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Arlecino [84]
3 years ago
13

Do the rock layers from the various locations that you investigated record evidence of an asteroid impact on Earth? Make a CLAIM

below: *​

Physics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

Drilling into the seafloor off Mexico, scientists have extracted a unique geologic record of the single worst day in the history of life on Earth, when a city-sized asteroid smashed into the planet 65 million years ago, wiping out the dinosaurs and three-quarters of all other life.

Their analysis of these new rock samples from the Chicxulub crater, made public Monday, reveals a parfait of debris deposited in layers almost minute-by-minute at the heart of the impact during the first day of a global catastrophe. It records traces of the explosive melting, massive earthquakes, tsunamis, landslides and wildfires as the immense asteroid blasted a hole 100 miles wide and 12 miles deep, the scientists said.

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What is true about energy that is added to a closed system?
Setler [38]

The correct answer is B

6 0
3 years ago
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You can enter compound units that are combinations of other units that are multiplied together.Torque can be calculated by multi
Paraphin [41]

Answer:

3.0 x 10¹ Nm

Explanation:

Torque = F x r

Where F is force applied and r is perpendicular distance from pivot point . r

is also called lever arm

Here F = 15 N and r = 2.0 m

Torque

= 15 N X 2.0 m

= 3.0 10¹ Nm.

4 0
3 years ago
a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q
Romashka-Z-Leto [24]

The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

  • A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
  • These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.

Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

#SPJ4

5 0
1 year ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
Which bar graph could represent the reaction rates of a reversible reaction<br> that has just begun?
Galina-37 [17]

The answer is A. Just did it.

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