Answer:
a) Frope= 71.7 N
b) Frope=6.7 N
Explanation:
In the figure the skier is simulated as an object, "a box".
a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:
∑F=0
Frope=w*sen6.8°
Frope=71.71N
Take into account that w is the weight that is calculated as mass per gravitiy constant:
w=m*g
b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:
F=m*a
F=61.8Kg*0.109m/s2
Frope=6.73N
Answer:
This is true,the rod with smaller elastic modulus will stretch more than larger elastic modulus.
Explanation:
σ=E*ε
ε=δ/L
σ=E*δ/L
δ=(σ*L)/E
σ=F/A
δ=(F*L)/(A*E)
As Force,Area and Length is same
δ∞1/E
From the expression as E increase δ will be small,so there will be more stretch for smaller elastic modulus.
Answer:
Options A, B, and C are all possible.
Explanation:
We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?
If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.
If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.
If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.
Answer:
12N to the right.
Explanation:
There is a force of 12N upwards and a force of 12N downwards: these cancel out.
Assign a negative value to forces towards the left, and a positive value to the forces towards the right: -3N and +15N
Combine them: -3N+15N = 12N
The net force has a magnitude of 12N, and since our answer was positive, it acts towards the right.
