Answer:
1. Wf= 627.9 J
2. Wg= 416.32J
3. Work done by normal force = 0 always
4. Wk = 192.3J
5. Wt = 19.28 J
Explanation:
The question have no questions specific so here some possible questions:
1) Calculate the work done on the suitcase by the force F
2) Calculate the work done on the suitcase by the gravitational force.
3) Calculate the work done on the suitcase by the normal force.
4) Calculate the work done on the suitcase by the friction force.
5) Calculate the total work done on the suitcase.
1)
work = F * d
Wf= 161N * 3.90m
Wf= 627.9 J
2)
work = mgh
Wg= 20kg*9.8m/s² * 3.9m*sin33º
Wg= 416.32J
3)
Work done by normal force = 0 always
4)
friction work = µmgdcosΘ
Wk = 0.30*20kg* 9.8m/s²*3.90m*cos33º
Wk = 192.3J
5)
total work = (627.9- 416.32- 192.3) J
Wt = 19.28 J
Answer:
your pretty no matter what others say
Explanation:
a) They collide and stick to one another
b) Inside of the star called Sun. On the earth we don't have such materials that would withstand very high temperatures.
c) As it commonly happens today: boil water, get high-temperature steam, rotate turbine connected to a generator and get electricity
d) Sorry, I don't see any charts.
Answer:
0.0953125 N
Explanation:
Applying,
F = kq'q/r²................. Equation 1
Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.
From the equation,
If both charges remain constant,
Therefore,
F = C/r²
C = Constant = product of the two charge(q' and q) and k
Fr² = F'r'²................ Equation 2
From the question,
Given: F = 6.10 N
Assume: r = x m, r' = 8x
Substitute these value into equation 2
6.1(x²) = F'(8x)²
F' = 6.1/64
F' = 0.0953125 N
Hence the new force will become 0.0953125 N
Answer:
d = 44.64 m
Explanation:
Given that,
Net force acting on the car, F = -8750 N
The mass of the car, m = 1250 kg
Initial speed of the car, u = 25 m/s
Final speed, v = 0 (it stops)
The formula for the net force is :
F = ma
a is acceleration of the car
Let d be the breaking distance. It can be calculated using third equation of motion as :
So, the required distance covered by the car is 44.64 m.
