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Black_prince [1.1K]
3 years ago
14

A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown?

Physics
2 answers:
marusya05 [52]3 years ago
7 0
KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s
anyanavicka [17]3 years ago
5 0

Answer:

Kinetic energy = 225 J

Explanation:

As we know that kinetic energy is given as

KE = \frac{1}{2}mv^2

now here we know that

m = 2 kg

v = 15 m/s

now from above formula we know

KE = \frac{1}{2}(2)(15)^2

now we have

KE = 225 J

so kinetic energy of the moving ball is 225 J

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anyanavicka [17]

At point E

  • the kinetic energy of the rollercoaster is small compared to the potential energy
  • the potential energy is greater than the kinetic energy
  • the total energy is a mixture of potential and kinetic energy

<h3>What is the energy of the roller coaster at point E?</h3>

The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.

Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.

At point E

  • the kinetic energy of the rollercoaster is small compared to the potential energy
  • the potential energy is greater than the kinetic energy
  • the total energy is a mixture of potential and kinetic energy

In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,

Learn more about potential and kinetic energy at: brainly.com/question/18963960

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5 0
2 years ago
FOR 50 POINTS! -- Your car is parked at the high school. After school you take off and travel for 27 km towards Iowa City. At th
Dovator [93]

Answer:

36km

Explanation:

Im pretty sure displacment is the start and finish in a straight line

6 0
3 years ago
Read 2 more answers
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after
Delicious77 [7]
\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D
5 0
3 years ago
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Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl
Trava [24]

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac =  q (  Va -  Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]  

= Wac - Wab

= Wac + 3Wac

= 4Wac

4 0
3 years ago
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