equal to the distance traveled. | |3. A bullet moving horizontally to the right (+x direction) with a speed of 500 m/s strikes a sandbag and penetrates a distance | |of 10.0 cm. What is the average acceleration‚ in m/s2‚ of the bullet?(Points : 5) | | [pic]-1.25 x 103
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highway. Assuming a deceleration of 7.00 m/s2‚ estimate the speed of the car just before braking. [32.4 m/s] 6. Determine the stopping distances for an automobile with an initial speed of 90 km/h and human reaction time of 1.0 s: (a) for an acceleration a = –4.0 m/s2; (b) for a = – 8.0 m/s2. [103 m; 64.1 m] 7. A ball is hit almost straight up into the air with a speed of about 20 m/s. (a) How high does it go? (b) How long is it in the air? [20.4 m; 4.08 s] 8. A helicopter is ascending
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IMPORTANT QUESTIONS Physics · How much force is needed to accelerate a trolley of mass 20g through 1 m/s2. · A force of 100N acts on a mass of 25 kg for 5 s .What velocity does it generate? · A bullet leaves a rifle with a velocity of 100m/s and the rifle of mass 2.5 kg recoils with a velocity of 1m/s. find the mass of the bullet? · Certain force acting on a mass of 15kg for 3s‚ gives it a velocity of 2m/s. Find the magnitude of force. · A cricket ball of mass 0.15 kg is moving with
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Kinematics is one of the topics under dynamics. Kinematics describes motion without regard to its causes. In this experiment‚ kinematics focuses in one dimension: a motion along a straight line. This kind of motion‚ involves velocity‚ displacement‚ and acceleration with regards to time. The objectives of the experiment are to draw the displacement versus time graphs and velocity versus time graphs for uniform motion and uniformly accelerated motion‚ and to determine one’s normal reaction time and his reaction
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1. A 1.08 x 103 kg car uniformly accelerates for 12.0s from rest. During this time the car travels 132 m north. What is the net force acting on the car during this acceleration 2. A net force of 12 N is exerted on and to cause it to accelerate at a rate of 0.03 m/s2. Determine the mass of the encyclopedia. HYPERLINK http//fc.codmanacademy.org/branches/physicsofdriving2/pushingbuses.jpg INCLUDEPICTURE http//fc.codmanacademy.org/branches/physicsofdriving2/images/pagemaster/pushingbuses
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(2) Ignoring air drag‚ a projectile will take the same time for its horizontal motion and its vertical motion. (2) Neglecting air drag‚ the horizontal velocity of a projectile will remain constant. (3) The vertical motion will have constant acceleration‚ g. Equations: sy = voyt + ½ g t2 [ vertical motion] R = voxt [horizontal motion] R= t= H= A hypothesis: The water rocket which is launched at different angles will fly into different heights in the sky and each time it will
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motion can be split into two separate dimensions of movement. The first being constant motion in the horizontal x axis witch neglecting air resistance should stay constant throughout the projectiles path. In the vertical y axis we have constant acceleration due to gravity toward the ground. These two motions are linked in time witch allows you observe the instantaneous characteristics of the projectile. Time is the connecter between the equations so you can salve for time in one equation and plug
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speed of 12 m/s‚ how far will it go before hitting the ground below? ? (a) 48m (b) 39m (c) 7.3m (d) 24m (e) 34m 4. If I throw a rock straight up in the air‚ and ignore air resistance‚ then at the top of its arc it will have an acceleration (in m/s2) of (a)
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relation x 2t3 6t2 10‚ where x is expressed in m and t in seconds. Determine the time‚ position‚ and acceleration when v 0. ( Ans. x 2m‚ a 12 m/s2 ) Q2. The motion of a particle is defined by the relation x 2t3 -15t2 24t 4‚ where x is expressed in meters and t in seconds. Determine (a) when the velocity is zero‚ (b) the position and the total distance traveled when the acceleration is zero. (Ans. (a) 1s ‚4s (b) 1.5m‚24.5m) Q3. A motorist is traveling at 54 km/h when
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Julie Kim Free Fall Lab Purpose: to use collected data and the kinematics equations to determine the value of local gravity Data: height 161 cm(1 m/100 cm) = 1.61 m mass of small ball 16.5 g mass of big ball 28.0 g 1 2 3 4 5 6 7 8 9 10 Average Small 0.585 sec 0.571 sec 0.567 sec 0.571 sec 0.571 sec 0.572 sec 0.571 sec 0.574 sec 0.576 sec 0.571 sec 0.573 sec Big 0.573 sec 0.568 sec 0.569 sec 0.569 sec 0.570 sec 0.569 sec 0.571 sec 0.563 sec 0.571 sec 0.570
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