Answer:
a) 20 nodes b) zero nodes
Explanation:
When we have standing waves in a bend we have nodes at the ends and the equation describes the number of possible waves in the string is
L = n λ/2
Where λ is the wavelength, L is the length of the string, in our case it would be D and n is an entered. We can strip the wavelength of this expression
λ = 2L / n
Let's calculate what value of n we have for a wavelength equal to D/10
λ = 2D / n
λ = D / 10
We match and calculate
2D / n = D / 10
2 / n = 1/10
n = 20
Perform them for λ = D / 20
λ = 2D / n
2D / n = D / 20
n = 2 20 = 40
Since n is an inter there should be a wavelength for each value of n in the bone period there should be 20 different wavelengths
B) for La = 10D
2D / n = 10D
1 / n = 5
n = 1/5 = 0.2
La = 20D
2D / n = 20D
1 / n = 10
n = 1/10 = 0.1
These numbers are not entered so there can be no wave in this period
Answer:
12.7 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 56.7 Km/hr
Maximum height (h) =..?
First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:
Initial velocity (m/s) = 56.7 x 1000/3600
Initial velocity (m/s) = 15.75 m/s
Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:
Initial velocity (u) = 15.75 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
v = u – gt (since the ball is going against gravity)
0 = 15.75 – 9.8 × t
Rearrange
9.8 × t = 15.75
Divide both side by 9.8
t = 15.75/9.8
t = 1.61 secs.
Finally, we shall determine the maximum height as follow
h = ½gt²
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 1.61 secs.
Height (h) =..?
h = ½gt²
h = ½ × 9.8 × 1.61²
h = 4.9 x 1.61²
h = 12.7 m
Therefore, the maximum height reached by the ball is 12.7 m
We can calculate the length of each spring by using the relationship:
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)
Re-arranging the equation, we have
The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
Therefore, the correct answer is
<span>
D.Spring A is 3 m longer than spring B because 15 – 12 = 3.</span>
Answer:
that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
Explanation:
Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.
Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.
In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.
If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
Answer:
The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.
Explanation:
