For a transverse wave, the velocity is known to be perpendicular to the direction of the propagation of the particular wave. To get the acceleration, what is necessary is to take the partial derivative as it compares to time and the velocity.

With this in mind, we would have

transverse acceleration

a(t) = d²y/dt² = -0.150*50.02sin(0.8x - 50t)

= 0.150*502 = 375 m/s2

Hence we can say that the maximum acceleration of the element on the string is given as 375 m/s²

Read more on acceleration here

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6 0

10 months ago

"Which particle builds a static electric charge when it is transferred from one object to another?

Anestetic [448]

When two things rub against each other and become charged, it is
electrons that been rubbed off of one thing and on to the other one.

Protons would do that too if they got moved from one thing to another
by rubbing, but they don't.

5 0

2 years ago

for an object that is speeding up in the positive direction what does the displacement vs. time graph look like

avanturin [10]

For an object that is speeding up in the positive direction, its positive displacement is greater every second, and the AMOUNT greater every second is greater than the AMOUNT greater was in the previous second.

So the graph of displacement vs. time is rising as time goes on, and the rise is becoming steeper as time goes on.

The graph is <em>curving upward</em> as time goes on.

3 0

2 years ago

Find the electric field at a point midway between two charges of +40.0 × 10−9 c and +60.0 × 10−9 c separated by a distance of 30

Anna [14]

The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
$E=k_e \frac{q}{r^2}$
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
$E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C$

While the electric field generated by the second charge is
$E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C$

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
$E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C$

3 0

2 years ago