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V125BC [204]
2 years ago
10

6. A lumberjack is standing on a log floating on a lake. She starts from rest, then runs along the log to the end, when she jump

s from the first log onto a second. After landing safely on the second log, she slows down and ends up standing on the second log. Both logs both have masses of 150 kg each and the mass of the lumberjack is 70 kg. The lumberjack reaches a speed of 7.0 m/s relative to the shore during her jump. What are the speeds of the two logs and of the lumberjack after she has stopped on the second log
Physics
1 answer:
scoray [572]2 years ago
4 0

Answer:

a) -3.267 m/s

b) 2.227 m/s

Explanation:

As per the conservation of momentum

m1v1 + m2v2=0

m1= mass of log

m2 = mass of lumber jack

v1 = velocity of log

v2 = velocity of lumber jack

a) Velocity of first log

-\frac{70*7}{150} = -3.267 m/s

b) m1v1 + m2v2 = m3v3

Velocity of log

= \frac{70*7}{150+70} \\2.227

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Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
2 years ago
One disadvantage of using proprietary licensed software is that
Vaselesa [24]

Answer:

its b

Explanation:

5 0
2 years ago
Read 2 more answers
High mass stars are more massive produce more energy burn brighter and have shorter life cycle than low mass stars
Zarrin [17]
Yes, they do all of those things.
5 0
3 years ago
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0
2 years ago
A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a
Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

v_i and v_f are the initial and final velocities.

With <em>m</em> = 0.175 kg,

E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}

The negative sign appears because energy is lost.

7 0
3 years ago
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