Answer:
4 m/s or 4 meters per second.
Explanation:
In order to calculate the speed of wave, you multiply the wavelength in meters and the frequency of the Wave in Hertz. 2 times 2 equals 4. The wave speed is always in m/s considering that the wavelength is also in meters.
The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
Explanation:
The attached figure shows data for the cart speed, distance and time.
For low fan speed,
Distance, d = 500 cm
Time, t = 7.4 s
Average velocity,

Acceleration,

For medium fan speed,
Distance, d = 500 cm
Time, t = 6.4 s
Average velocity,

Acceleration,

For high fan speed,
Distance, d = 500 cm
Time, t = 5.6 s
Average velocity,

Acceleration,

Hence, this is the required solution.
There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be
<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²
Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.
From another perspective: recall that
<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>
where
• <em>v</em>₀ = initial velocity
• <em>v</em> = final velocity
• <em>a</em> = acceleration
• ∆<em>y</em> = displacement
At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.
So we have
0² - (40 m/s)² = -2<em>g </em>(160 m)
but this reduces to
(40 m/s)² = 2 (9.8 m/s²) (160 m)
1600 m²/s² ≠ 3136 m²/s²