1. A force of 25.0 Newtons is applied so as to move a 5.0 kg mass a distance of 20.0 meters. How much work was done?
Ans. W = F × d = 25 N × 20.0 m = 50 J
2. A force of 120 N is applied to the front of a sled at an angle of 28.0 above the horizontal so as to pull the sled a distance of 165 meters. How much work was done by the applied force?
Ans. W = F •d cos = 120 N • 165 m cos 28.0 = 17,482.36 J
3. A sled, which has a mass of 45.0 kg., is sitting on a horizontal surface. A force of 120 N is applied to a rope attached to the front of the sled such that the angle between the front of the sled and the horizontal is 35.0o. As a result of the application of this force the sled is pulled a distance of 500 meters at a relatively constant speed. How much work was done to this sled by the applied force?
Ans. W = F• d cos = 120 N • 500 m • cos 35.0 = 49,149.12 N
4. A rubber stopper, which has a mass of 38.0 grams, is being swung in a horizontal circle which has a radius of R = 1.35 meters. The rubber stopper is measured to complete 10 revolutions in 8.25 seconds.
a. What is the speed of the rubber stopper?
Ans. v = v = • v = .•(. ) v = 10.29 m/s .
b. How much force must be applied to the string in order to keep this stopper moving in this circular path at a constant speed?
Ans. The Centripetal Force, Fc = • = . • (. /)= 29.76 N .
c. How far will the stopper move during a period of 25.0 seconds?
Ans. d = vt = 10.29 m/s25 s = 257.25 m
d. How much work is done on the stopper by the force applied by the string during 25.0 seconds?
Ans. 0 J. The force is towards the center and the distance is around the perimeter of the circle. They are at right angles to each other. For work to be done, the force has to be in the same direction as the displacement.
5. How much work would be required to lift a 12.0 kg mass up onto a table 1.15 meters high?
Ans. W = F• d = mg d = 12 kg 9.8 m/s2 1.15 m = 135.24 J
Answer:
a) A = 0.07129 m²
b) A / A ’= 1.77
Explanation:
In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.
width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)
a = 0.2159 m
length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)
l = 0.3302 m
The area of a rectangle is
A = l a
A = 0.3302 0.2159
A = 0.07129 m²
b) we have a second sheet with reduced dimensions
a ’= 3/4 a
l ’= ¾ l
Let's find the area of this glossy sheet
A ’= l’ a ’
A ’= ¾ l ¾ a
A ’= 9/16 l a
To find the factor we divide the two quantities
A / A ’= l a 16 / (9 l a
A / A ’= 1.77
Answer:
3 quarters of the world likes pizza. (75%) and the remaining quarter of people (25%) dont like pizza.
The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,
1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.
It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.
Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.
A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.
So, in analyzing the four choices given, we look for low P and high T.
A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.
B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.
We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
