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ss7ja [257]
3 years ago
12

Where does each one go ??

Physics
1 answer:
spayn [35]3 years ago
6 0

Sublimation, Evaporation, and Melting are all in the section of 'Thermal Energy Added'.


Freezing, Condensation, and Deposition are all in the section of 'Thermal Energy Removed'.


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A ladybug has a mass of 0.01 g and a velocity of 0.6 m/s. What is his momentum?
Alex

0.000006‬kgm/s

Explanation:

Given parameters:

Mass = 0.01g

Velocity = 0.6m/s

Unknown:

Momentum = ?

Solution:

The momentum of a body is the product of its mass and velocity. It is subject to the impact a body in motion will have .

  Momentum = mass x velocity

The unit is in kgm/s

 We need to convert the mass to kilograms:

       1000g = 1kg

      0.01g = x kg ;  0.00001‬kg

Input the parameter:  

     Momentum = 0.00001 x 0.6 = 0.000006‬kgm/s

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Momentum brainly.com/question/2990238

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How would the earth move if the sun suddenly disappeared
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Just a guess without asking Google, that without the sun's gravity holding us in orbit, we would end up drifting whatever direction we were headed as it disappeared?
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In this lab, you will use a magnet and a simple circuit to examine the concepts of electricity and magnetism. Think about how th
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Sample Response: How can magnetic and electric fields be demonstrated?


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Read 2 more answers
A(n) ____ like rubber or plastic has high electrical resistance.
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insulator

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2 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
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