1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
inn [45]
3 years ago
9

If a rock weighing 2,200 N is dropped from a height of 15 m, what is its kenetic energy just before it hits the ground

Physics
1 answer:
ycow [4]3 years ago
7 0

Kinetic energy of the rock just before it hits the ground=KE=33000 J

Explanation:

Weight= 2200N

mg=2200

m(9.8)=2200

m=224.5 kg

initial velocity=0

final velocity =V

using kinematic equation V²=Vi²+2gh

V²=0+2 (9.8)(15)

V=17.1 m/s

now kinetic energy= 1/2 mV²

KE= 1/2 (224.5)(17.1)²

KE=33000 J

Thus the kinetic energy of the rock just before it hits the ground=33000 J

You might be interested in
What is the frequency of this wave<br> 1<br> 2<br> 3<br> 4
viva [34]
The frequency of this wave is 3
3 0
3 years ago
Three resistors of resistances, R1=10Ω, R2=5Ω, R3=20Ω are connected in a circuit in such way that same amount of current flows t
Nezavi [6.7K]

Answer

Explanation:

As the three resistors are connected in series, the expression to be used for the  

calculation of RT equivalent resistance

is:  

RT = R1 + R2 + R3

We replace the data of the statement in the previous expression and it remains:  

5 10 15 RT + R1 + R2 + R3 + +

We perform the mathematical operations that lead us to the result we are looking for:  

RT - 30Ω

5 0
3 years ago
The higher the temperature of an object the
Allushta [10]

 higher temp = higher energy = higher frequency = shorter wavelength

4 0
3 years ago
Which of the following is not a physical change?
Alekssandra [29.7K]
C because of galvination is sized
5 0
4 years ago
Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri
Elanso [62]

First we can say that since there is no external force on this system so momentum is always conserved.

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}

0.48= 0.6*v_{1f} + 1.2*v_{2f}

0.8  = v_{1f} + 2v_{2f}

now by the condition of elastic collision

v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6

v_{2f} = 0.533 m/s

also from above equation we have

v_{1f} = -0.267 m/s

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

8 0
4 years ago
Other questions:
  • What 3 simple machines are in scissors
    6·1 answer
  • Andre is studying snails and how fast they travel. He observes three groups of snails containing four snails each. Here are the
    12·1 answer
  • TRUE OR FALSE
    12·1 answer
  • 2. How much work does gravity do in causing a 6 kg hammer to fall to the ground from a
    9·1 answer
  • The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radiu
    7·1 answer
  • A hollow Spherical conductor of radius 12 cm is
    11·1 answer
  • Please help!!!!!!!!!!!!!!!!
    7·1 answer
  • Ewxrdctfvgybhjnkml,;kjffff
    14·1 answer
  • Fill in the blank with the correct response.
    12·1 answer
  • A child pulls a toy wagon with a force of 25.0 N for a distance of 8.5M. How much work did the child do on the wagon?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!