B) Weather changes day to day, while climate changes region to region.
Climate is the weather in a certain area. It's usually the average weather over a long period of time
Weather is in shorter terms then climate
Hope this helped!
~Just a girl in love with Shawn Mendes
Because of internal friction between layers of mud particles called viscosity. When you walk, your foot exerts a force on the mud; and according to Newton, the mud also (is supposed to) exert an equal opposite force, which leading to an equal net resultant force in your direction, propelling you forward.
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
Answer:
3rd picture straight line going up right
Explanation:
3rd picture
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>