Practice Problems for TEST II, PHY2020



TestII PHY2020 Formula sheet

g=9.8 m/s2 100 cm=1 m 39.37 inches=1 m 1 km=1000 m

12 inches=1 foot 5280 feet=1 mile 1 kg = 1000 g

1 hour = 3600 s G=6.7 10-11 Nm2/kg2 1 J = 1N*m 1 Watt=1 J/s

Comparison between metric and English units of Force: 1 lb (pound) = 4.45 N

Useful formulas:

Linear: Espring = ½ kx2 Egrav. potential = mgh (h is the height)

EKinetic Energy = ½ mv2 F=ma

p(the momentum) = mv (remember, p and v are vectors, so p has both size and – if it’s important – direction)

If there are no external forces (→ momentum is conserved) and the collision is elastic (→ energy is conserved) if mass m1 hits a stationary mass m2 then

v1final = v1initial (m1-m2)/( m1+m2) and v2final = v1initial (2m1)/( m1+m2)

Rotational: s=rθ vlinear=(r alinear=(r Remember, use radians for θ, (, and ( in these and all formulas below.

1 rotation = 360 o ( #4 = 300 N

__________________ N

5. The density of gold is 19.3 g/cm3. Gold currently costs about $50/g. What volume does $1,000,000 (1 million dollars) worth of gold occupy?

$1 x 106/($50/g) means that the gold has a mass of 20,000 g. The volume occupied by this mass (since the density, (, =M/V we calculate V=M/() is 20,000 g/(19.3 g/cm3)=1036 cm3. This is about 1 liter (=1000 cm3)

_______________________cm3

[pic]

6. A piece of wood of some density is floating half submerged (see picture) in fresh water, where the density of the water is 1 g/cm3. (The level of the water is marked by dashed lines. The total volume of the block is 100 cm3 and, to repeat, half of the block (i. e. a volume of 50 cm3) is submerged. What is the density of the wood?

___________________________g/cm3

From Archimedes’ principle, the buoyancy force = the weight of the volume of water displaced, or 50 cm3 * density of water * g = 50 cm3 * 1 g/cm3 * 9.8 m/s2 = 490 g m/s2 = 0.49 kg m/s2 = 0.49 N. So the whole block is in equilibrium, i. e. the buoyancy force of 0.49 N balances the weight of the block. Since the volume of the block is 100 cm3, its mass is M=(V, where ( is the sought-for wood density.

So (=M of block/100 cm3. The 0.49 N force up matches the weight of the block down, of Mg, so M(block)=0.49 N/g = 50 g, so ( = 50 g/100 cm3 = 0.5 g/cm3

7. What temperature in oC is -400 oF?

_____________________oC

-400 oF is 400+32 oF =432 oF below freezing, which is -432/1.8 oC or -240 oC. Or, using the formula you’re used to, temperature, T, in oC = (T in oF – 32)/1.8=

(-400 oF – 32)/1.8 = -240 oC

8. An ideal gas is warmed up from 100 oF to 190 oF, in a container where the pressure is kept constant, and the gas either expands or contracts to change the volume occupied by the gas at different temperatures. What is the new volume V2 (V2 is the volume that the warmer gas occupies) in terms of V1?

______________________V1

PV1=nRT1 ; PV2=nRT2 so P=nRT1/V1 = nRT2/V2 so, canceling the common factor of nR, we have T1/V1 = T2/V2. Now we need T in units of Kelvin:

First convert 100 oF to (100-32)/1.8 = 37.8 oC and 190 oF to (190-32)/1.8 = 87.8 oC.

Convert from Celsius to Kelvin: 37.8 oC=37.8 +273.15=310.95 K and 87.8 oC=87.8+273.15=360.95 K So V2=(T2/T1)V1 = (360.95/310.95)V1 = 1.16 V1

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download