Question

At time t,  r→ = 8.60t2 î - (4.90t + 6.10t2) ĵ gives the position of a...

At time t,  r→ = 8.60t2 î - (4.90t + 6.10t2) ĵ gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( r→ is in meters and t is in seconds). (a) Find the torque acting on the particle relative to the origin at the moment 4.50 s (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

The answer for above problem is explained below.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
At time t,  r→ = 1.70t2 î - (7.60t + 7.80t2) ĵ gives the position of a...
At time t,  r→ = 1.70t2 î - (7.60t + 7.80t2) ĵ gives the position of a 3.0 kg particle relative to the origin of an xycoordinate system ( r→ is in meters and t is in seconds). (a) Find the torque acting on the particle relative to the origin at the moment 6.50 s (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?
At time t, the vector ~r = 4t2i − (2t + 6t 2)ˆj gives the position...
At time t, the vector ~r = 4t2i − (2t + 6t 2)ˆj gives the position of a 3 kg particle relative to the origin of an xy coordinate system (~r is in meters and t is in seconds). What is the torque (in Newton-meters) acting on the particle relative to the origin? ANSWER IS 48t, PLEASE EXPLAIN IN DETAIL HOW YOU GOT IT
Force F→=(5.50N⁢)î-(5.35N⁢)k̂ acts on a pebble with position vector r→=(8.84m⁢)ĵ-(5.23m⁢)k̂, relative to the origin. What is...
Force F→=(5.50N⁢)î-(5.35N⁢)k̂ acts on a pebble with position vector r→=(8.84m⁢)ĵ-(5.23m⁢)k̂, relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (6.30 m, 0, -7.07 m)?
A 3.65 kg particle with velocity v→=(8.97m/s)î-(4.43m/s)ĵ is at x = 7.38 m, y = 5.02...
A 3.65 kg particle with velocity v→=(8.97m/s)î-(4.43m/s)ĵ is at x = 7.38 m, y = 5.02 m. It is pulled by a 2.09 N force in the negative x direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?
A particle of mass 2.00 kg moves with position r(t) = x(t) i + y(t) j...
A particle of mass 2.00 kg moves with position r(t) = x(t) i + y(t) j where x(t) = 10t2 and y(t) = -3t + 2, with x and y in meters and t in seconds. (a) Find the momentum of the particle at time t = 1.00 s. (b) Find the angular momentum about the origin at time t = 3.00 s.
A) a particle is located at r=(4.0i+3.5j+6.0k) meters has a force F=(6.5j-3.2k) Newtons acting on it....
A) a particle is located at r=(4.0i+3.5j+6.0k) meters has a force F=(6.5j-3.2k) Newtons acting on it. Find the torque with reference to the origin. B) the origin of a coordinate system is at the center of a wheel which rotates in the xy plane about its axle which is the z axis. A force F=215N acts in the xy plane at a 33 degree angle to the x axis (33 degrees north of east) at the point x=28cm, y=33.5cm. Determine...
A 4.71 kg mass moving in space according to v= (6.00t2 - t)i + (15.0t2)j +(t3...
A 4.71 kg mass moving in space according to v= (6.00t2 - t)i + (15.0t2)j +(t3 + 3.14t)k (relative to the origin), with v in meter/second and t in seconds. At t= 1.57s (a) what are the magnitude and direction of the force acting on the mass (b) what is the angle between the acceleration and velocity vector (c) what is the average velocity? (d) What is the mass angular momentum relative to the origin? (e) What is the torque...
A particle of mass m=0.2kg moves in the xy plane subject to a force such as...
A particle of mass m=0.2kg moves in the xy plane subject to a force such as that its position as a function of time is given by the vector r(t)= (3.0m/s2)t*2i+[12.0m-(2.0m/s*3)t*3]j what is the magnitude of the torque on the particle about the origin at the moment when the particle reaches the x axis?
A 3.00-kg particle starts from the origin at time zero. Its velocity as a function of...
A 3.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v = (3t^2) i+ (2t) j where v is in meters per second and t is in seconds. (a) Find its position at t = 1s. (b) What is its acceleration at t = 1s ? (c) What is the net force exerted on the particle at t = 1s ?   (d) What is the net torque about the origin...
The position of a particle for t > 0 is given by →r (t) = (3.0t...
The position of a particle for t > 0 is given by →r (t) = (3.0t 2 i ^ − 7.0t 3 j ^ − 5.0t −2 k ^ ) m. (a) What is the velocity as a function of time? (b) What is the acceleration as a function of time? (c) What is the particle’s velocity at t = 2.0 s? (d) What is its speed at t = 1.0 s and t = 3.0 s? (e) What is...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT