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3 years ago

7

Where did scientists like Einstein, Bethe, Fermi, Teller, and Bloch emigrate in order to escape persecution and pursue scientifi

c research?
a. Italy
b. France
c. Germany
d. the United States

1 answer:

enot [183]3 years ago

6 0

Those particular individuals that are mentioned in the question
all emigrated to the USA.

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The tenth pair of cranial nerves is the longest in the body and leaves by way of the jugular foramen. These nerves are called th

tekilochka [14]

Answer:

Vagus nerves

Explanation:

The vagus nerves are the tenth pair of cranial nerves

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4 years ago

What way would the balls fall? Top<br> Left<br> Right<br> Bottom

stira [4]

Each ball, when released, will fall toward the center of the big ball that everybody is standing on.

Another way to describe it: Each ball, when released, will fall in the direction parallel to a line from the boy's head to his feet.

-- The ball held by the boy on top will fall toward the bottom of the picture.

-- The ball held by the boy on the right will fall toward the left side of the picture.

-- The ball held by the boy on the bottom will fall toward the top of the picture.

-- The ball held by the boy on the left will fall toward the right side of the picture.

In each case, the direction toward the center of the big ball that they're all standing on is the direction that each person calls "down".

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4 years ago

A 35 kg object has -450 kgm/s of momentum. Calculate its velocity.

shtirl [24]

Hi there!

Recall the equation for momentum:

$\large\boxed{p = mv}\\$

p = momentum (kgm/s)

m = mass (kg)

v = velocity (m/s)

We can plug in the given values to solve for velocity:

$-450 = 35v\\\\v = \frac{-450}{35} = \boxed{12.86 m/s}$

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2 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

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3 years ago

Two wires, each of length 1.2m are stretched between 2 fixed supports. On wire A there is a second harmonic standing wave whose

emmainna [20.7K]

Explanation:

Below is an attachment containing the solution

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4 years ago