Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
Before the impact, let the velocity of the baseball was v m/s.
After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.
a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.
Answer:
A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.
Explanation:
Answer:
31.905 ft/s²
Explanation:
Given that
Mass of the pilot, m = 120 lb
Weight of the pilot, w = 119 lbf
Acceleration due to gravity, g = 32.05 ft/s²
Local acceleration of gravity of found by using the relation
Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)
119 = 120 * a/32. 174
119 * 32.174 = 120a
a = 3828.706 / 120
a = 31.905 ft/s²
Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²
Option A overweight
HOPE IT HELPS!!