<p>sry these are better questions than the ones before...help please? :) with explanations ^^</p>
<ol>
<li><p>A sphere rolls without slipping down an incline plane 1.8 m high. Its speeed at the bottom is:</p></li>
<li><p>A rocket is fired at an angle of 45 degrees above the horizontal and accelerates the projectile at 5.4 m/s^2. The initial magnitude of the vertical acceleration is approx:</p></li>
<li><p>A construction worker pushes a wheelbarrow 5 m with a horizontal force of 50.0 N. How muc hwork is done by the worker on the wheelbarrow? </p></li>
<li><p>Two perpendicular forces, one of 45 N directed upward and the other of 60 N directed to the right, act simultaneously on an object wit ha mass of 35 kg. What is the magnitude of the resultant acceleration of the object?</p></li>
<li><p>How much work is requried to accelerate a 1000kg car from 20m/s to 30m/s?</p></li>
<li><p>How much elastic potential energy is stored in a bungee cord with a spring constant of 10 N/m when the cord is stretched 2m? </p></li>
</ol>
<p>THANKS! :D</p>
<ol>
<li><p>mgh = 1/2mv^2</p></li>
<li><p>Gravity is -9.8 m/s^2 90 degrees below the horizontal. The sum of that vector plus the vertical component of the acceleration (5.4sin30) is your initial vertical acceleration.</p></li>
<li><p>W = f * d</p></li>
<li><p>sumF = ma; find the magnitude of the resultant force, divide by mass, and you’ll get resultant acceleration.</p></li>
<li><p>Work-energy theorem</p></li>
<li><p>U = 1/2 kx^2; k = 10 N/m, x = 2m</p></li>
</ol>
<p>These are all straight formula problems for the most part. If you need me to go deeper or need more hints, please post.</p>
<p>o ok thanks :]
my teacher hasn’t reallly been in school so I don’t really know what we’re doing lol.</p>
<p>i still don’t know how to use the work-energy theorem :/</p>
<p>i’m really sorry but could i send you a picture question somehow? b/c i don’t know how to do it…thanks. appreciate your help :)</p>
<h1>5: The work done is just another name for the change in kinetic energy. So, find the KE when v=30 and KE when v=20. Subtract them and that gives you the work.</h1>
<p>The problems you wrote are very common and I understand what you’re asking. For #5:</p>
<p>W = change in kinetic energy.</p>
<p>1/2 (m)(v-final)^2 - 1/2(m)(v-initial)^2 = W</p>
<p>Hope it helps - if you need me to elaborate more here or anywhere else, please ask.</p>
<ol>
<li><p>Two perpendicular forces, one of 45 N directed upward and the other of 60 N directed to the right, act simultaneously on an object with a mass of 35.0 kg. What is the magnitude of the resultant acceleration of the object?</p></li>
<li><p>An ant on a picnic table travels 30 cm eastward, then 25 northward, and finally 15 cm westward. What is the magnitude of the ant’s displacement relative to its original position?</p></li>
<li><p>force is applied downward at an angle of 15 degrees above the horizontal to a 12 kg box on a horizontal floor causing it to accelerate at 1.6 m/s^2, even tho the coefficient of frictio nis .18. The work done to move the box 6.5 m is</p></li>
<li><p>A vector A in a Cartesian coordinate system has the components A (sub x) = -6, A (sub y) = 8. The angle the vector makes with the positive x-axis is:</p></li>
<li><p>A race car finishes the final lap of a race in 1 min and 2s. If the track is 4.34 km long what was its average speed during the final lap in m/s?</p></li>
<li><p>A bullet is fired horizontally. HTe muzzle speed of the rifle is 400 m/s. The bullet strieks a target that is 100 m away. Neglect aerodynamic effects. Where does the bullet hit the target? </p></li>
<li><p>A ball is thrown straight upward from the ground and lands 5 s later. What was the initial speed of the ball?</p></li>
</ol>
<p>thanks again :)</p>
<p>anybody please? ^-^</p>
<ol>
<li><p>This is the same as #4 above. </p></li>
<li><p>displacement is basically the shortest distance between 2 points. It goes 30 east, 25 north, and 15 west, so it’s basically 15 east and 25 north. Find the hypotenuse of the triangle to get the displacement.</p></li>
<li><p>W = F * d. sumF = ma. Find the net force then multiply by distance.</p></li>
<li><p>(theta) = tan-1 (Ay/Ax)</p></li>
<li><p>avg. speed = m/s. 4.34 km * 1000 m/1km * 1/62 s. It’s a unit conversion problem basically.</p></li>
<li><p>What do you mean where? What height is it fired at? It will travel 100 m in 0.25 seconds. gravity will make it fall y = 1/2 gt^2, where t is time and g is 9.8 m/s^2</p></li>
<li><p>vf^2 = v1^2 + 2a(yf - yi) is standard but you don’t have a height. This is a common problem but I’m drawing a blank right now :(</p></li>
</ol>
<p>7: Since it lands on the ground in 5s, it reaches the top of its path at 2.5s, where vfinal (vf) is 0.</p>
<p>vf: 0
vi: ?
a: -10 (g)
y: who cares?
t: 2.5</p>
<p>vf=vi+at</p>
<p>Plug in and solve for vi.</p>