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2 years ago

6

Ian and Dane are twins. Ian loves art and music and work after school. Dane loves sports and would do anything to avoid having a

job. This is an example of how development is:

1 answer:

Brilliant_brown [7]2 years ago

5 0

Answer:

Explanation:

Pliability

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An object attached to an ideal spring oscillates with an angular frequency of 2.81 rad/s. the object has a maximum displacement

NikAS [45]

Ω = 2.81
A = 0.232
k = 29.8

x = A cos(ωt + Ф)

at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0

at t = 1.42, with Ф = 0:
x = A cos(ωt)

U = 1/2 k x² = 1/2 k [A cos(ωt)]²

4 0

2 years ago

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

r-ruslan [8.4K]

Answer:

2406 miles

Explanation:

Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

$\angle ABC =(180-10) \textdegree=170\textdegree$

#Distance traveled in 1.5hrs is;

$c=690x1.5\\=1035mi$

#Distance traveled in next two hrs:

$a=690\times 2\\=1380mi$

#Now using the Cosine Rule:

$b^2=a^2+c^2-2ab\cos B\\\\=1380^2+1035^2-2(1380)(1035)cos170\textdegree\\\\b^2=5788.83\\\\b\approx 2406.00 \ mi$

Hence, the pilot is 2406 miles from her starting position.

4 0

2 years ago

Lighting from the sun travels through space to earths atmosphere.

Lady bird [3.3K]

Yes it does do all that

3 0

2 years ago

What is the net force on this object?

Sergio039 [100]

200N

Explanation:
600N-400N = 200N

6 0

2 years ago

Suppose that the distance an aircraft travels along a runway before takeoff is given by Upper D equals (5 divided by 3 )t square

Dahasolnce [82]

Answer:

a. Time=25seconds

b.distance=1041.67m

Explanation:

a.The equation for $D$ in terms of m/s is $\frac{250}{3}m/s$ after conversion.

To find when speed reaches 300km/hr=83.33m/s, we find $D\prime$ and solve for $t$

$D=\frac{5}{3}t^2\\D\prime=\frac{5}{3}(2t)=\frac{10}{3}t=\frac{250}{3}\\t=25sec$

b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.

#To find distance travelled in that time , we substitute for $t=25$ in our distance equation:

$D(25)=\frac{5}{3}(25)^2\\=1041.67m$

Hence the distance of the plane before it gets airborne is 1041.67m

4 0

2 years ago