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:
0 degrees
Explanation:
Let 
are two forces. The resultant of two forces acting on the same point is given by :
Where 
is the angle between two forces
When 
i.e. when two forces are parallel to each other,
When 
i.e. when two forces are parallel to each other,
When 
i.e. when two forces are parallel to each other,
It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.
Answer:
m = 63 grams
Explanation:
ω = 10 cycles/s(2π radians/cycle) = 20π rad/s
ω = √(k/m)
m = k/ω² = 250/(20π)² = 0.06332... kg
It will be
d=m/v
=10/2
=5g/ml
Answer:
The magnitude of gravitational force between two masses is 
.
Explanation:
Given that,
Mass of first lead ball, 
Mass of the other lead ball, 
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :
So, the magnitude of gravitational force between two masses is . Hence, this is the required solution.
