Question

Notify Me Bookmark

A proton initially has v→ = (−2.00)i^ + (5.40)j^ + (7.80)k^ and then 4.30 s later has v→ = (−17.0)i^ + (5.40)j^ + (11.0)k^ (in meters per second). (a) For that 4.30 s, what is the proton's average acceleration a→ avg in unit vector notation, (b) in magnitude, and (c) the angle between a→ avg and the positive direction of the x axis? (a) Number i^ k^ Units (b) Number Units (c) Number Units

A proton initially has v→ = (−2.00)i^ + (5.40)j^ + (7.80)k^ and then 4.30 s later has v→ = (−17.0)i^ + (5.40)j^ + (11.0)k^ (in meters per second). (a) For that 4.30 s, what is the proton's average acceleration a→ avg in unit vector notation, (b) in magnitude, and (c) the angle between a→ avg and the positive direction of the x axis? (a) Number i^ k^ Units (b) Number Units (c) Number Units

Image text
A proton initially has v → = ( − 2.00 ) i ^ + ( 5.40 ) j ^ + ( 7.80 ) k ^ and then 4.30 s later has v → = ( − 17.0 ) i ^ + ( 5.40 ) j ^ + ( 11.0 ) k ^ (in meters per second). (a) For that 4.30 s , what is the proton's average acceleration a → avg in unit vector notation, (b) in magnitude, and (c) the angle between a → avg and the positive direction of the x axis? (a) Number i i ^ i ◻ k ^ Units ◻ (b) Number i ◻ Units ◻ (c) Number i ◻ Units ◻

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

The acceleration of a particle moving only on a horizontal xy plane is given by a→ = 5ti^ + 6tj^, where a→ is in meters per second-squared and t is in seconds. At t = 0, the position vector r→ = (27.0 m)i^ + (36.0 m)j^ locates the particle, which then has the velocity vector v→ = (7.40 m/s)i^ + (2.20 m/s)j^. At t = 2.80 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis? (a) Number i^ + j^ Units (b) Number Units
Solution
0 0

The acceleration of a particle moving only on a horizontal xy plane is given by a→ = 6ti^ + 7tj^, where a→ is in meters per second-squared and t is in seconds. At t = 0, the position vector r→ = (27.0 m)i^ + (39.0 m)j^ locates the particle, which then has the velocity vector v→ = (8.00 m/s)i^ + (1.90 m/s)j^. At t = 2.60 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis? (a) Number i^ + j^ Units (b) Number Units
Solution
0 0

Tarzan, who weighs 870 N, swings from a cliff at the end of a 22.2 m vine that hangs from a high tree limb and initially makes an angle of 25.2∘ with the vertical. Assume that an x-axis points horizontally away from the cliff edge and a y-axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is 787 N. Just then, what are (a) the force from the vine on Tarzan in unit-vector notation, and (b) the net force acting on Tarzan in unit-vector notation? What are (c) the magnitude and (d) the direction (measured counterclockwise from the positive x-axis) of the net force acting on Tarzan? What are (e) the magnitude and (f) the direction of Tarzan's acceleration just then? (a) Number i^ + j^ Units (b) Number i^ + j^ Units (c) Number Units (d) Number Units (e) Number Units (f) Number Units
Solution
0 0