does the velocity change with distance downstream? 3. How does channel efficiency change with distance downstream? Figure 1: Bradshaw Model shows the changes of river from Upstream to Downstream According to the Bradshaw Model I would expect. Discharge increases downstream because tributaries join the main river and increase amount of water in Main River. Cross-Section will increase as the river goes downstream‚ channel depth and width of the river increases due to abrasion. Velocity will
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moving has momentum. Momentum is equal to the objects mass times its velocity. Momentum is conserved‚ which means that “momentum before an event equals momentum immediately after‚ or pi=pf”. Since pi=pf‚ then pai+ pbi = paf+ pbf and (ma* vai)+ (mb* vbi)= (ma* vaf) + (mb * vbf). Having velocity simply means that an object has a speed and direction. Using the formula “(ma * vai) + (mb * vbi) = (ma* vaf) + (mb * vbf‚)” the final velocity of two carts after they collide can be found. The first cart is
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introductory height of two meters would be far too great causing the egg to break. Therefore‚ lab participants needed to find a way to either elongate the duration of the impact‚ or find a way to slow down the egg’s normal final velocity when it strikes the ground. Velocity is described as‚ “the displacement divided by the time interval during which the displacement occurred” (Serway & Faughn‚ 2002‚ p. 43). Prior to the trials at the set heights‚ there seemed to be multiple structural designs
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collisions. This background knowledge is essential in understanding the experiment‚ resulting calculations‚ and analysis. The main objective of this experiment was to determine the initial velocity of a ball shot from a spring loaded gun‚ into a receptacle which traveled up a ramp. As well as finding the initial velocity of the ball we also wanted to determine the spring constant of the spring used in the spring loaded gun. Lastly‚ we wanted to analyze what speed would theoretically be reached if no
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car starting from rest‚ accelerates for 15.0 min until it’s velocity is 20 m/s. It then moves at constant velocity for another 20.0 min before it slow down and finally stopped in another 10.0 min. Find (a) acceleration during the first 15 min‚ (b) the deceleration during the last 10 min of its motion‚ (c) the distance traveled during the last minute‚ and the (d) total displacement. (e) Draw the displacement versus time graph and velocity versus time graph for the motion of the car. Given:
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high speed and covers a relatively large distance in a short amount of time. * Velocity: Velocity is a vector quantity that refers to "the rate at which an object changes its position." When evaluating the velocity of an object‚ one must keep track of direction. It would not be enough to say that an object has a velocity of 55 mi/hr. One must include direction information in order to fully describe the velocity of the object. * Distance: Distance is a scalar quantity that refers to "how
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07X1VW5 I‚ 2013 SUMMATIVE ASSESSMENT – I‚ 2013 / SCIENCE IX / Class – IX 3 90 Time Allowed : 3 hours Maximum Marks : 90 General Instructions : The question paper comprises of two Sections‚ A and B. You are to attempt both the sections. All questions are compulsory. All questions of Section-A and all questions of Section-B are to be attempted separately. 1 3 Question numbers 1 to 3 in Section-A are one mark questions. These are to be answered in one word or in one sentence. 4 6 30-30 Question
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inclined plane. To measure the instantaneous velocity and to determine the acceleration of the cart from the slope of the velocity-time graph. Theoretical Background A cart moving down a smooth incline speeds up. This is a simple case of a uniformly accelerated motion in one dimension. The rate of change of velocity is constant or uniform. The rate of change of velocity is called acceleration. To determine the acceleration‚ one needs to measure the velocity at two different points along the incline
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after which there is no interference apart from gravity. The initial velocity If the projectile is launched with an initial velocity v0‚ then it can be written as \mathbf{v}_0 = v_{0x}\mathbf{i} + v_{0y}\mathbf{j}. The components v0x and v0y can be found if the angle‚ ϴ is known: v_{0x} = vgh_0\cos\theta‚ v_{0y} = v_0\sin\theta. If the projectile’s range‚ launch angle‚ and drop height are known‚ launch velocity can be found by V_0 = \sqrt{{R^2 g} \over {R \sin 2\theta + 2h \cos^2\theta}}
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FORM 7 NOTES ‚ EXAMPLES ‚ PROBLEMS & SOLUTIONS Table of Content Chapter Topic Page 1. Mechanics …………………. 2 2. Gravitation …………………. 61 3. Direct Current ………………….. 71 4. Electrostatics …………………. 90 5. Simple Harmonic Motion ……. 116 6. Waves ………………………… 128 7. Magnetism …………………. 156 8. Electromagnetic Induction …… 162 9. Atomic Physics …………. 184 10. Answers …………. 194
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