What is the wavelength of a wave traveling with a wave speed of 1348 m/s

Physics

1 answer:

saw5 [17]2 years ago

5 0

Answer: f = 4 Hz i think

Explanation:

sorry i dont rember (cant type lol)

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Consider the concepts of kinetic energy (KE) and gravitational potential energy (GPE) as you complete these questions. A ball is

Lena [83]

Answer:

When the ball is held motionless above the floor, the ball possesses only GPE energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.

Explanation:

If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.

If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.

$K = \frac{1}{2}mv^2\\U = mgh$

3 0

2 years ago

The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a

max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

$q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C$