At which point will the riders experience centripetal acceleration?

Adding heat to a liquid causes which of the following physical changes?

SashulF [63]

In almost every case in nature, adding heat to a liquid
causes the density of the liquid to decrease. That is,
when the liquid gets warmer, it expands and occupies
more space.

The one big exception to this rule is water !

Starting with a block of ice at zero°C (32°F), as the ice melts,
becomes water at zero°C, and all the way to 4°C (about 39°F),
its density increases all the way. That is, it shrinks and occupies
less volume as it goes from ice at zero°C to water at 4°C.

This sounds like an interesting but insignificant quirk ... until
you realize that if water didn't do this, then life on Earth would
be impossible !

8 0

3 years ago

The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started

bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

$W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J$

The change in electrostatic potential energy $\Delta U$ , of one point charge q is defined as the product of the charge and the potential difference.

$\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V$

5 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .