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3 years ago

A car has positive acceleration. What information can you infer from this?

Physics

1 answer:

mario62 [17]3 years ago

6 0

Answer: B: The car is speeding up in the same direction as it is traveling.

Explanation:

I took the test and got it right

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I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second

Degger [83]

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

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4 years ago

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I need help ASAP. This is for 15 points

Sladkaya [172]

Answer:

the Prime Meridian is the planets line of 0 degrees longitude sometimes called the Greenwich Meridian or the international Meridian

the great circle of Earth with the latitude of 0 degrees

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3 years ago

Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all

Alexandra [31]

Answer:

$\Delta h=0.0364\ m=3.64\ cm$

Explanation:

Given:

change in stiffness constant of the spring on replacing the original springs, $\delta k=5355\ N.m^{-1}$

mass of the car, $m=1455\ kg$

initial length of the original car-spring before compression, $l_i=12\ cm=0.12\ m$

final length of the original car-spring after compression, $l_f=8.55\ cm=0.0855\ m$

So, weight of the car:

$w=m.g$

$w=1455\times 9.81$

$w=14273.55\ N$

Now the spring constant of original spring:

$w=4k_o.(l_i-l_f)$ (since 4 springs are in parallel)

$14273.55=4k_o\times (0.12-0.0855)$

$k_o=103431.522\ N.m^{-1}$

So the stiffness constant of the new springs:

$k_n=k_o-\delta k$

$k_n=103431.522-5355$

$k_n=98076.522\ N.m^{-1}$

Now the height lowered:

$w=k_n.4\Delta h$ (since 4 springs are in parallel)

$14273.55=4\times98076.522\times \Delta h$

$\Delta h=0.0364\ m=3.64\ cm$

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4 years ago

White light (400–700 nm) is incident on a 600 line/mm diffraction grating. What is the width of the first-order rainbow on a scr

Tema [17]

Answer:

Δy=0.431m

Explanation:

Diffraction grating with split space d,to find the fringe position ym,we must to find the angle from

dSinα=mλ

A grating with N slits or lines per mm has slit spacing of

d=1/N

d=(1/600mm)

d=1.67×10⁻³mm

For 400nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{400*10^{-9} }{1.67*10^{-6}} )\\ \alpha =13.910^{o}$

And the position of first order lowest wavelength fringe on the screen is:

$y_{1}=Ltan\alpha_{1}\\y_{1}=2tan(13.910)\\ y_{1}=0.49445m$

For 700nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{700*10^{-9} }{1.67*10^{-6}} )\\ \alpha_{2} =24.83^{o}$

And the position of first order highest wavelength fringe on the screen is:

$y_{2}=Ltan\alpha_{2}\\y_{2}=2tan(24.83)\\ y_{2}=0.925595m$

The difference between the first order lowest and highest wavelength fringe is

Δy=(0.925595 - 0.49445)m

Δy=0.431m

6 0

3 years ago

What is the meaning of oxidation

MA_775_DIABLO [31]

Answer: the process or result of oxidizing or being oxidized.

Explanation:

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3 years ago

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