4 years ago

Why is water known as the "Universal Solvent"?

1 answer:

4 0

<span>Water is known as the universal solvent because it is capable of dissolving a variety of substances, more than any other liquid.

Hope this helps!</span>

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Using the formula for calculating Pascal’s Principle, solve the following. Pressure(Pascal)=force(Newton)/area(meter2)

Verdich [7]

P= 5/30—-> P= .16667

Does that help?

8 0

4 years ago

Consider two copper wires. one has twice the length and twice the cross-sectional area of the other. how do the resistances of t

Anestetic [448]

Answer:

R = ρ L / A where R is the resistance and ρ the resistivity:

R2 / R1 = (L2 / A2) / (L1 / A1) = L2 * A1 / (L1 * A2)

R2 / R1 = (2 L * A) / (L * 2 A) = 1

Their resistances should be the same.

5 0

2 years ago

A ball is dropped from the rest from a high window of a tall building and called for 4 seconds. Neglecting air resistance, the f

diamong [38]

Answer:

-39.2m/s

Explanation:

Given that :

t = 4secs

g = -9.8m/s^2

v = ?

u = 0m/s ( since it was at rest )

V = u +at............. 1

Where v is the final velocity

a = -g = -9.8m/s^2 since the ball was dropped from a height which will eventually make it move against gravity

t = 4secs

Substitute the values into 1

v = 0 - 9.8×4

v = -39.2m/s

3 0

4 years ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

3 years ago

1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee

Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

7 0

4 years ago