Answer: A projectile is any object in which the only force is gravity
Explanation: Equations on how to calculate projectile velocity is stated below:
The initial velocity Vo being a vector quantity, has two componentsVox and Voy
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The acceleration A is a also a vector with two components Axand Ay given
Ax = 0 and Ay = - g = - 9.8 m/s2
Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant
Vx = Vocos(θ)
Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g
Vy = Vo sin(θ) - g t
Along the x axis the velocity Vx is constant and therefore the component x of the displacement is
x = Vocos(θ) t
Along the y axis, the motion is of uniform acceleration and the y component of the displacement is
y = Vo sin(θ) t - (1/2) g t2
So, first you find your acceleration which is 3m/s^2, using the acceleration formula.
Now set up your equation, F=ma, so put in the stuff, F=80kg·3m/s^2. Then solve your equation by multiplying, and you get F=240N, since newtons are your measurement.
Hope this helps
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
Think of the cell membrane as a net and the nutrients are the perfect fit to fall through it. Where the waste is not the right size and will not fit through the holes of the net.
Answer: 95.7 N
Explanation: The tension on the chandelier is equal to the its weight.
Weight is the force directed downwards and tension is the force exerted of the opposite side.
T = W of the chain + W of the chandelier
= m1( g) + m2 (g)
= 0.37 kg ( 9.8 m/s2) + 9.4 kg (9.8 m/s2)
= - 95.7 N it means the force is directed downwards
since tension is equal weight
T = 95.7 N
