Answered: : A metal wire of length 2.5 m and area… | bartleby

Related questions

Question

: A metal wire of length 2.5 m and area of
cross section 1.5 x 106 m² is stretched through 2
mm. Calculate the work done during the
stretching. [Y= 1.1 x 10¹¹ Pa]

Expert Solution

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

7. A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° T with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate- incline system owing to friction. (c) How much work is done by the 100-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m?
The force required to compress a non-standard spring as a function of displacement is given by the equationF(x) = -Asin(bx) + kx,where A = 11 N, b = 11 rad/m, and k = 78 N/m. Part (b) Calculate the work done in joules as the spring is compressed from x = 0 to x1 = 32 cm. Part (c) Calculate the work done in joules as the spring is compressed from x1 = 32 cm to x2 = 69 cm
During terrible weather at a ski-flying world championship, a ski jumper tried to adjust for bad conditions midway down the ramp but completely lost control and tumbled wildly off the side at its bottom into a scattering crowd. The 73-kg jumper left the ramp about 45 m lower than his starting position. After hitting the ground, the jumper crashed wildly and descended down the hill another 20 vertical meters before coming to rest. Here, friction cannot be ignored. Calculate the total work done on the ski jumper to bring him to rest. Round the final answer to the nearest whole number.
An elevator having a mass of 585 kg is raised a distance of 32 m. Determine the work done, neglecting frictional effects and the work required for acceleration.
Timothy weighs 515 N. he runs up a 16.5m tall building and jumps off. When he reaches the ground, he is travelling at 16.4 m/s. What is the work done by air resistance? Use formulas mgh and 1/2m(vf^2) - 1/2m(vi^2) if possible.
Let W = (3.40 x 10° J) In(V/V). Solve for the unknown quantity in each of the following cases. (a) the final volume V, is nine times the initial volume V, %3= (b) the initial volume is 1.00 m and the work W is 3.70 x 10° ) 31 V; =