Solving Kinetic Energy: Mass, Velocity, and Jogger's Speed

AI Thread Summary
To solve for the mass of the continent, the volume is calculated as 4200 km on each side and 32 km deep, leading to a mass of approximately 3.58 x 10^14 kg using the given density of 2680 kg/m^3. The kinetic energy (KE) of the continent is then computed using the formula KE = 0.5 * m * V^2, resulting in a value of about 8.6 x 10^12 J. For the jogger, with a mass of 75 kg and the same kinetic energy as the continent, the velocity is determined to be approximately 6.78 m/s. The discussion emphasizes the importance of understanding the relationship between mass, velocity, and kinetic energy in solving these physics problems. Overall, the calculations illustrate the application of fundamental physics equations to real-world scenarios.
anglum
Messages
275
Reaction score
0
kinetic energy?

a continent is 4200 km on a side and 32 km deep

its mass density is 2680 kg/m^3
it is moving at an avg speed of 4.8cm/year

what is its mass? answer in units of J

what is its kinetic energy? answer in units of J

a jogger with mass 75kg has the same kinetic energy as the continent... what is his velocity?? answer in units of m/s


i know the KE equation is KE = .5 * m * V^2

i am stuck on what my steps are with this problem... i know if i solve the first part the 2nd and 3rd parts will be easy becuz i plug them into the KE equation
 
Physics news on Phys.org
to find the mass of the contintent i would take

Density = mass/volume

2680kg/m^3 = mass / (42000000m * 4200000m * 32000m)

and then solve for mass and that gives me the mass in kg?
 
yes...exactly...then plug in for KE equation
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Kinetic Energy of a Cylinder Rolling Without Slipping
Replies
21
Views
4K
Work done during a collision -- Change in Kinetic Energy & change in Momentum
Replies
1
Views
2K
Resistance force changing the velocity of an object
Replies
8
Views
1K
Internal energy of a gas and kinetic energy, "typical velocity"
Replies
5
Views
760
Conservation of Energy with Gravitational Potential Energy
Replies
6
Views
2K
Finding kinetic energy and initial velocity of a cart over time
Replies
6
Views
2K
Kinetic energy in center of mass reference frame
Replies
3
Views
1K
Conservation of Momentum Problem - Mechanical and Kinetic Energies
Replies
2
Views
903
Confused about a conservation of energy problem
Replies
4
Views
2K
Kinetic Energy of Colliding Protons
2
Replies
54
Views
10K

Hot Threads

Recent Insights

Back
Top