All categories

3 years ago

One method used to determine the volume of solids and liquids is displacement.

Physics

2 answers:

ivann1987 [24]3 years ago

5 0

Answer:

true

Explanation:

My teacher went through the quiz w us

andreyandreev [35.5K]3 years ago

4 0

A. True. You can use displacement to determine the volume of solids and liquids.

You might be interested in

the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed

iragen [17]

Answer:

I dont know bro

Explanation:

Ask an expert

8 0

3 years ago

Read 2 more answers

What is the kinetic energy of an object that has a mass of 50.0kg and a velocity of 18 m/s?

Rom4ik [11]

<h2>Answer </h2>

The kinetic energy is 8100 J.

<u>Explanation</u>

Mass is 50.0kg and velocity is 18 m/s, the kinetic energy is:

As we know the formula of kinetic energy which is K.E = ½ ( mv ^ 2 ),

mass = m = 50.0kg

velocity = v = 18 m / s,

by putting values in the formula,

K.E = ½ ( mv ^ 2 ),

K.E = ½ ( 50kg ) . ( 18 m / s ) ^ 2

K.E = ½ ( 50kg ) . ( 324 ),

=> K.E = 1/2 ( 16200 ),

=> K.E = 16200 / 2,

=> K.E = 8100J.

Hence, the kinetic energy ( K.E ) is 8100 joule ( J ).

7 0

3 years ago

Tool used to measure time

charle [14.2K]

Answer:

clock

Explanation:

personal experience

6 0

3 years ago

Read 2 more answers

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.