Define scalar quantity
2 answers:
You might be interested in
In order to find out if the exponential function represents a growth or a decay, let's look at the number that is base to the exponent x.
If the number is greater than 1, so the function represents a growth, and if the number is less than 1, the function represents a decay.
Since the number is 1.075, the function represents a growth.
To find the % increase, first let's convert the number to percentage, and then subtract 100%:
So the percent increase is 7.5%.
(a) 69.3 J
The work done by the applied force is given by:
where:
F = 25.0 N is the magnitude of the applied force
d = 3.20 m is the displacement of the sled
is the angle between the direction of the force and the displacement of the sled
Substituting numbers into the formula, we find
(b) 0
The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.
(c) 69.3 J
According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:
where
is the change in kinetic energy
W is the work done
Since we already calculated W in part (a):
W = 69.3 J
We therefore know that the change in kinetic energy of the sled is equal to this value:
(d) 4.9 m/s
The change in kinetic energy of the sled can be rewritten as:
(1)
where
Kf is the final kinetic energy
Ki is the initial kinetic energy
m = 5.50 kg is the mass of the sled
u = 0 is the initial speed of the sled
v = ? is the final speed of the sled
We can calculate the variation of kinetic energy of the sled, , after it has travelled for d=3 m. Using the work-energy theorem again, we find
And substituting into (1) and re-arrangin the equation, we find
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