The result is -15.625 m/s².
Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:
Where:
vf = final velocity
vi = initial velocity
t = time
Let’s examine the information provided in your query:
Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.
However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.
The time can be calculated using the kinematics equation:
We derive the equation by substituting the known values first.
The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.
It is important to note that the acceleration is negative, indicating the vehicle slowed down.
The full sentence states:
In a third class lever, the distance between the effort and the fulcrum is LESS than the distance between the load/resistance and the fulcrum.
In a third class lever, the fulcrum is positioned on one end of the effort, while the load/resistance is on the opposite side, placing the effort somewhere in between. Consequently, the distance from the effort to the fulcrum is less than that from the load to the fulcrum.
Response:
The ball remained airborne for 3.896 seconds
Explanation:
Given that
g = 9.8 m/s², representing gravitational acceleration,
If the angle of launch is 45°, the horizontal range will be maximized.
Both horizontal and vertical launch velocities are equal, each equating to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The duration to reach maximum height is half of the flight time.
v = u + at ∵ v = 0 (at maximum height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The total time in the air equals twice the time to reach maximum height
2 t₁ = 3.896 s
The horizontal distance covered is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in the air for 3.896 seconds
In terms of light energy, a higher frequency corresponds to increased energy within the light.
We establish that frequency is essentially the inverse of wavelength:
frequency = 1 / wavelength
Calculating frequencies:
f UVA = 1/320 to 1/400
f UVA = 0.0031 to 0.0025
f UVB = 1/290 to 1/320
f UVB = 0.0034 to 0.0031
Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.
