How do scientist determine how old the world is and the fossils inside of it?

4 answers:

8 0

Yeah they usually see how far down in the ground a fossil is and if it’s deep down they know it’s very old if it’s not deep down then it ain’t very old

valkas [14]2 years ago

6 0

The layers of the fossil the oldest is usual the bottom layer,and the top layer is the newest layer

yaroslaw [1]2 years ago

5 0

Yeah they usually see how far down in the ground a fossil is and if it’s deep down they know it’s very old if it’s not deep down then it ain’t very old

Mazyrski [523]2 years ago

5 0

Yeah they usually see how far down in the ground a fossil is and if it’s deep down they know it’s very old if it’s not deep down then it ain’t very old

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Why are household wall sockets alternating current (AC) instead of direct current (DC)? Select all that apply.

Bond [772]

Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

8 0

1 year ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$