Answer:
For a velocity versus time graph how do you know what the velocity is at a certain time?
Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.
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How do you know the acceleration at a certain time?

Hence,
By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.
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How do you know the Displacement at a certain time?
Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.
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To solve this problem we will apply the concepts related to the Doppler effect. This is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically this is given as,

Here,
v = Speed of the waves in the middle
= Speed of the receiver in relation to the medium (Positive if the receiver is moving towards the transmitter or vice versa)
= Speed of the source with respect to the medium (Positive if the source moves away from the receiver or vice versa)
Our values are given as,




Replacing,

Solving for the velocity of the source,

Therefore the speed of the other train is 26.1m/s
Assuming that’s a right triangle, in this case A^2 + C^2 = B^2 … (16)^2 + C^2 = (25)^2 … C = 19.2 N
Answer:
4 the plant is leaning towards the window
Explanation:
that is due to the plant inhabit to grow towards the sunlight so the plants tries its best to get the sunlight by doing so and growing
Answer:
φ = B sin (2π n/a x)
Explanation:
In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is
φ = A cos kx + B sin kx
we must place the boundary conditions to determine the value of the constants A and B.
In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2
therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains
φ = B sin kx
the wave vector is
k = 2π /λ
now let's adjust the period, in the border fi = 0 therefore the sine function must be zero
φ (a /2) = 0
0 = A sin (2π/λ a/2)
therefore the sine argument is
2π /λ a/2 = n π
λ= a / n
we substitute
φ = B sin (2π n/a x)