The pressure in a container will stay the same if you —

An object of mass 82kg is accelerated upward at 3.2m/s/s. what force is required

vampirchik [111]

F = m · a

In order to accelerate 82 kg upward at the rate of 3.2 m/s², a NET upward force of (82kg · 3.2m/s²) = 262.4 Newtons is required.

But if the object is on or near the surface of the Earth, then there's a downward force of (82kg · 9.8m/s²) = 803.6 N already acting on it because of gravity.

So you need to apply (803.6N + 262.4N) = <em>1,066 Newtons UPward</em>, in order to cancel its own weight and accelerate it upward at that rate.

6 0

3 years ago

How many feet does a No. 2 pencil last?

Ray Of Light [21]

Answer:45,000

Explanation:

:TIL the average pencil holds enough graphite to draw a line about 35 miles long or to write roughly 45,000 words

6 0

2 years ago

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

Kamila [148]

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

${v}^{2} - {u}^{2} = 2gh$

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

${30}^{2} - {0}^{2} = 2 \times 10 \times h$

$= > 20h = 900$

$= > h = \frac{900}{20} = 45$

5 0

3 years ago

What is intermolecular forces and how is it related to phases of matter

MakcuM [25]

Intermolecular forces are forces that keep molecules together. For example, the forces between two water molecules. The stronger the intermolecular forces are, the more "solid" is the matter going to be, meaning that the intermolecular forces are the strongest in solids and weakest in gases.
Make sure not to confuse intERmolecular forces (forces between *molecules*) and intRAmolecular forces (forces between *atoms* that make up a molecule).

6 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

3 years ago

The pressure in a container will stay the same if you ——?