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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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Can you list the offensive position on a flag football team?
Alik [6]

Answer:

yes u can flag football has everything that pad football has so you can enlist on being offensive position but you have to play like you want that position

Explanation:

6 0
2 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i
gregori [183]

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

8 0
3 years ago
Assume you are in the car and the car is moving at a certain speed to
Setler79 [48]
You are at rest with respect to the car.
You are in motion with respect to the School.

6 0
3 years ago
The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st
Rudiy27

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

a=\frac{\textup{Change in speed}}{\textup{Time}}

thus,

a=\frac{26.8224 - 0 }{2.70}

or

a = 9.934 m/s²

6 0
3 years ago
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