A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip

Increasing the frequency of a wave does what to its period

Ivenika [448]

If the frequency is large then the period is short and if the frequency is small then the period is long

5 0

4 years ago

A ball accelerates from rest down a ramp at 2.4 m/s^2. Write an equation that could be used to determine the balls finals positi

Alik [6]

Answer:

x=2.4t+4.9t^2

Explanation:

This equation is one of the kinematic equations to solve for distance. The original equation is as follows:

X=Xo+Vt+1/2at^2

We know that the ball starts at rest meaning that its initial velocity and position is zero.

X=0+Vt+1/2at^2

Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.

X=Vt+4.9t^2

Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.

Now, substitute in your velocity value.

X=2.4t+4.9t^2

8 0

2 years ago

Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1

zheka24 [161]

Answer:

$v_{f}=1721.1m/s$

Explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

$dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\ (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s$