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atroni [7]
2 years ago
6

Which statements can be inferred from the Paleozoic era time scale? There was volcanic activity during the Paleozoic era. Dinosa

urs existed in the middle part of the Paleozoic era. Marine organisms existed in the early part of the Paleozoic era. Ammonites existed before the trilobites did during the Paleozoic era. Erosion and deposition were processes that occurred during the Paleozoic era.
Physics
2 answers:
Alchen [17]2 years ago
5 0

Answer:

A,C,E

Explanation:

Orlov [11]2 years ago
5 0

Answer: the answers are A,C and E

Explanation:

I took the test and got it right

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The connections of many simple pieces in the brain is evidence of the:
olga55 [171]

Brian’s Complexity Brian’s Complexity Brian’s Complexity Brian’s Complexity

6 0
2 years ago
Read 2 more answers
A pipe contains water at 500,000 Pa above atmospheric pressure. If you patch a 4.00 mm diameter hole in the pipe with a piece of
IRISSAK [1]

Answer: 6.284N

Explanation:

Pressure is the ratio of force exerted to cross sectional area of the material.

Pressure = Force/Area

Pressure = 500,000Pa

Area = Πd²/4 where d is the diameter of the hole.

If d = 4mm = 0.004m

Area = Π×0.004²/4

Area = 1.26×10^-5m²

Force = Pressure×Area

Force = 500,000× 1.26×10^-5

F = 6.284N

The gum must be able to withstand 6.284N force

7 0
2 years ago
A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how
storchak [24]

Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.69}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}

b) Expression for rate law for first order kinetics is given by:

A=A_0e^{-kt}

where,

k = rate constant  

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

A_0 = initial activity  = 600, 000

A=600000\times e^{-0.346\times 6}

A=75258

Thus there will be 75258 nuclei left at 6 pm.

7 0
2 years ago
A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied
erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

\to \sigma y  =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\

Calculating Shear Transverse:

\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}

        = \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi

= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\  in

\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\

       = [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi

8 0
2 years ago
A boy is standing motionless on a skateboard. He throws a basketball forward. Describe his motion.
weqwewe [10]

Answer:

Random motion

Explanation:

If the boy throws the basketball forward while at a position on the skateboard, the motion of the ball will be a random motion since we are not told if the ball is moving on a straight line when thrown forward.

In this case, the boy will tend to move in the direction of the ball. Since the ball is moving in a random manner, the motion of the boy will also be a random motion.

A random motion is a motion of a body in a zig zag manner. It is also known as Brownian motion e.g motion of a buzzing mosquito, motion of a smoke coming out of a chimney etc.

8 0
2 years ago
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