one is able to grow muscle faster and larger than the othe

vitfil [10]

I believe the answer is A

7 0

3 years ago

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Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

3 years ago