College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- Please Asaparrow_forward51. A nucleus, initially at rest, decays radioactively. In the process, it emits an electron horizontally to the east, with momentum 10.0x 10-21 kg m/s and a neutrino horizontally to the south, with momentum 5.0 x 10-2¹ kgm/s. (a) In what direction does the residual nucleus move (draw diagram, find angle )? (b) What is the magnitude of its momentum? (c) If the mass of the residual nucleus is 4.0x 10-25 kg, what is its recoil velocity?arrow_forwardA 2.10 kg "particle" traveling with velocity = (3.7 m/s )i collides with a 4.70 kg "particle" traveling with velocity V = (1.7 m/s )j. The collision connects the two particles. What then is their velocity in (a) unit-vector notation and as (b) a magnitude and (c) a positive (counterclockwise) angle measured from the +x axis? (a) Number i+ į Units (b) Number Units (c) Number Unitsarrow_forward
- The figure below shows a bullet of mass 210 g traveling horizontally towards the east with speed 420 m/s, which strikes a block of mass 2.5 kg that is initially at rest on a frictionless table. After striking the block, the bullet is embedded in the block and the block and the bullet move together as one unit. (a) What are the magnitude (in m/s) and direction of the velocity of the block/bullet combination immediately after the impact? magnitude m/s direction ---Select--- (b) What are the magnitude (in N· s) and direction of the impulse by the block on the bullet? magnitude N.S direction ---Select--- (c) What are the magnitude (in N: s) and direction of the impulse from the bullet on the block? magnitude N.s direction ---Select-- (d) If it took 3 ms for the bullet to change the speed from 420 m/s to the final speed after impact, what is the average force (in N) between the block and the bullet during this time? (Enter the magnitude.)arrow_forwardFor the situation in the figure below, use momentum conservation to determine (a) the magnitude and (b) the direction of the final velocity of ball 1 after the collision. The angle = 56.0⁰. (a) Number (b) Number i 101 = 0.900 m/s m₁ = 0.150 kg i ¹02=0.540 m/s m₂ = 0.260 kg Units +v Units ly +V (a) UF1x (b) (a) Top view of two balls colliding on a horizontal surface. (b) This part of the drawing shows the x and y components of the velocity of ball 1 after the collision. 35.0* 2 = 0.700 m/s ·+xarrow_forwardA soccer ball with a mass of 0.427 kg approaches a player horizontally with a speed of 14.0 m/s. The player kicks the ball with her foot, which causes the ball to move in the opposite direction with a speed of 22.3 m/s. (a) What magnitude of impulse (in kg · m/s) is delivered to the ball by the player? kg · m/s (b)What is the direction of the impulse delivered to the ball by the player? - In the same direction as the ball's initial velocity - Opposite to the ball's initial velocity - Perpendicular to the ball's initial velocity - The magnitude is zero. (c) If the player's foot is in contact with the ball for 0.0600 s, what is the magnitude of the average force (in N) exerted on the player's foot by the ball?arrow_forward
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