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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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Answer:

<em>Total momentum is conserved</em>

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<u>Conservation of Momentum </u>

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p=m_1v_1+m_2v_2=(10)(10)+(10)(-50)=50\ kg.m/s

When both bears stick together, the total mass is 20 kg, and the new momentum is

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
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Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

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α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

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For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

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t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

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