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kaheart [24]
3 years ago
13

How are eras and periods of the geologic time scale named?

Physics
2 answers:
gogolik [260]3 years ago
5 0

Answer: the answer is A, They are named for the places where geologists first described rocks or fossils from that time.

Explanation: i took the test and I got it right

Natalka [10]3 years ago
3 0

Answer:

mark me brainliest plz

Explanation:

In the early 1800's a system for naming geologic time periods was devised using four periods of geologic time. They were named using Latin root words, Primary, Secondary, Tertiary and Quaternary. ... Keep in mind that this chart is focused on geologic time periods. There are also geologic Eons, Eras, and epochs.

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16
ycow [4]
We assign the variables: T as tension  and x the angle of the string
 The  <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx. 
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<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

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4 0
2 years ago
Read 2 more answers
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
Initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity
stiks02 [169]

Answer:

<em>The final velocity is 20 m/s.</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

v_f=v_o+at

The provided data is: vo=10 m/s, a=5\ m/s^2, t=2 s. The final velocity is:

v_f=10~m/s+5\ m/s^2\cdot 2\ s

v_f=20\ m/s

The final velocity is 20 m/s.

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Joe has a mass of 40 kg, and Sally has a mass of 25 kg. They sit in class with a separation of 6 meters. Find the gravitational
goldfiish [28.3K]

Answer:

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Explanation:

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