Difference between acceleration and decceleration short notes

Physics

2 answers:

Vadim26 [7]3 years ago

8 0

Acceleration is spiking up and decceleration is spiking down

Ghella [55]3 years ago

3 0

Answer:

Acceleration is considered to describe an increase or positive change of speed or velocity. On the other hand, deceleration is considered to describe a decrease or negative change of speed or velocity.

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The plates of a spherical capacitor have radii 6.25 cm and 15.0 crn. The space between the two spheres is filled with a material

aleksklad [387]

Answer:

Capacitance is 0.572×10⁻¹⁰ Farad

Explanation:

Radius = R₁ = 6.25 cm = 6.25×10⁻² m

Radius = R₂ = 15 cm = 15×10⁻² m

Dielectric constant = k = 4.8

Electric constant = ε₀ = 8.854×10⁻¹² F/m

ε/ε₀=k

ε=kε₀

$Capacitance\ (C)=\frac{4\pi k\epsilon_0 R_1\times R_2}{R_2-R_1}\\\Rightarrow C=\frac{4\pi 4.8\times 8.854\times 10^{-12}\times 15\times 10^{-2}\times 6.25\times 10^{-2}}{15\times 10^{-2}-6.25\times 10^{-2}}\\\Rightarrow C=0.572\times 10^{-10}\ Farad$

∴ Capacitance is 0.572×10⁻¹⁰ Farad

3 0

3 years ago

If blue light hits a red filter, what kind of light comes through the filter?

kotegsom [21]

It can be either C or B

Reasons it can be C: Red and Blue together(if I'm correct in art) is the combined color of two of the 3 primary colors to get a purple/violet color and if said filter is see through or just too dense for the light to even penetrate the said filter(in theory) but all in all purple is the answer with the two primary colors blue and red.

But also, it depends on what kind of filter it is, if the filter is like a screen filter then it will just come out in blue with the slightly different colors of again purple but in a darker tone then the actual eye can see.

Or it can be just C again cause the filter can be a film but that's a bit too far and to complex for right now so I believe it is B

8 0

3 years ago

In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top pla

lianna [129]

Answer: See below

Explanation:

<u>Given:</u>

The potential between plates, V = 240 V

Distance between plates, d = 0.02 m

The mass of drop, m = 2x10^-11

Charge on electron, e = 1.6x10^-19

Part (a)

The free-body diagram is attached below

Part (b)

The electric field is given by,

$E=\frac{V}{d}$

On applying force balance, the force on oil drop is equal to the weight of the oil,

$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$

Substituting the given values in the above equation,

$\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}$