Answer:
D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force
Explanation:
For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.
8+6.5+2x=52
52/2=8+6.5+x
26=8+6.5+x
26-8-6.5=x
11.5=x
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½
)/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Answer:
10N right
Explanation:
First, we need to find the base of the triangle to find the force acting on the right of the object.
- We have been given the hypotenuse and we need to find the adjacent, so we are using cos
- SOH, CAH, TOA
- cos(θ) = adjacent ÷ hypotenuse
Next, sub in the values (we don't know the adjacent so i've called it x):
To find x, we need to multiply both sides by 60
Now we know all the forces, we can work out the net force:
- Both of the 40N cancel out because they are opposite forces
- The force on the right of the object is 10N stronger than the force on the left (30N - 20N), so the object would move to the right
