When the crests of 2 identical waves meet, what is the amplitude of the resulting wave?

slader A jet is circling an airport control tower at a distance of 15.9 km. An observer in the tower watches the jet cross in fr

liubo4ka [24]

Answer:

y = 138.96 m

Explanation:

The angle subtended by the moon is the mean of the angle of the arc between the two most extreme points of the moon, we can see that the angle is very small, so we can approximate this arc to a straight line and then use the trigonometric relationships

sin θ = y / L

where L = 15.9 10³ m and θ = 8.74 10⁻³ rad

y = L sin θ

y = 15.9 10³ sin (8.74 10⁻³)

y = 15.9 10³ 0.0087399

y = 138.96 m

4 0

3 years ago

4. Think back to Coulomb's Law. Two coins with identical charges are placed on a lab table 1.35 m apart.

FinnZ [79.3K]

A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:

F=K(q1xq2)/r²

F: The magnitud of the force between the charges. (F=2.0 N).
K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
q1 and q2: Electrical charges.
r: The distance between the charges (r=1.35 m).

We have the values of F, K and r, so we can calculate q1xq2, because both coins have identical charges:

q1xq2=(r²xF)/K
q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
q1xq2=3x10^-10 C
q1=q2=(3x10^-10 C)/2

Then, the charge of each coin, is:

q1=1.5x10^-10 C

q2=1.5x10^-10 C

B) Would the force be classified as a force of attraction or repulsion?

It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.

5 0

3 years ago

Read 2 more answers

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$