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lukranit [14]
2 years ago
15

A 3.0-kg mass moving in the positive x direction with a speed of 10 m/s collides with a 6.0-kg mass initially at rest. after the

collision, the speed of the 3.0-kg mass is 8.0 m/s, and its velocity vector makes an angle of 35° with the positive x axis. what is the magnitude of the velocity of the 6.0-kg mass after the collision?
Physics
1 answer:
OverLord2011 [107]2 years ago
6 0
Sorry i dont know the answer 
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One car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. During their per
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They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

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Alejandro has collected over 35 studies focusing on the impact of vitamin C on a person's cold duration. He hopes to combine the
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Imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sa
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A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo
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Read 2 more answers
A lake of water is at a temperature of 60∘F. The air temperature drops to 30∘F. Assume that Newton's law of cooling applies to t
inna [77]

Answer:

time = 2.7 days

Explanation:

Given:

Water temperature = 60°F

Air temperature drops = 30°F

1st day drop is = 10°F

Now according to Newtons law of cooling,

\frac{dT}{dt}=-k.(T-T_{a})

where dT is change in temperature

           dt is change in time

           k is coefficient of cooling

           T is temperature

           T_{a} is ambient temperature = 30°F

∴\frac{dT}{T-30}=-k.dt

Integrating we get

ln(T-30) = -k.T+C -----------------------------(1)

Now when t = 0, T = 60°F

        when t = 1, T = 60-10

                              = 50°F

∴ln(60-30) = -k.0+C

  ln 30  = C

Now putting the value of C in (1)

ln (T-30) = -k.T+ln30

ln ( T -30) - ln 30 = -kt

ln ( t-30) / (30) = -k.T

Now at t - 1

ln (50-30) / 30 = -k x 1

ln ( 20/30 ) = -k

ln 2/3 = -k

∴ln\left | \frac{(T-30)}{30}\right |= ln\left | \frac{2}{3} \right |\times t

Let T = 40°F

ln\left | \frac{(40-30)}{30}\right |= ln\left | \frac{2}{3} \right |\times t

ln\left | \frac{1}{3}\right |= ln\left | \frac{2}{3} \right |\times t

\frac{ln\left | \frac{1}{3} \right |}{ln\left | \frac{2}{3} \right |}= t

t = \frac{-1.0986}{-0.4054}

 = 2.7 days

4 0
2 years ago
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