Archive for April, 2008

Physics projectile formula

Tuesday, April 8th, 2008

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)

Measure of voltage

Monday, April 7th, 2008

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.

Capacitor circuits for energy storage?

Sunday, April 6th, 2008

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

Gravitational Potential Energy

Sunday, April 6th, 2008

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

Elastic collision with the earth

Sunday, April 6th, 2008

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

Bcd seven segment driver control pic

Sunday, April 6th, 2008

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

Mutual inductance in parallel

Sunday, April 6th, 2008

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

Archimedes Principle for kids

Sunday, April 6th, 2008

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.

Ozone and Greenhouse

Sunday, April 6th, 2008

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.

What is a diode?

Sunday, April 6th, 2008

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)