A bullet with a mass m, = 11.5 g is fired into a block of wood at velocity v, = 245 m/s. The block is attached to a spring that has a spring constant k of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass mw of the wooden block. т, тy — 304.6 kg Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time, when they're not sleeping or eating, joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as 0.0137 kg, 0.0245 kg, 0.0155 kg, and 0.0213 kg. At this very instant, the x- and y-components of the mice's velocities are, respectively, (0.769 m/s, -0.393 m/s), (-0.919 m/s, -0.889 m/s), (0.797 m/s, 0.219 m/s), and (-0.793 m/s, 0.317 m/s). Calculate the x- and y-components of Bryce's mice's total momentum, p, and py. kg-m/s kg-m/s Py =

The kinetic energy (T) of an object with mass m traveling at a speed v is defined as T = \frac{1}{2}mv^2T=21​mv2.  What is the kinetic energy (in J) of an object of mass 41 g traveling a velocity of 37 miles per hour?  (1 mile = 1.609 km)  Round your answer to the tenths (0.1) place.

Find the kinetic energy for two masses, m1=(64±3)Kg and m2=(136±4)Kg, traveling together at a velocity v=(5.0±1.5)m/s. Hint: the equation for kinetic energy is E = 1/2mv^2, where m is the total mass (m1+m2).

While doing an experiment a student uses a triple beam balance to measure the mass of a cart and gets 273.9 g. While calculating the mass they get the values: 268.5 g. 288.7 g, and 270 g. What would the GAV for the cart be? 268.5 g 288.7 g 275.7 g 274.8 g 273.9 g

In an effort to exact the most severe capital punishment upon a rather unpopular prisoner, the execution team at the Dark Ages Penitentiary search for a bullet that is ten times as massive as the rifle itself. What type of individual would want to fire a rifle that holds a bullet that is ten times more massive than the rifle? Explain.

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time, when they're not sleeping or eating, joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as 0.02450.0245 kg, 0.01470.0147 kg, 0.02390.0239 kg, and 0.01230.0123 kg. At this very instant, the x‑ and y‑components of the mice's velocities are, respectively, (0.515 m/s,−0.719 m/s), (−0.857 m/s,−0.215 m/s), (0.395 m/s,0.963 m/s), and (−0.743 m/s,0.759 m/s). Calculate the x‑ and y‑components of Bryce's mice's total momentum, px and py.

n/course.html?courseld=17544544&OpenVellumHMAC=a830981ff68f36618e84254ba22c5db5#10001 Aktiv Chemistry DZL Homepage - Vande... ▼ Lobby | Top Hat Part A Pe - Pb Submit Part B A cardinal (Richmondena cardinalis) of mass 4.20x10-2 kg and a baseball of mass 0.150 kg have the same kinetic energy. What is the ratio of the cardinal's magnitude pe of momentum to the magnitude p of the baseball's momentum? ► View Available Hint(s) Km Kw = V—| ΑΣΦ Submit ? Provide Feedback M Gmail A man weighing 750 N and a woman weighing 440 N have the same momentum. What is the ratio of the man's kinetic energy Km to that of the woman Kw? ► View Available Hint(s) 195| ΑΣΦ ? Handshake 4 VYES - Vanderbilt - S... Course Home P Pearson Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions | Contact Us | Other TACTION | VOISIUMIS Next > 545 DI

Calculate the mass in kilograms of a 1.06 kg object when its velocity, relative to us, is each of the following values. (Notice the progression of these numbers toward the velocity of light, 3.00 × 10° ms1.) velocity = 3.35 × 107 m s1 i kg m = velocity = 2.64 × 10° m s1 m = i kg velocity = 2.99 × 10° ms-1 i kg m =

Let's say we have a M1 and M2. Let's just say. Let's examine this hypothetical situation. The first mass would be 1.50 kg and the second mass would be 2.00 kg. These two masses would then be separated by a length or we should say a distance of this L = 2.50 m. Let's say we want to place a third mass (immaterial mass) in the middle of the two masses, such that there would be no net force. Find this specific place where this would occur, and to make it easy, find the distance from the first mass towards this third mass

D Gm₁m₂ Fg KE = mv², Ug = - 2πr , ac = =²₁, v = ²7₁ T Gm₁m₂ GM g = G, Vesc = 2GM R , E = KE + Ug, G = 6.674 x 10-¹1 Nm²/kg² Problem 1: You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its radius to be 9 x 106 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.2 x 10¹¹ m from its star with a period of 402 days (3.473 x 107 sec). Once on the surface you find that the free-fall acceleration is 12.2 m/sec². a) What is the mass of the planet? Answer: 1.5 x 1025 kg. b) What is the mass of the star? Answer: 5.2 x 1030 kg.