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:
Explanation:
mass of charged particle m = 6 x 10⁻³ kg .
speed of particle v = 4 x 10³ m /s
speed of particle perpendicular to magnetic field = v sin37
= 4 x 10³ sin37
= 2.41 x 10³ m / s
Force on charged particle
= B q v , B is magnetic field , q is charge on particle and v is velocity perpendicular to B
Force = ma
= 6 x 10⁻³ x 8
= 48 x 10⁻³
Force = Bqv
48 x 10⁻³ = 5 x 10⁻³ q x 2.41 x 10³
q = 48 x 10⁻³ / (5 x 2.41)
= 3.98 x 10⁻³C.
Answer:
The speed is maximum and the acceleration is zero
Explanation:
- The speed of the mass in simple harmonic motion can be found by using the law of conservation of energy. In fact, the total mechanical energy of the mass-spring system is sum of kinetic energy and elastic potential energy:
where
m is the mass
v is the speed
k is the spring constant
x is the displacement
As we can see, when the displacement is zero (x=0), the term representing the kinetic energy is maximum, so v (the speed) is also maximum.
- The acceleration of the mass in simple harmonic motion is proportional to the restoring force acting on the mass, which is given by Hook's law
where
k is the spring constant
x is the displacement
When x = 0, F = 0, so the net force acting on the mass is zero. Therefore, this also means that the acceleration of the mass is also zero: a = 0.
Answer: The partial pressure of 
if the total pressure of the mixture is 3.9 atm is 0.975 atm
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
where, x = mole fraction of nitrogen in solution = 0.25
= partial pressure of nitrogen = ?
= Total pressure = 3.9 atm
Putting in the values :
The partial pressure of is 0.975 atm
