Answer:
240cm
Explanation:
Speed = 6cm/s
Time = 40 seconds
Speed= distance/time
6cm/s = distance/40seconds
Distance = 6×40
Distance = 240cm
Hence, in 40 seconds, the ball must have rolled the distance of 240cm
Answer:
-21.6 m/s^2
Explanation:
First of all, we need to convert the initial and final velocities into m/s:
is the initial velocity
is the final velocity
The acceleration is given by

where
t = 90 s is the time elapsed
Substituting the numbers, we find

and the negative sign means the rocket is decelerating.
Answer:
a) a = 1 [m/s²]
b) a = 0.25 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
a)
Between 0 and 2 seconds
where:
Vf = final velocity = 2 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
t = time = 2 [s]
Note: the positive sign of acceleration in the previous equation allows us to appreciate that the velocity is increasing
2 = 0 + a*2
a = 2/2
a = 1 [m/s²]
b)
Between 2 and 4 seconds
where:
Vf = final velocity = 2.5 [m/s]
Vo = initial velocity = 2
a = acceleration [m/s²]
t = time = (4 - 2) = 2 [s]
2.5 = 2 + a*2
2.5 - 2 = 2*a
0.5 = 2*a
a = 0.25 [m/s²]
Answer:
The width of the central bright fringe will increase
Explanation:
In a single-slit diffraction pattern, the distance of the nth-minimum from the central maximum on the screen is given by

where
is the wavelength of the wave
D is the distance of the screen from the slit
d is the width of the slit
The width of the central bright fringe is equal to twice the value of y=1 (first minimum), so:

And we see that it is inversely proportional to the width of the slit: therefore, if the width of the slit is reduced, the width of the central brigh fringe will increase.
Answer:i have none but good luck also this is a learning site ;-;
Explanation: