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4 years ago

Most geologists accept radiometric dating techniques as valid because

2 answers:

7 0

Most geologists accept radiometric dating techniques as valid because radioactive elements decay at a constant and measurable rate.

Answer: Option C

<u>Explanation:</u>

Scientists prefer radioactive dating to carbon dating because it is more accurate in measuring. The analysis depends upon the radioactive decay of radioactive isotopes of any matter in a given rock or soil.

The parent atoms and daughter atoms are compared while studying, and hence age can be calculated easily. Radioactive decay depends upon the given half-life of the atom, which is a constant and is known. So, it would be very easy to calculate the number of progeny atoms and parent atoms and find out their age.

DanielleElmas [232]4 years ago

6 0

The correct answer is C. Radioactive elements decay at a constant and measurable rate.

Explanation:

Radiometric dating is a technique used by geologists to determine how old some materials or elements are, for example, rocks or fossils. This technique focuses on measuring the decay of radioactive elements in the materials. This is possible because most materials contain traces of radioactive elements and decay at a constant rate over time. which has been widely studied and measured. Due to this, a certain rate of decay is equivalent to a specific number of years and through this, geologists can now the age of rocks and other elements. Thus, this technique is accepted as valid because "radioactive elements decay at a constant and measurable rate".

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QUESTION 1

frez [133]

Go90 day would work better if they could just pay ate a little bit more than we

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3 years ago

The rings of Saturn occupy the region inside Saturn's Roche limit.

Oliga [24]

I suppose this is a true or false question and that sentence is true.

7 0

3 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

dimulka [17.4K]

Answer:

The power output of this engine is $P = 17.5 W$

The the maximum (Carnot) efficiency is $\eta_c = 0.7424$

The actual efficiency of this engine is $\eta _a = 0.46$

Explanation:

From the question we are told that

The temperature of the hot reservoir is $T_h = 1250 \ K$

The temperature of the cold reservoir is $T_c = 322 \ K$

The energy absorbed from the hot reservoir is $E_h = 1.37 *10^{5} \ J$

The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

The power output is mathematically represented as

$P = \frac{W}{t}$

Where t is the time taken which we will assume to be 1 hour = 3600 s

W is the workdone which is mathematically represented as

$W = E_h -E_c$

substituting values

$W = 63000 J$

$P = \frac{63000}{3600}$

$P = 17.5 W$

The Carnot efficiency is mathematically represented as

$\eta_c = 1 - \frac{T_c}{T_h}$

$\eta_c = 1 - \frac{322}{1250}$

$\eta_c = 0.7424$

The actual efficiency is mathematically represented as

$\eta _a = \frac{W}{E_h}$

substituting values

$\eta _a = \frac{63000}{1.37*10^{5}}$

$\eta _a = 0.46$

7 0

4 years ago

How far (in meters) above the earth's surface will the acceleration of gravity be 85.0 % of what it is on the surface?

suter [353]

Answer:

X = 6910319.7 m

Explanation:

let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.

on the surface of the earth, the gravitational acceleration is given by:

g1 = GM/(r^2) = [(6.67408×10^-11)(5.972×10^24)]/[(6371×10^3)^2] = 9.82 m/s^2

at X meters above the earth's surface, g2 = 85/100(9.82) = 8.35m/s^2

then:

g2 = GM/(X^2)

X^2 = GM/g2

X = \sqrt{GM/g2}

= \sqrt{(6.67408×10^-11)(5.972×10^24)/ 8.35

= 6910319.7 m

Therefore, the acceleration of gravity becomes 85% of what it is on the surface at 6910319.7 m .

4 0

3 years ago

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0

4 years ago