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Semenov [28]
4 years ago
5

Radioactive isotopes are used to a. determine the absolute age of rocks. c. determine how rocks are formed. b. determine the rel

ative age of rocks. d. determine how rocks are changed. Please select the best answer from the choices provided A B C D
Physics
2 answers:
PolarNik [594]4 years ago
8 0
Hello <span>Mr1guy24 

Question: </span><span>Radioactive isotopes are used to?

Answer: Determine the absolute age of rocks (A)

Reason: If we want to know the absolute age of rocks, then we use radioactive isotopes to find this out.

Hope This Helps!
-Chris</span>
grandymaker [24]4 years ago
3 0

The answer is A: Radioactive isotopes are used.

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mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

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3 years ago
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A motorboat heads due east at 16 m/s across a river that flows due south at 9.0 m/s. a.) What is the resultant velocity of the b
alukav5142 [94]

Answer:

a)V=18.35 m/s (South -East)

b) t =7.41 m/s

c)D= 66.70 m

Explanation:

Given that

Velocity of boat in east direction = 16 m/s

Velocity of river = 9 m/s

a)The resultant velocity V

V=\sqrt{16^2+9^2}\ m/s

V=18.35 m/s (South -East)

b)

We know that

Distance = Velocity x time

Lets t time takes to cross the river

136 = 18.35 x t

t =7.41 m/s

c)

   The distance covered downstream  

We know that

Distance = Velocity x time

t= 7.41 s

D= 7.41 x 9 m

D= 66.70 m

3 0
3 years ago
The modern periodic law, according to Moseley, classifies elements based on arranging elements in ___ order
Daniel [21]
D because it makes sense
4 0
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Which is more dangerous to living things: gamma rays or X-rays?
Dima020 [189]

Answer:

gamma

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quizlet

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6 0
2 years ago
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The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure
krok68 [10]

(a) 1.15\cdot 10^9 N

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

p=\frac{I}{c}

where

I is the intensity

c is the speed of light

In this problem, I=1360 W/m^2; substituting this value, we find the radiation pressure:

p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

A=2pi R^2

where

R=6.37\cdot 10^6 m

is the Earth's radius. Substituting,

A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2

And so the force exerted by the sunlight is

F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N

(b) 3.2\cdot 10^{-14}

The gravitational force exerted by the Sun on the Earth is

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=1.99\cdot 10^{30}kg is the Sun's mass

m=5.98\cdot 10^{24}kg is the Earth's mass

r=1.49\cdot 10^{11} m is the distance between the Sun and the Earth

Substituting,

F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}

6 0
4 years ago
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