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3 years ago

The total resistance of resistors in parallel is always _________ the sum of the resistors.

Physics

2 answers:

docker41 [41]3 years ago

7 0

A. less than

In fact, the resistance will be less than that of the lowest resistance resistor.

pashok25 [27]3 years ago

7 0

A.) The total resistance of resistors in parallel is always "Less than" <span>the sum of the resistors.

Hope this helps!</span>

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The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

Time = $\frac{\textup{0.016}}{\textup{23}}$

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

q = 9.98 × 10⁻⁹ C

4 0

3 years ago

A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

$V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm$

So, the thickness of the film is equal to 3.1 cm.

8 0

3 years ago

Why the effect of gravitational force is more in liquid than in solid?

LekaFEV [45]

Gravitational force depends only on mass and distance, not on the state of matter.
The forces of attraction between molecules in matter are electromagnetic in nature, not gravitational.
These attractive forces are stronger in a solid than in a liquid than in a gas.
Gravitational forces between molecules is completely negligible compared to the em forces.

So, key answer is inter-molecular forces of solids is stronger than liquids.

4 0

3 years ago

Read 2 more answers

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively: