Answer:
<u><em>1) if they are moving away from each other it will take 1.43 secs</em></u>
<u><em>2) if they are moving towards each other then it will take 1.11 secs</em></u>
Explanation:
Distance between them is 10 m
Speed ( if they are moving towards each other)= distance/time
time = 10/8+1
time = distance / speed= 10/9= 1.11 secs
if they are moving away from each other than it will take
time = 10/8-1= 10/7= 1.43 secs
Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
Answer:
in pretty sure the last one
Answer:
the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Explanation:
Given that;
final velocity v = 0
initial velocity u = 15m/s
time taken t = 4 s
acceleration a = ?
from the equation of motion
v = u + at
we substitute
0 = 15 + a × 4
acceleration a = -15/4 = - 3.75 m/s²
the negative sign tells us that its a deacceleration so the sign can be ignored.
Deacceleration due to friction a = μ × g
we substitute
3.75 = μ × 9.8
μ = 3.75 / 9.8 = 0.3826 ≈ 0.38
Therefore the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
In order to calculate the unknown reaction force, we need to know that the sum of forces pointing down is equal the sum of the forces pointing up, so all forces will be in equilibrium.
So we have:
