Answer:
Explanation:
To solve this, we start by using one of the equations of motion. The very first one, in fact
1
V = U + at.
V = 0 + 0.8 * 3.4 = 2.72 m/s.
2.
V = 0 + 0.8 * 4.3 = 3.44 m/s.
3.
d = ½ * 0.8 * 4.3² + 3.44 * 12.9
d = 7.396 + 44.376
d = 51.77 m.
4.
d = 62 - 51.77 = 10.23 m. = Distance
traveled during deceleration.
a = (V² - Vo²) / 2d.
a = (0² - 3.44²) / 20.46
a = -11.8336 / 20.46 = -0.58 m/s²
5.
t = (V - Vo)/a =(0 - 3.44) / -0.58
t = -3.44/-.58 = 5.93 s
= Stop time.
T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total
time the hare was moving.
6.
d = Vo * t + ½ * a * t² = 62 m.
0 + 0.5 * (23.13)² * a = 61
267.5a = 61
a = 61/267.5
a = 0.23 m/s²
<span>Electric field repulsive for objects of like charge and attractive for opposite type of charges and for a magnet you can say that like poles repel and unlike attracts so D makes sense</span>
Answer:
A) A warm wire
Explanation:
A warm wire has the most resistance. Heating the metal wire causes atoms to vibrate more, which in turn makes it more difficult for the electrons to flow, increasing resistance. Heating the wire increases resistivity.
Answer:
The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.
Explanation:
It is given that,
A high jumper jumps over a bar that is 2 m above the mat, h = 2 m
We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

g is acceleration due to gravity

v = 6.26 m/s
So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.