Uses chemical and generates electrical
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, 
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
The bottom of the cup passes the sound waves to the string, and so on to the other cup. You can hear surprisingly far using a string telephone if help the right way! If the string is kept tight, the sound waves will travel.
Explanation:
