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finlep [7]
3 years ago
14

A ball of clay falls freely to a concrete floor. The ball does not bounce noticeably, and it quickly comes to rest. What has hap

pened to the mechanical energy the ball had during its fall?
Physics
1 answer:
omeli [17]3 years ago
7 0

Answer:

The mechanical energy is converted to potential energy while the kinetic energy is zero

Explanation:

mechanical energy is the sum of potential energy and kinetic energy. It is the energy associated with the motion and position of an object. The total mechanical energy is the sum of these two forms of energy.

The Law of Conservation of Energy: Energy cannot be created or destroyed, but is merely changed from one form into another. This means that potential energy can become kinetic energy, or vice versa, but energy cannot “disappear”.

The mechanical energy is converted to potential energy while the kinetic energy is zero

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EleoNora [17]
The correct answer is the reverse wave I took the test
6 0
3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
4 years ago
Complete this sentence. If the mass of object A is __________ than object B, then the weight of object A will be _______________
AfilCa [17]

Answer:

Complete this sentence. If the mass of object A is greater than object B, then the weight of object A will be heavier the weight of object B.

Explanation:

If the mass of Object A is more/greater than object B then it would be "heavier"

4 0
3 years ago
Read 2 more answers
Viewed through a spectroscope, the spectral profile of a yellow street lamp has a narrow line in the yellow region of the visibl
muminat

Answer:

discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source

Explanation:

Bulbs can emit light in several ways:

* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.

* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.

Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type

6 0
3 years ago
A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa
Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф =  angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the  circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

5 0
3 years ago
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