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nordsb [41]
3 years ago
11

intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr

avelled​
Physics
2 answers:
V125BC [204]3 years ago
4 0

Answer:

v= u+at

0=45×1000×60×60+3a

_3a=16200045

a=5400015m per sec

Usimov [2.4K]3 years ago
4 0

Answer:

Acceleration=12.5/3m/s^2=4.17m/s^2

Explanation:

Initial velocity u = 45 km/h = 12.5 m/s; Final velocity v = 0 m/s; time t = 3 s, acceleration a.

[a=v-u/t] : a=0-12.5/3=-12/3m/s^2

(Retaradation=-(-12.5/3)=12.5/3.

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated
sp2606 [1]

Answer:

611.8 nm

Explanation:

n_{oil} = Index of refraction of soap bubble  = 1.33

t = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

\lambda = wavelength of light = ?

m = order = 0

For reflection , the necessary condition is

2 n_{oil} t = (m + 0.5) \lambda

2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda

\lambda = 6.118\times 10^{-7}

\lambda = 611.8 nm

7 0
3 years ago
A 42-kg crate rests on a horizontal floor, and a 52-kg person is standing on the crate. (a) determine the magnitude of the norma
Lemur [1.5K]
Hope this helps you.

6 0
3 years ago
Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do
Kaylis [27]

Answer:

Explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work  is 25 % .

Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.

4000 Cals = 4.2 x 4000 = 16800 J  .

Work done in one jump = kinetic energy while jumping

= 1/2 m v²

= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

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4 0
3 years ago
ASAP pls answer right and I will mark brainiest
Ronch [10]
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3.D:All of the above
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7.A:Is bigger than Pluto

I hope this helps
6 0
3 years ago
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