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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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if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t
VikaD [51]

Answer:

100years later

Explanation:

Because the lights will arrive at world after 100 years later.

7 0
3 years ago
A horizontal compass is placed 21 cm due south from a straight vertical wire carrying a 36 a current downward. in what direction
Anit [1.1K]

 <span>
The needle of a compass will always lies along the magnetic field lines of the earth. 
A magnetic declination at a point on the earth’s surface equal to zero implies that 
the horizontal component of the earth’s magnetic field line at that specific point lies along 
the line of the north-south magnetic poles. </span>

The presence of a current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field. Since magnetic 
<span>fields are vector quantities, therefore the magnetic field of the earth and the magnetic field of the vertical wire must be combined vectorially. </span></span>

<span>
Where:</span>

B1 = magnetic field of the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T

B2 = magnetic field due to the straight vertical wire along the y-axis

We can calculate for B2 using Amperes Law:

B2 = μ₀ i / [ 2 π R ]

B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>

The angle can be calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>

θ = 53°

<span>
<span>The compass needle points along the direction of 53° west of north.</span></span>

8 0
3 years ago
The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow
grin007 [14]

Answer:

420m

Explanation:

Given parameters:

Time  = 5minutes

Average speed  = 1.4m/s

Unknown:

Distance covered  = ?

Solution:

Speed is the rate of change of distance with time.

 Mathematically;

              Speed  = \frac{distance}{time}  

    Distance  = speed x time

 

Insert the parameters and solve;

   Convert the time to seconds;

              1 minute = 60s

               5 minute = 5  x 60 = 300s

So,

   Insert the parameters and find the distance;

     Distance  = 300 x 1.4  = 420m

3 0
2 years ago
What force is most responsible for the path of an object during projectile motion?
Levart [38]
The gravitational force.

In fact, the motion of the projectile is composed by two independent motions:
- on the horizontal direction, it is a uniform motion (with constant speed)
- on the vertical direction, it is a uniformly accelerated motion, where the vertical acceleration g is given by the gravity exerted by the Earth on the projectile.

For this reason, the composition of the two motions results in a parabolic trajectory.
7 0
3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0
3 years ago
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