Recent content by Peto

Sorry for the late response I was caught up with other work, anyways, so I drew the free-body diagram as suggested in part a, For part b, F_BCy = 20000N down F_BCx = 0 F_BAy = 20000sin40 = 12855N down F_BAx = 20000cos40 = 15320N left Since in equilibrium, the boom must counteract these...

Sorry the length is 0.1m so that would give a force of 1.5N and an acceleration iof 1.5/0.4=3.75m/s^2 in which direction ?

The expression is F=BIl so F=(3.0T)(5.0A)(1.0m)= 15N. Then using F=ma the acceleration would be a=F/m=(15N)/(0.40kg)=37.5m/s^2 so if that is the acceleration, then Which way would it accelerate? Towards the battery or away?

A 0.40kg metal slider is sitting on smooth (frictionless) conducting rails as shown below. What is the magnitude and direction of the acceleration of the slider? Given, B = 3.0T I = 5.0A m = 0.40kg R = either 0.1m or 0.05m, I am not sure if you half it because...

Referring to the bridge circuit of fig P90, if I=6A, I2=4A, and I3 = 0, find I1, I4, and I5. I did, I1=2A I4=4A I5=2A My logic being, since I is 6A, and I2 is 4A, I1 must be the difference, 2A. since I4 is coming from I2 and I3 it should be the sum, (0 + 4A = 4A). and I thought because I5 is...

The Crane in P19 (Attached) supports a 20.0-kN load.. to simplify things take the bales to be represented by two separate bundles: Cable-AB and cable-BC. (a) Draw a free-body diagram of point-B putting in the forces exerted by the boom, cable-AB, and cable -BC. (b) determin the horizontal and...

yes, v = sqrt(MG/r) so figuring out the difference in total energy, would that give me the amount of energy needed to boost the space station to its new height? what I mean is, does difference in total energy = amount of energy required to boost to new height?

Hmmm, so I would have to find the total energy, Et = Ep + Ek = G Mm/r + .5mv^2 ok I think I got it now.

The International Space Station, with a mass of 370,000 kg, is orbiting the Earth at a height 335 km and needs to be boosted to an orbit of 352 km. Calculate the energy needed to boost the ISS to its new height. m = 370,000 kg M = 5.98 x 10^24 kg G = 6.67 x 10^-11 Nm^2/kg^2 Initial...