Answer:
The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.
Explanation:
Given that,
Acceleration of the car, 
Initial speed of the car, u = 24.6 m/s
Final speed of the car, v = 26.8 m/s
We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :
t = 0.84 seconds
So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα= 
Senα= 
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
Answer:
0.82 mm
Explanation:
The formula for calculation an 
bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength 
= 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength 
= 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m 
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
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