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Ivanshal [37]
2 years ago
13

Which times and distances are represented by the function? select three options. the starting distance, at 0 hours, is 300 miles

. at 2 hours, the traveler is 725 miles from home. at 2.5 hours, the traveler is still moving farther from home. at 3 hours, the distance is constant, at 875 miles. the total distance from home after 6 hours is 1,062.5 miles.
Physics
1 answer:
Naya [18.7K]2 years ago
4 0

The times and distances are represented by the function are

  • at 2 hours, the traveler is 725 miles from home.
  • at 3 hours, the distance is constant, at 875 miles
  • the total distance from home after 6 hours is 1,062.5 miles.

<h3>What is distance?</h3>

The distance is the length of path travelled by a body when moving with some speed and taking time t.

Given function is

D (t) = 300t +125 for 0≤t<2.5

D (t) = 875  for 2.5 ≤t≤3.5

D(t)= 75t +612.5 for 3.5<t≤6

According to the given function, for 2 hours, the traveler will travel from home,

D (2) = 300x2 +125 = 725 miles.

For time between 2.5 to 3.5 hr, the distance is constant i.e. 875 miles.

For total time 6hr, the distance travelled will be

D(6) = 75 x 6 +612.5 = 1062.5 miles.

Thus, the correct options are picked up.

Learn more about distance.

brainly.com/question/15172156

#SPJ1

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A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
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Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

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Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

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v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

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3 years ago
Which of the following intermolecular forces explains why iodine (I2) is a solid at room temperature?
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A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
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Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

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4 0
3 years ago
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
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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

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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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