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a_sh-v [17]
3 years ago
5

Overlearning is

Physics
1 answer:
scoundrel [369]3 years ago
7 0
A method for improving your memory…
You might be interested in
Can you think of a scenario when the kinetic and gravitational potential energy could both be zero ? Describe or draw how this c
Inga [223]

Both kinetic and gravitational potential energy can become zero at infinite distance from the Earth.

Consider an object  of mass <em>m </em>projected from the surface of the Earth with a velocity <em>v. </em>

The total energy of the body on the surface of the Earth is the sum of its kinetic energy \frac{1}{2} mv^2and gravitational potential energy -\frac{GMm}{R^2}.

here, <em>M</em> is the mass of the Earth, <em>R</em> is the radius of Earth and <em>G</em> is the universal gravitational constant.

The gravitational potential energy of the object is negative since it is in an attractive field, which is the gravitational field of the Earth.

The energy of the object on the surface of the earth is given by,

E_i=\frac{1}{2} mv^2-\frac{GMm}{R^2}

As the object rises upwards, it experiences deceleration due to the gravitational force of the Earth. Its velocity decreases and hence its kinetic energy decreases.

The decrease in kinetic energy is manifested as  an equal increase in potential energy. The potential energy becomes less and less negative as more and more kinetic energy is converted into potential energy.

At a height <em>h</em> from the surface of the Earth, the energy of the object is given by,

E_h=\frac{1}{2} mv_h^2-\frac{GMm}{(R+h)^2}

The velocity v_h is less than <em>v</em>.

When h =∞, the gravitational potential energy increases from a negative value to zero.

If the velocity of projection is adjusted in such a manner that the velocity decreases to zero at infinite distance from the earth, the object's kinetic energy also becomes equal to zero.

Thus, it is possible for both kinetic and potential energies to be zero at infinite distance from the Earth. In this case, kinetic energy decreases from a positive value to zero and the gravitational potential energy increases from  a negative value to zero.


7 0
3 years ago
The four living things seen here, dog, flower, algae, and bacteria, are all made of cells. A - C are multicellular, but D is uni
djverab [1.8K]
B) They are all made up of at least one cell
6 0
3 years ago
Read 2 more answers
A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at
docker41 [41]

Answer:

y = 4.36

Explanation:

Let the height of the ladder be L

L = 10

Also:

  • Let x = distance\ from\ the\ base\ of\ the\ ladder
  • Let y = height\ of\ the\ base\ of\ the\ ladder

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

L^2 = x^2 + y^2 --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

x = 9

Substitute 9 for x and 10 for L in L^2 = x^2 + y^2

10^2 = 9^2 + y^2

100 = 81 + y^2

Make y^2 the subject

y^2 = 100 - 81

y^2 = 19

Make y the subject

y = \sqrt{19

y = 4.36

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0
3 years ago
How are waves that are created by a musical instrument and waves that
Wittaler [7]

Answer:

both caused  by physical vibrations

7 0
3 years ago
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0
1 year ago
Read 2 more answers
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