a dogs tail hit a 1.3kg flower vase, knocking it over. the net force on the flower vase over time is shown below

what is the bases speed at t=20 ms?
round answer to two significant figures

note that time is in ms.

Answer:

Its a wave

Explanation:

THis is one

Answer: a light wave

Explanation: a pex

Answer:

I'm not sure but, this video's pretty cool: gestyy.com/eyRkH2

Explanation:

pretty scary gameplay

Answer:

After five monthly payments, Quiana’s loan amount changes by $1,237.15.

Explanation:

Answer:

convection

Explanation:

I hope this helps

Answer:

If a coil of wire is placed in a changing magnetic field, a current will be induced in the wire. This current flows because something is producing an electric field that forces the charges around the wire. (It cannot be the magnetic force since the charges are not initially moving). ... that determines the induced current.

Answer:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.

~Hoped this helped~

~Brainiliest?~

Answer:

We describe motion in terms of velocity and acceleration. Velocity: The rate of change of displacement of an object (displacement over elapsed time) is velocity. Velocity is a vector since it has both magnitude (called speed) and direction. ... Acceleration: The rate of change of velocity is acceleration.

Explanation:

Displacement is a vector which points from the initial position of an object to its final position. ... Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity.

Displacement is the vector difference between the ending and starting positions of an object. Velocity is the rate at which displacement changes with time. ... The average velocity over some interval is the total displacement during that interval, divided by the time.

Hope this helps      :)

Answer:

The third charge needs to be placed at [tex]x \approx 0.57\; \rm m[/tex].

Explanation:

Both [tex]q_1[/tex] and [tex]q_2[/tex] would attract [tex]q_3[/tex].

These two electrostatic attractions need to balance one another. Hence, they need to be opposite to one another. Therefore, [tex]q_1[/tex] and [tex]q_2[/tex] need to be on opposite sides of [tex]q_3[/tex]. That is possible only if [tex]q_3 \![/tex] is on the line segment between [tex]q_1 \![/tex] and [tex]q_2 \![/tex].

Assume that [tex]q_3[/tex] is at [tex]x\; \rm m[/tex], where [tex]0 < x < 3.72[/tex] (in other words, [tex]q_3 \![/tex] is on the line segment between [tex]q_1[/tex] and [tex]q_2[/tex], and is [tex]x\; \rm m \![/tex] away from [tex]q_2 \![/tex].)

Let [tex]k[/tex] denote Coulomb's constant.

The magnitude of the electrostatic attraction between [tex]q_1[/tex] and [tex]q_3[/tex] would be:

[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}}[/tex].

Similarly, the magnitude of the electrostatic attraction between [tex]q_2[/tex] and [tex]q_3[/tex] would be:

[tex]\displaystyle \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].

The magnitudes of these two electrostatic attractions need to be equal to one another for the resultant electrostatic force on [tex]q_3[/tex] to be [tex]0[/tex]. Equate these two expressions and solve for [tex]x[/tex]:

[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}} = \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].

[tex]\displaystyle \frac{q_1}{(3.72 - x)^{2}} = \frac{q_2}{x^{2}}[/tex].

[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1}[/tex].

[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1} = \frac{1}{30}[/tex].

By the assumption that [tex](0 < x < 3.72)[/tex], it should be true that [tex](x > 0)[/tex] and [tex](3.72 - x > 0)[/tex]. Therefore, [tex]\displaystyle \frac{x}{(3.72 - x)} > 0[/tex].

Take the square root of both sides of the equation [tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{1}{30}[/tex].

[tex]\displaystyle \sqrt{\frac{x^2}{(3.72 - x)^{2}}} = \sqrt{\frac{1}{30}}[/tex].

[tex]\displaystyle \frac{x}{3.72 - x} = \frac{1}{\sqrt{30}}[/tex].

[tex]\sqrt{30}\, x = 3.72 - x[/tex].

Therefore:

[tex]\left(1 + \sqrt{30}\right)\, x = 3.72[/tex].

[tex]\displaystyle x = \frac{3.72}{1 + \sqrt{30}} \approx 0.57[/tex].

Hence, [tex]q_3[/tex] should be placed at [tex]x \approx 0.57\; \rm m[/tex].