Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.
k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
To solve this problem it is necessary to apply the concepts related to Force of Friction and Torque given by the kinematic equations of motion.
The frictional force by definition is given by
Our values are here,
Replacing,
Consider the center of mass of the body half its distance from the floor, that is d = 0.85 / 2 = 0.425m. The torque about the lower farther corner of the refrigerator should be zero to get the maximum distance, then
Re-arrange for x,
Then we can conclude that 1.42m is the distance traveled before turning.
The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
a) for y=0.200m, r1=0.200m and r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.200m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D228391.8N%2FC)
b) for y=0.600m, r1=0.600m, r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.600m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D-59345.91N%2FC)
