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garik1379 [7]
2 years ago
13

Zoe is setting up a track for a toy car. The track has a ramp that is 32° above horizontal. If Zoe wants the car to travel as a

projectile for 1.0 seconds, how fast does the toy car need to be moving as it leaves the ramp?
4.9 m/s
1.0 m/s
9.2 m/s
7.4 m/s m
Physics
1 answer:
jolli1 [7]2 years ago
8 0

Answer:

Explanation:

Not enough information.

IF we ASSUME she wants the car to be at LAUNCH LEVEL after 1 second of flight.

THEN

The highest point will have zero vertical velocity and will have taken ½ second to get there. This means that the initial vertical velocity was

v = gt

vy₀ = 9.8(0.5)

vy₀ = 4.9 m/s

vsinθ = vy₀

v = vy₀/sinθ

v = 4.9/sin32

v = 9.2466...

v = 9.2 m/s

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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,

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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

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