If a rock has a weight of 30 N on Earth, would its weight be more or less if it was on Jupiter (gravity on Jupiter = 25 m/s2)?

Why is it generally easy to filter out particles in suspension?

sattari [20]

Answer: A suspension is a heterogeneous mixture in which some of the particles settle out of the mixture upon standing. The particles in a suspension are far larger than those of a solution, so gravity is able to pull them down out of the dispersion medium (water).

Explanation:

7 0

3 years ago

Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.

lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, $q=3\times 10^{-9}\ C$

Distance of the given point from the point charge is, $d=3\ cm=0.03\ m$

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

$E=\frac{kq}{d^2}$

Plug in $k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m$ . Solve for 'E'.

$E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C$

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0

3 years ago

Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch

Natasha2012 [34]

There are no choices on the list you provided that make such a statement,
and it's difficult to understand what is meant by "the following".

That statement is one way to describe the approach to 'forces of gravity'
taken by the theory of Relativity.

4 0

3 years ago

four equal charges of +3microcolumb are placed at the four corners of a square that is 40cm on a side. Find the force on any one

Reil [10]

Answer:

i can see your question

pls fast

6 0

3 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

Readme [11.4K]

Answer:

(a)0.0675 J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0

3 years ago