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QveST [7]
3 years ago
9

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

he nearest star is 4.1×1016m 4.1 × 10 16 m away. Suppose you had a spacecraft that could accelerate at 1.5 g g for two thirds of a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.)How long would it take you to reach the nearest star to earth?
Physics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

It would take 8.22037 hrs away. Wouldn't it?

Explanation:

Because

    4.11016

    4.11016

            15

= 8.22037

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Which of the following is the correct relationship among the acceleration, velocity, and position?
Damm [24]

Answer:1

Explanation:

If

s=displacement

v=velocity of particle

a=acceleration of particle

acceleration can be written as rate of change of velocity

so a=\frac{\mathrm{d} v}{\mathrm{d} t}

multiply and divide by ds

a=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{ds}{ds}

a=\frac{\mathrm{d} v}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ads=vdv

option 1 is correct

5 0
3 years ago
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What is your hypothesis (or hypotheses) for this experiment?
mariarad [96]

Answer:

mr or ms please type ur question fully please

7 0
3 years ago
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A net force of 10.0 N causes an object to accelerate at 2.00m/s^2. What is the mass of the object?
madam [21]

5.00kg

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4 0
3 years ago
Which of the following is a field (action-at-a-distance) force?
Kobotan [32]

Answer:

F_{grav} is a field force, as gravity does not physically apply force and does not require proximity.

Explanation:

3 0
3 years ago
A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.
miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0
1 year ago
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