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2 years ago

What makes astronomers think that impact rates for the Moon must have been higher earlier than 3.8 billion years ago?

1 answer:

3 0

There is strong evidence indicating that 3.8 billion years ago there was a higher impact rate. This deduction starts from comparing the number of craters in the lunar highlands with those of the Mary. If this comparison is made, it will be observed that there are 10 times more craters in the highlands than in a similar area of Mary. It should be borne in mind that through radioactive dating processes the samples indicate that there is a slightly greater antiquity in the highlands than those of Maria. This allows us to deduce that if the impact rates had been constant, the highlands would have been 10 times older. They would have to be formed 38 billion years ago, long before the universe itself began.

Therefore one of the most obvious reasons is there are ten times more craters on the older highlands than the Younger Maria.

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In 1896, S. Riva-Rocci developed the prototype of the current sphygmomanometer, a device used to measure blood pressure. When it

Anastasy [175]

Answer:

Explanation:

A pressure that causes the Hg column to rise 1 millimeter is called a torr. The term 1 mmHg used can replaced by the torr.

1 atm = 760 torr = 14.7 psi.

120 mmHg

Psi:

760 mmHg = 14.7 psi

120 mmHg = 14.7/760 * 120

= 2.32 psi

Pa:

1mmHg = 133.322 Pa

120 mmHg = 120 * 133.322

= 15998.4 Pa

80 mmHg

Psi:

760 mmHg = 14.7 psi

80 mmHg = 14.7/760 * 80

= 1.55 psi

Pa:

1mmHg = 133.322 Pa

80 mmHg = 80 * 133.322

= 10665.6 Pa

4 0

3 years ago

What are the thee major wind systems

astra-53 [7]

Trade winds, prevailing westerlies, polar easterlies

8 0

3 years ago

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

5 0

3 years ago

The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason

kap26 [50]

Overloading

Explanation:

The reason reason why the mcb of rupa's room keeps tripping is due to the fact that excessive current has being supplied to his room.

MCB stands for a miniature circuit breaker.

A miniature circuit breaker opens up the electrical circuit by switching it off when there is overloading or faulty connections.

A MCB helps to control the flow of current and it is designed to hand certain limits of electrical voltage.
If the MCB keeps tripping, it suggests a surge in current supplied, overloading or probably a faulty connection.

learn more:

Circuit brainly.com/question/12472944

#learnwithBrainly

4 0

3 years ago

By moving a 20 nC charge from point A to point B, you determine that the electric potential at B is 100 V . Part A What would be

ivolga24 [154]

Answer:

Potential at B would be 100V

Explanation:

The electric potential is defined as the work done to bring a unit positive charge from infinity to some point in the field.

We always determine the potential with respect to some reference point. Let the potential at A be zero. If the potential at B is V, then work done to bring charge q from A to B = qV

which is the electric potential energy.

If instead we use some charge Q, the electric potential <em>energy</em> will be QV, but the electric potential will always be V.

4 0

2 years ago