A-11 polar easterlies
b-8 winds blowing between the equator and 30° N and south
c-10
d-9
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
![2gh = v_f^2 - v_i^2](https://tex.z-dn.net/?f=2gh%20%3D%20v_f%5E2%20-%20v_i%5E2)
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
![(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\](https://tex.z-dn.net/?f=%282%29%289.8%5C%20m%2Fs%5E2%29h%20%3D%20%280%5C%20m%2Fs%29%5E2-%284%5C%20m%2Fs%29%5E2%5C%5C)
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
![v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20gt_1%5C%5C0%5C%20m%2Fs%20%3D%204%5C%20m%2Fs%20%2B%20%28-9.8%5C%20m%2Fs%5E2%29t_1%5C%5Ct_1%20%3D%200.41%5C%20s)
(c)
Now we will consider the downward motion and use the third equation of motion:
![2gh = v_f^2-v_i^2](https://tex.z-dn.net/?f=2gh%20%3D%20v_f%5E2-v_i%5E2)
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
![2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\](https://tex.z-dn.net/?f=2%289.8%5C%20m%2Fs%5E2%29%282.62%5C%20m%29%20%3D%20v_f%5E2%20-%20%280%5C%20m%2Fs%29%5E2%5C%5C)
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
![v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20gt_1%5C%5C7.17%5C%20m%2Fs%20%3D%200%5C%20m%2Fs%20%2B%20%289.8%5C%20m%2Fs%5E2%29t_2%5C%5Ct_2%20%3D%200.73%5C%20s)
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
Thinking the small glass bead as a single point charge, the electric field generated by it is given by
![E(r) = k_e \frac{Q}{r^2}](https://tex.z-dn.net/?f=E%28r%29%20%3D%20k_e%20%20%5Cfrac%7BQ%7D%7Br%5E2%7D%20)
where
![k_e = 8.99 \cdot 10^9 N m^2 C^{-2}](https://tex.z-dn.net/?f=k_e%20%3D%208.99%20%5Ccdot%2010%5E9%20N%20m%5E2%20C%5E%7B-2%7D%20)
is the Coulomb's constant
![Q=8.0 nC=8.0 \cdot 10^{-9} C](https://tex.z-dn.net/?f=Q%3D8.0%20nC%3D8.0%20%5Ccdot%2010%5E%7B-9%7D%20C)
is the charge of the bead
![r=3.0 cm=0.03 m](https://tex.z-dn.net/?f=r%3D3.0%20cm%3D0.03%20m)
is the distance at which we calculate the field.
Using these data, we find:
Answer: Stars are in space for very long time, much longer than that one night. You are looking back in time because those stars have been there for so long that it’s like looking back in time, to when those stars were there.
Explanation:
May I please have brainlest