Meter #2 is more precise.
There's no information here that tells us which meter is more accurate.
Answer:
Mass, m = 6.18 kg
Explanation:
Given the following data;
Frequency, F = 10 Hz
Spring constant, k = 250 N/m
We know that pie, π = 22/7
To find the mass, we would use the following formula;
F = 1/2π√(k/m)
Where;
F is the frequency of oscillation.
k is the spring constant.
m is the mass of the spring.
Substituting into the formula, we have;
10 = 1/2 * 22/7 * √250/m
10 = 22/14 * √250/m
Cross-multiplying, we have;
140 = 22 * √250/m
Dividing both sides by 22, we have;
140/22 = √250/m
6.36 = √250/m
Taking the square of both sides, we have;
6.36² = (√250/m)²
40.45 = 250/m
Cross-multiplying, we have;
40.45m = 250
Mass, m = 250/40.45
Mass, m = 6.18 kg
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .
If the machine is less than 100% efficient,
then the MA is more than 9 .