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A 0.11 kg ball is thrown straight up into the air with an initial velocity of 26 m/s
Find the momentum of the ball halfway to its maximum height on the way up.

V0 = 26 m/s
g=9.8 m/s^2

Max Height = V^2/2g
= 26^2 / 2*9.8
= 676 /19.6
= 34.489 meters

Halfway up is 1/2 this number or 17.2445 meters

Use the equation for height in terms of time
and solve for t

Height = Vt – 1/2 gt^2
17.2445 = 26t -1/2 (9.8)t^2
4.9t^2-26t+17.2445=0

Use the quadratic formula

( -b +- SQRT(b^2-4ac))/2a

to solve for roots

Roots are 0.777(ball on way up) and 4.52(ball on way down)

Calculate the velocity at the point when the ball is halfway to top

Use equation for velocity
V = V0 -gt
V= 26 – 9.8(0.7777)
V= 18.3854 m/s

Momentum is the product of velocity and mass

Mball x Velocity = 0.11Kg x 18.3854 m/s = 2.022 Kg *m/s

The ball will finally hit the ground 5.306 seconds after the toss
The maximum height is reached at 2.6485 seconds where the velocity is zero

Because the ball is moving fastest when first tossed, it reaches the halfway point in only 0.77 seconds but takes until 2.6 seconds to reach the maximum height.

Newtons equations of motion

How do you measure voltage in a circuit?

Yes, you need to use a voltmeter or a multimeter with a voltage scale.

Voltage is always measured with respect to 2 points in an electrical circuit.

Most DC circuits use a common output of the power supply(usually the negative) for one of the points and call this ground (abbreviated GND). The black terminal of your meter should be attached to this point, while the red lead is used to measure voltages in the rest of the circuit with respect to this ground point.

AC circuits also use a common ground point but the polarity of the leads is not as critcal since AC voltage doesn’t have a “polarity”.

If you have an Ultracapacitor with the below specs…

Boostcap® Ultracapacitor
3000.0 Farad ± 20%; 2.7V
How many would you need to store 1KWH of energy? ( 1,000 watts for an hour)

The energy in Joules stored in a capacitor can be calculated with the following formula

E=(1/2) C V^2

for C=3000F V=2.7 Volts

E=10935 Joules

1Kwh = 3,600,000 Joules

Therefore you would need

3600000/10935 = 329.2 or rounding up

330 capacitors to supply 1Kwh of energy

Modern ultracapacitors are now approaching the energy density of the best batteries and have the advantage of very rapid charge and discharge currents

The device shown at the following link is very similar to the calculated result shown above
Capacitor Bank

These links are very useful for these sorts of calculations:
Energy storage in a capacitor

Energy equivalents

A total of 10^4 kg of water per second flows over a waterfall 25m high!
If half of the power this flow could be converted into electricity, how many 100W light bulbs could be supplied?

This is a classic gravitational potential energy problem where we are asked for electrical power.

The formula for the potential energy of the water is
P=mgh

where
m = 1×10^4 Kg
g=9.8m/s^2
h=25m

the power is simply the amount of energy available in one second or:

P= 2.45 x 10^6 J/s

With 50% efficiency

Power = 1.225 x 10^6 Watts
divide by 100 watts/bulb
12,250 light bulbs

A 500g rubber ball is dropped from a height of 10m and undergoes a perfectly elastic collision with the earth.

(a) What is the Earth’s velocity after the collision, assuming the Earth is at rest just before the collision.
(b) How many years would it take the Earth to move 1.0mm at this speed?

Well, the second assumption in statement (a) is incorrect, but let’s do the math anyway:

(a)First calculate time for ball to drop 10m
s=1/2 g t^2

s=10m
g=9.8 m/s^2 (the accelaration due to gravity)

solve for t

t= sqrt(2s/g) = sqrt (2.04) or about 1.43 seconds

(2)Next calculate velocity of ball at impact = t multiplied by g
va= 14.01 m/sec

(3)Calculate velocity of Earth after impact
Use equation at top of page of following link

Formulas for elastic collisions

with ma=0.5kg mb=5.9742 × 10^24 kilograms
vb=0
va=velocity calculated above

Vb’ = velocity of earth after impact in meters/second

vb’ = 2 * ma * va / ( ma+mb)
= 2*14.01*0.5 /(0.5+5.97*10^24)

Mass of ball can be ignored in the denominator part of this calculation

= 14.01 / 5.97*10^24
= 2.34 * 10^-24 m/s

(4)Calculate time to move 1.0 mm = s=1×10^-3 meters

T= s/vb’ = 1×10^-3 divided by 2.34 *10^-24
= 0.42 * 10^21 = 4.2 * 10^20

Result will be in seconds

Seconds in one year= 60*60*24*365= 3.15 x 10^7 seconds

Divide T by the number of seconds in one year to get the answer in years.

Answer = 4.2 x 10^20 divided by 3.15 x 10^7
= 1.33 x 10^ 13 years

Also look here for more information:
Elastic collision
Frame of reference

Now let’s redo the calculation but not ASSUME the Earth is at rest when the ball hits:

This is actually a far deeper question than it first appears because we are also dealing with gravity as well as elastic collisions.

(1)At the moment the ball is dropped, the momentum of the ball (and the earth) can be considered zero.

(2)At the moment the ball hits the ground, the ball is moving at:

(a)First calculate time for ball to drop 10m
s=1/2 g t^2

s=10m
g=9.8 m/s^2

solve for t

t= sqrt(2s/g) = sqrt (2.04) or about 1.43 second

(b)Next calculate velocity of ball at impact = t multiplied by g=
va= 14.01 m/sec

(3)However, gravity is a two way street, and the ball is also attracting the earth

The gravitational force of the Earth on the ball= Force of ball on the earth
F=mg = 0.5kg x 9.8 m/s^2 = 4.9 N

Mearth = 5.9742 x 10^24 Kg
Therefore, acceleration of the earth =
F/Mearth = 4.9N divided by 5.9742 x 10^24 Kg = 8.2 x 10 ^-25 m/s^2

Before impact, the earth is moving toward the ball at

Vearth = Aearth x t = 8.2 x 10 ^-25 m/s^2 x 1.43 seconds

Vearth = -1.17 x 10 ^-24 m/s (toward the ball)

(4) To calculate the Velocity after impact use the formula on this page, but first we need
to change reference frame for Mb (the earth) so it it’s velocity is zero.

Since we are interested in the velocity at the time of impact, we can ignore the fact that we have
an accelerating reference frame (due to gravity).

To correct the reference frame we simply subtract Vearth from both objects (Ball and the earth)

http://ac.wwu.edu/~vawter/PhysicsNet/Top…

with ma=0.5kg mb=5.9742 × 10^24 kilograms
vb=0 va=velocity of ball calculated above – Vearth

Vb’ = velocity of earth in meters/second

vb’ = 2 * ma * va / ( ma+mb)
= 2*14.01*0.5 /(0.5+5.97*10^24)

Mass of ball and reference frame correct can be ignored in calculation

= 14.01 / 5.97*10^24
= 2.34 * 10^-24 m/s

now we correct for the change in reference frame by adding
Vearth(before collision) = -1.17 x 10 ^-24 m/s (toward the ball)

Therefore
Vearth(after collision) = 2.34 * 10^-24 m/s -1.17 x 10 ^-24 m/s
= 1.17 x 10 ^-24 m/s (away from the ball)

(5)Calculate time to move 1.0 mm = s=1×10^-3 meters

T= s/vb’ = 1×10^-3 divided by 1.17 *10^-24
= 0.854 * 10^21 = 8.54 * 10^20

Result will be in seconds

Seconds in one year= 60*60*24*365= 3.15 x 10^7 seconds

Divide T by the number of seconds in one year to get your answer.

Answer = 8.54 x 10^20 divided by 3.15 x 10^7
= 2.71 x 10^ 13 years

However, the Earth will NOT keep moving for this period!
It will actually stop as soon as the ball reaches zero velocity at the top of its next bounce.

I’m making an encoder and have to add a PIC to transfer the binary result to BCD-to-7 segment encoder. However, i don’t know how to program a PIC.

Try this site:
BCD to 7 segment with a PIC microcontroller

This site also claims to have the code you need as part of a frequency counter project but you need to provide your email:
Scroll down for the links

Pic frequency counter

Two inductors having inductances L1= 5H and L2=2H and are connected in parallel. The mutual inductance M between the two inductors is 1H. Determine the equivalent inductance (in H) for this system.

With mutual inductance the combined inductance is more complicated to calculate

M=1H

1/Ltotal =( L1 + L2 – 2(M)) / (L1L2 – M^2)
1/Ltotal = (7-2) / (10-1)
1/Ltotal = 5/9

Ltotal = 1.8 H

See also this page for future reference
Series and parallel circuits

What is the connection between people putting bricks in their toilets and Archimedes principle?

Archimedes’ principle is better known as the law of buoyancy.
It states:
Something submerged(partially or fully) in water experiences a force that tries to push it up (and out of the water).

The amount of that force is exactly equal to the WEIGHT of the WATER that the object is displacing.

Back to the submerged bricks…

Assuming a brick is 4″ x 2″ x 8″ you can determine the buoyant force on the brick as follows:

(1)Make a box the exact same size as the brick.
(2)Fill it with water
(3)Weigh the water.

Let’s say that the water weighs 1 kilogram.
That means that every brick that is put into the toilet is experiencing an upward buoyant force of 1Kilogram.

Assuming the brick weighs 2 Kilograms, which is greater than the buoyant force, the brick will stay on the bottom of the toilet.

If we try to use a block of wood the same size, that weighs 0.5Kg instead of the brick, that block will be pushed out of the water and float.

How far will the wood block be pushed out of the water?

Until the Volume below the water line is equal would be 0.5kg of Water!

Based on 1Kg for the weight of water that a brick displaces,
we would expect that it would float exactly halfway submerged.

What is the relationship between the ozone layer and greenhouse gases?

The ozone layer and greenhouse gases are not directly related.

The ozone layer is an area in the atmosphere where more ozone (also known as O3) compared to regular oxygen (O2) collects.
The Ozone layer is very good at blocking Ultraviolet light from the Sun from reaching the surface.

Some other byproducts of pollution (such as Nitrous Oxides from car exhaust) do react with the Ozone in this layer and reduce the amount present.

Greenhouse gasses are very good at trapping heat in the atmosphere by preventing Infrared energy (heat) from radiating back into space.

One of the strongest greenhouse gasses is Methane, which is released in the Oceans as a byproduct of biological decay.

Carbon Dioxide is another greenhouse gas that is released:

(1)When you breath
(2)When an animal breathes
(3)By plants at night
(4)When something is burned

The primary purpose of diodes is to allow electrical current to flow in only one direction.

The primary purpose for power diodes is as components of power supplies that are used to convert AC (Alternating Current) line voltage (from the wall socket) into DC (Direct Current) to power an electronic device.

Diodes are typically rated based on 2 numbers:

(1)Maximum current they can carry
(2)The maximum reverse voltage they can withstand (also called PIV or Peak Inverse Voltage)