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4 years ago

Waves on a pond are an example of which kind of wave?

Physics

1 answer:

KATRIN_1 [288]4 years ago

6 0

Waves on a pond are an example of which kind of wave?

B. surface waves

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n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat

musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0

3 years ago

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t

viktelen [127]

Answer:

a) F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ), b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

F = $k \frac{q_2q_2}{r^2}$

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron r₁ = x

right side -electron r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

F_net = k q Q ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$ )

F _net = kqQ ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

let's substitute the values

F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given

x (m) F (N)

0.05 27.0 10-16

0.10 8.10 10-16

0.15 5.76 10-16

0.20 8.10 10-16

0.25 27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m