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liubo4ka [24]
3 years ago
14

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? assume that a rocket is fired from rest at a

space station in deep space, where gravity is negligible. assume that c = 3.00 108 m/s.
Physics
1 answer:
steposvetlana [31]3 years ago
3 0

 

I’ve answered this problem before and there were 2 parts in this problem.

The solution would be like this for this specific problem:

<span>A.    </span><span>Vf = Vi + Vex*ln(Mi / Mf) </span><span>
<span>0.002 * 3e8m/s = 0 + 2000m/s * ln(Mi / Mf) </span>
<span>300 = ln(Mi / Mf) </span>
<span>1.9e130 = Mi / Mf </span></span>

 

<span>B.    </span><span>4000m/s = 2000m/s * ln(Mi / Mf) </span><span>
<span>2 = ln(Mi / Mf) </span>
<span>7.389 = Mi / Mf </span>
<span>Mf = Mi / 7.389 = 0.135*Mi<span> </span></span></span>

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The sun warms equatorial regions of Earth ______ than the poles. This cause an area of ________ pressure to develop near the equ
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Answer:

more

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Explanation:

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3 years ago
Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®
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C. Kevlar

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An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther
jeka94

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = \frac{m}{\frac{I_o}{r}  +mr} \ r v_o

let's calculate

       v = \frac{0.020}{\frac{1.4}{0.6  } + 0.020 \ 0.6  } \ 0.6 \ 4

       v = \frac{0.020}{2.345} \ 2.4

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

6 0
3 years ago
Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

brainly.com/question/14033058

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