Answer:
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Answer:
Explanation:
initial velocity, u = 0
final velocity, v = 60 mph = 26.8 m/s
time t = 10 s
Let a be the acceleration and s be he distance traveled.
Use first equation of motion
v = u + a t
26.8 = 0 + a x 10
a = 2.68 m/s
Use second equation of motion
s = ut + 1/2 at²
s = 0 + 0.5 x 2.68 x 10 x 10
s = 134 m
As, 1 m = 3.28 ft
So, s = 134 x 3.28 ft
s = 439.6 ft
Answer:
Speed of the boat, v = 4.31 m/s
Explanation:
Given that,
Height of the bridge, h = 32 m
The model boat is 11 m from the point of impact when the key was released, d = 11 m
Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

Here, u = 0 and a = g


t = 2.55 seconds
Let v is the speed of the boat. It can be calculated as :


v = 4.31 m/s
So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.
Answer:
The width of the central bright fringe on the screen is observed to be unchanged is 
Explanation:
To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

Where,
w = width
wavelength
m is an integer, m = 1, 2, 3...
We here know that as
as w are constant, then

We need to find
, then

Replacing with our values:


Therefore the width of the central bright fringe on the screen is observed to be unchanged is 