First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
Answer:
c
Explanation:
a, b, and d are examples of moving forward
while c is moving backwards.
Answer:
Explanation:
Hello,
In this case, considering that the acceleration is computed as follows:
Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:
Regards.
Answer:
For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive
Explanation:
Solution
Given that:
From the question stated, Anna drew a diagram to compare forces that are strong and weak.
Now,
We are to find which labels are grouped in areas marked as X, Y, Z respectively.
Thus,
For X, Y, Z it is marked as:
X: Always attractive or attractive only
Y: Very small range
Z: Repulsive and attractive
<span>A: It is not an exact representation of the atom, but is close enough to be very useful.
Hope this helps!</span>