What are some examples of static potential energy?

a bullet of mass 4g when fired with a velocity of 50m/s can enter a wall upto a depth of 10cm how much will be the average resis

jok3333 [9.3K]

Mass, m = 4g = 0.004 kg
Velocity, = 50cm/s = 0.5m/s
Distance, 10cm = 0.1m
The wall would have to resist the energy acquired by the bullet.

Kenetic Energy of bullet = Resistance offered by the wall.

1/2 mv² = Resistance Force * Distance

(1/2) * 0.04 * 0.5 * 0.5 = F * 0.1

0.5 * 0.04 * 0.5 * 0.5 = F * 0.1

0.5 * 0.04 * 0.5 * 0.5/0.1 = F

0.05 = F

Therefore, Resistance offered by the wall = 0.05 N

8 0

3 years ago

Thinking about an analogue clock, calculate the angular displacement in revolutions, radians and degrees for the following: • A

Inessa [10]

Answer:

1.047 rad

Explanation:

A secondhand makes a complete revolution (360 degrees) in 60 seconds, so it displaces 6 degrees for every second of elapsed time (360/60 = 6).

And in 10 seconds, it will make a displacement of 10 degrees (6 * 10 = 60).

Finally converting the result into Radians by multiplication with π/180.

5 0

3 years ago

Bromium has two naturally occurring isotopes: 79br, with an atomic weight of 78.918 amu, and 81br, with an atomic weight of 80.9

saul85 [17]

The two different isotopes have weights :

w1 = 78.918 amu

w2 = 80.916 amu

average weight w3 = 79.903 amu

The mixing of two components can be modeled as

let the fraction of w1 be 'x'

hence $w1. x + w2.(1-x) = w3$

now this is a linear equation in 'x'. Substituting the values we get

x = 0.507

hence the percentage of Br79 = 50.7% and the percentage of BR81 = 49.3%

8 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Scrat [10]

Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

$E_k = E_p$

$mv^2/2 = mgh$

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

$v^2 = 2gh = 2*9.81*0.35 = 6.867$

$v = \sqrt{6.867} = 2.62 m/s$

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

$\omega = v / r = 2.62 / 0.175 = 14.97 rad/s$