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kogti [31]
3 years ago
8

Why are some isotopes used as tracers?

Physics
1 answer:
sineoko [7]3 years ago
6 0
<span>C. They decay at a predictable rate. IE carbon 14 decay to carbon 12 to understand how long ago something was alive.</span>
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Janet recently read an article about the role that the brain plays in processing emotions. Janet comes over to your house for di
Ira Lisetskai [31]
Amygdala helps in processing emotions
I hope I could answer part of your question
3 0
3 years ago
A coin is dropped from a height and reaches the ground in 2 seconds. Neglecting air resistance, from what height (in meters) was
Kamila [148]

Answer:

Coin is dropped from a height of 19.62 m

Explanation:

We have given time t = 2 sec

As coin is drop means its initial velocity u = 0 m/sec

We have to find the height from which coin is dropped

From second equation of motion we know that

h=ut+\frac{1}{2}gt^2

So height h=0\times2+\frac{1}{2}\times 9.81\times 2^2=19.62m

So coin is dropped from a height of 19.62 m

3 0
3 years ago
When the northern hemisphere is tilted toward the sun that part of the earth is closer to the sun and thus hotter? True or false
Ratling [72]

<u>False</u>

<u>Explanation:</u>

Most of us have a belief that Earth is very close to the sun in summer season and so it is hotter. Also, since Earth is far away from the sun in winter season, it is colder.  Though this idea looks true, it is incorrect.

The orbit of the Earth isn't a perfect circle. It is quite tilted. During some months of the year, Earth is very close to the sun than at other times. We have winter when the Earth is very close to the sun and summer when it is far away in the Northern Hemisphere. So when we compare the distance of the Sun from the Earth, this change in Earth's distance throughout the year does not affect our weather much.

8 0
3 years ago
A cannon shoots an artillery shell with a initial velocity of 400 meters/second at an indirect fire angle of 60 on level ground
notsponge [240]

Answer: 14139.19 m

Explanation:

This situation is related to parabolic motion and can be solved using the following equations:

x=V_{o}cos \theta t (1)

y=y_{o}+V_{o} sin \theta t+\frac{g}{2}t^{2} (2)

Where:

x is the horizontal distance (where the artillery shell lands)

V_{o}=400 m/s is the initial velocity

\theta=60\° is the angle

t is the time

y=0 m is the final height

y_{o}=0 m is the initial height

g=-9.8 m/s^{2} is the acceleration due gravity, always directed downwards

So, let's begin by isolating t from (2):

0=V_{o} sin \theta t+\frac{g}{2}t^{2} (3)

t=-\frac{2 V_{o}sin \theta}{g} (4)

Substituting (4) in (1):

x=V_{o}cos \theta (-\frac{2 V_{o}sin \theta}{g}) (5)

Rewriting (5) and taking into account sin(2\theta)=2 sin \theta cos \theta:

x=-\frac{V_{o}^{2}sin(2\theta)}{g} (6)

x=-\frac{(400 m/s)^{2}sin(2(60\°))}{-9.8 m/s^{2}} (7)

Finally:

x=14139.19 m

6 0
3 years ago
A) An electron is moving with a speed of 3.5 x105 m/s when it encounters a magnetic field of 0.60T. The direction of the magneti
kirza4 [7]

Given Information:  

v = 3.5x10⁵ m/s

θ = 60°

Magnetic field = 0.60 T

no. of electrons = 6.242x10¹⁸

radius = 0.750 m

Required Information:  

Magnetic force = F = ?

Magnetic field = B = ?

Answer:

Part A) F = 2.909x10⁻¹⁴ N

Part B) B = 5.34x10⁻⁷ T

Explanation:

Part A:

The magnetic force can be found using

F = qvBsin(θ)

Where q = 1.60x10⁻¹⁹ C and v is the speed of electron, B is the magnetic field and  θ is the angle between v and B.

F = 1.60x10⁻¹⁹*3.5x10⁵*0.60sin(60°)

F = 2.909x10⁻¹⁴ N

Part B:

From the Ampere's law, the magnetic field can be found using

B = μ₀I/2πr

Where μ₀ = 4πx10⁻⁷ is the permeability of free space, I is the current, and B is the magnitude of the magnetic field produced.

1 ampere of current has 6.242x10¹⁸ electrons per second so

I = 1.25x10¹⁹/6.242x10¹⁸

I = 2.0032 A

B = 4πx10⁻⁷*2.0032/2π*0.750

B = 5.34x10⁻⁷ T

6 0
4 years ago
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