Can you help me please the question says

iVinArrow [24]

Frequency is no. of times this wave vibrates or complete its one cycle per unit time. As it vibrates 30000 times per second its frequency is 30000 per second or 30000 Hz [1 Hz = once per second].

6 0

3 years ago

You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el

nordsb [41]

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

$F_{g} = F_{c} (1)$

where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:

$F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)$

Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:

$F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)$

since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

$G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)$

Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

$Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)$

Since both charges are the same, the charge on each sphere is just the square root of (5):
Q = 2.066* 10⁻¹³ C.

2)

Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
$n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)$

4 0

3 years ago

Calculate the acceleration of a galloping horse going from 2 m/s to 12 m/s in 2 seconds.

olchik [2.2K]

A)
5m/s^2

(12m/s-2m/s)
__________ = 5m/s^2
2s

8 0

4 years ago

A hockey player swings her hockey stick and strikes a puck. According to Newtons 3rd law of motion which of the following is a r

Korvikt [17]

<u>Complete Question:</u>

A hockey player swings her hockey stick and strikes a puck. According to Newton’s third law of motion, which of the following is a reaction to the stick pushing on the puck?

A. the puck pushing on the stick .

B. the stick pushing on the player .

C. the player pushing on the stick .

D. the puck pushing on the player.

<u>Correct Option:</u>

According to Newton’s third law of motion the puck pushing on the stick is a reaction to the stick pushing on the puck.

<u>Option: A</u>

<u>Explanation:</u>

As when the hockey exert force on the puck (which is a flat ball basically used in ice hockey) then this action by hockey will receive equal and opposite reaction by puck. Thus when the stick is pushing on the this flat ball, then puck also push the stick. This is understood by newton's third law pf motion, where action and reaction forces are subject of discussion, displaying their is pair of forces applied among the interacting objects. This form is observed more practically in life and very frequent.

3 0

4 years ago

A truck is traveling 20m/s accelerates 3 m/s^2 for 4 seconds how far did it travel while it was accelerating. Using guess method

stellarik [79]