Uranium 235 can be split if it is hit by:

Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring

arlik [135]

This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

$E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J$

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

$E_k = \frac{1}{2}mv^2$

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

$E_p = 2E_k=mv^2$

From this we can determine the speed of the mass:

$E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}$

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

4 0

3 years ago

A tow truck drags a stalled car along a road. The chain makes an angle of 30???? with the road and the tension in the chain is 1

My name is Ann [436]

Answer: work = 1,305kJ

Explanation:

angle= 30°

force= 1,500N

distance= 1,000m

The formula for work is : Work= force x distance, however there is an angle of 30° between the direction of force applied and the direction of motion, therefore force must be decomposed to its value on the horizontal axis which is the direction of motion by using the cosine of the very angle.

W= F×cos(α)×D

W= 1,500×cos (30)×1,000

W= 1,305kJ ( kilojoules)

3 0

3 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$