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aleksklad [387]
2 years ago
12

What is wrong with the following diagram?

Physics
1 answer:
slega [8]2 years ago
5 0

What is wrong with the diagram is : ( 4 )

  • The spot with the "maximum kinetic energy" should be on the lowest point of the roller coaster at the far left side on the bottom.

<h3>Gravitational potential energy vs kinetic energy</h3>

The maximum gravitational potential energy is located at the highest peak which the roller coaster attains while as the roller coaster comes down from the peak the gravitational potential energy decreases while the kinetic energy increases.

The point at which the roller coaster should attain it maximum kinetic energy should be at the lowest point ( i.e far left side of the bottom ).

Hence we can conclude that error in the diagram is the spot with the "maximum kinetic energy" should be on the lowest point of the roller coaster at the far left side on the bottom.

Learn more about Kinetic energy: brainly.com/question/25959744

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the temperature of the water is less than the temperature of its surroundings. Explain in terms of the particles of the water, w
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Metal conducts heat quicker than wood or air so when you touch it heat is taking away from you body faster than when you touch the wood desk. When heat is taken from our bodies our brain interprets this as cold.
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3 years ago
Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be loca
Inga [223]

Answer:

1) an observer in B 'sees the two simultaneous events

2)observer B sees that the events are not simultaneous

3)  Δt = Δt₀ /√ (1 + v²/c²)

Explanation:

This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems

1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events

Consequently an observer in B 'sees the two simultaneous events

2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.

Consequently observer B sees that the events are not simultaneous

3) let's calculate the times for each event

        Δt = Δt₀ /√ (1 + v²/c²)

where t₀ is the time in the system S' which is at rest for the events

8 0
3 years ago
Is the Florida scrub a forest or grassland
PIT_PIT [208]
It is a forest. Hope this helps. Much luck
4 0
3 years ago
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If
katrin2010 [14]

A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If the disk starts from rest and is then brought in contact with a spinning rubber wheel, we observe that the disk gradually begins to rotate too. If after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 4.0 rad/s, find the average angular acceleration that the disk experiences. (Assume the positive direction is in the initial direction of rotation of the disk. Indicate the direction with the sign of your answer.)

Assume after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 11.0 rad/s.

Answer:

385 rad

Explanation:

The expression for the angular acceleration of a disk that is in contact with a spinning wheel can be given as:

\alpha = \frac{\delta \omega}{\delta t}

where \delta\omega = \omega_f - \omega_i

\alpha = \frac{ \omega_f-\omega_i}{\delta t}

\alpha = \frac{ 4.0 rad/s-0 rad/s}{35}

\alpha =0.14 rad/s^2

Angular displacement of a disk can be calculated by using the formula:

\theta = \omega t

substituting 11.0 rad/s for \omega and t = 35 s ; we have:

\theta = 11.0 rad/s * 35 s

\theta = 385 rad

4 0
3 years ago
Read 2 more answers
Please show work : A particle with mass 2.00 μg and a charge of – 200 nC has a velocity of 3000 m/s in the x-direction. There is
irga5000 [103]

Answer:

 x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

     m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg

     q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

     F = ma

     F = q E

     a = qE / m

     a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

     a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

     Vf² = Vo² -2 a x

     0 = Vo² - 2 a x

     x = Vo² / 2a

     x = 3000²/ 2100

     x =4.5 10⁴ m

This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

      F -W = 0

      F = w = mg

      E q = mg

      E = mg / q

      E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

      E = 9.8 10⁵ C

       

The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E

7 0
4 years ago
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