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2 years ago

6

imagine you cut something in half. Then, you cut each half in half and continue doing so. Could you keep cutting the pieces in h

alf forever?

1 answer:

algol [13]2 years ago

6 0

You can't keep cutting because the object will be gone

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What features of the organism are preserved in Amber

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I got no earthly idea! Sorry........

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2 years ago

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The acceleration of an object would increase if there was an increase in:

kolezko [41]

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Its either B or A

Explanation:

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2 years ago

What is ther atomic number of an element that has 2 electrons, 3 protons,<br> and 4 neutrons?

Bingel [31]

Answer:

9

Explanation:

A atomic number is the addition of every number

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3 years ago

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago

The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

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5 0

3 years ago