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uysha [10]
3 years ago
15

By what order of magnitude is something that runs in nanoseconds faster than something that runs in milliseconds?

Physics
1 answer:
maksim [4K]3 years ago
5 0

To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.

A nanosecond is one billionth of a second while and a millisecond is one millionth of a second

\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6

Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds

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Newton’s laws do not apply to small objects?
Soloha48 [4]

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

7 0
3 years ago
How do you change current in a circuit without changing the voltage?
Sunny_sXe [5.5K]

Decrease the reactance is the correct answer i believe

Explanation:

4 0
3 years ago
Which defintion of heat is most scientifically accurate
lukranit [14]
Hi there

Definition of heat

the quality of being hot; high temperature.
6 0
3 years ago
Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you ex
ollegr [7]

Answer:

angular resolution = 0.07270° = 1.269 × 10^{-3} rad

greatest distance from the camera = 118.20 m = 0.118 km

Explanation:

given data

diameter = 0.50 mm = 0.5 × 10^{-3} m

distance apart = 15 cm =  15× 10^{-2} m

wavelength λ = 520 nm = 520 × 10^{-9} m

to find out

angular resolution and greatest distance from the camera

solution

first we expression here angular resolution that is

sin θ = \frac{1.22* \lambda }{D}   .......................1

put here value λ is wavelength and d is diameter

we get

sin θ = \frac{1.22*520*10^{-9}}{0.5*10^{-3}}

θ = 0.07270° = 1.269 × 10^{-3} rad

and

distance from camera is calculate here as

θ = \frac{I}{r}    .................2

I = \frac{15*10^{-2}}{1.269*10^{-3}}

I = 118.20 m = 0.118 km

3 0
3 years ago
Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall
alexandr1967 [171]

Answer:

Explanation:

Given that

Force constant k=8.6N/m

Weight =64g=64/1000=0.064kg

Extension is 45mm=45/1000= 0.045m

It will have it highest spend when the Potential energy is zero

Therefore energy in spring =change in kinetic energy

Ux=∆K.e

½ke² = ½mVf² — ½mVi²

Initial velocity is 0, Vi=0m/s

½ke² = ½mVf²

½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²

0.0087075 = 0.032 Vf²

Then, Vf² = 0.0087075/0.032

Vf² = 0.2721

Vf=√0.2721

Vf= 0.522m/s

The time it will have this maximum velocity?

Using equation of motion

Vf= Vi + gr

0.522= 0+9.81t

t=0.522/9.81

t= 0.0532sec

t= 53.2 milliseconds

5 0
3 years ago
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