Recent content by eoneil

Young’s Modulus, E= tensile stress/tensile strain= (F/Ao)/(∆L/Lo) F= force applied to object= ma= mg= (12kg)(9.81m/s2)= 117 N Ao= cross-sectional area through which force is applied= πr2= 2.8 x 10^-8 ∆L= amount by which length of object changes= 0.008m Lo= original length of the object=...

pythagoras' theorem, so 300^2+(-490.5^2)= c^2 c= 575m

Solving for vertical displacement, v = 0, a= g, ∆y= yx0t- ½ gt2 ∆y= (0)(10s)- ½ (9.81m/s2)(10s)2= -490.5m Now that displacement is solved on both axes, how do I obtain initial position? I know the particle, at 10s, is 300m ahead on the x axis, and -490.5m on y. Do I combine the two...

As its only horizontal velocity we just need to take into account the horizontal initial component, and the vertical and coefficient of restitution are irrelevant in this case. Is that what you're saying danago? Solving for horizontal displacement at time: ∆x= vx0t, where vx0 is initial...

I think the answer lies somewhere in the 'heat gained' step. (Specific heat)(50g)(80 C)= (1.00cal/g deg)(92g)(6 deg) Since the water sits in the calorimeter, it must be at the same temperature. The calorimeter heats up alongside the water, but their specific heats are different. How do I...

I agree that the calorimeter values should be incorporated. As for the specific heat given, no units included.

I'm confused as to where to include specific heat of the calorimeter metal, 0.095. And is it included alongside the net weight of 92g water + 5g calorimeter?

Homework Statement A piece of metal of weight 50 grams is heated to a temperature of 100°C and then introduced into a calorimeter containing water at 14°C, the maximum temperature reached being 20°C. If the weight of the calorimeter is 5 grams, the specific heat of the metal of which it is...

I thought the receiver was the component responsible for pressure onto the piston, not the inverse. I guess I'm having trouble conceptualizing it. I even looked for a couple piston animations online. Do you mean to adjust the calculation to make the piston 3x its original volume?

Ok, got it, (60)(60)s, giving rad/s^2. The converting angle to degrees = angle in radians x 180 / Pi So for the hour hand, 2π /12(60)(60)s= 1.45 x 10-4 rad/s2= sqrt 1.45 x 10-4 rad/s2 final value is 1.20 x 10-4 rad/s, or 0.0069 deg/s

Homework Statement If the cross-section of the barrel of a pump is 5 square centimeters, the stroke of the piston is 20 centimeters, and the volume of the receiver, in which the air is drawn from, is 1 liter, calculate the ratio of the pressure of the air in the receiver to the original...

One revolution of the hour hand is 12hrs. a) Va Second Hand= 2π/60s= 0.1047 rad/s= 6 deg/s b) Va Minute Hand= 2π/3600s = 0.001745 rad/s= 0.1 deg/s c) Va Hour Hand= 2π /12(60s)(60s)= 1.45 x 10-4 rad/s= 8.33 x 10-3 deg/s

Homework Statement Calculate the angular velocity of the second, minute, and hour hand of a watch. Homework Equations Assuming the watch is functioning normally, a second hand must travel the entire 360 degrees in 60 second. The formula ω= ɵ/t would be used, where ω is angular velocity...

I made an erroneous final calculation, whereby I must've omitted a digit or two. This goes to show for anyone how even a simple problem can be all kinds of wrong if numbers aren't double-checked. So, and for the last time I believe: s=rɵ= (377793km)(0.009199rad)= 3475.3km Thank you for the help.

I converted radians to degrees and miles to km in order to get SI units. I realized soon after posting that radians are already the SI units for angular measure. So indeed that was a useless step. We then end up with s=rɵ= (377793km)(0.009199rad)= 2152.5. This is not the final value, as the...