Answer:
note:
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
If your accelerator gets stuck down, do the following: Shift to neutral. Apply the brakes. Keep your eyes on the road and look for a way out.If your accelerator gets stuck down, do the following:
Shift to neutral.
Apply the brakes.
Keep your eyes on the road and look for a way out.
Warn other drivers by blinking and flashing your hazard lights.
Try to drive the car safely off the road.
Turn off the ignition when you no longer need to change direction.
Answer:
import numpy as np
import time
def matrixMul(m1,m2):
if m1.shape[1] == m2.shape[0]:
t1 = time.time()
r1 = np.zeros((m1.shape[0],m2.shape[1]))
for i in range(m1.shape[0]):
for j in range(m2.shape[1]):
r1[i,j] = (m1[i]*m2.transpose()[j]).sum()
t2 = time.time()
print("Native implementation: ",r1)
print("Time: ",t2-t1)
t1 = time.time()
r2 = m1.dot(m2)
t2 = time.time()
print("\nEfficient implementation: ",r2)
print("Time: ",t2-t1)
else:
print("Wrong dimensions!")
Explanation:
We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.
Answer:
293 kg
Explanation:
Let's say the tension in each cable is Tb, Tc, and Td.
First, find the length of cable AD:
r = √(2² + 2² + 1²)
r = 3
Using similar triangles:
Tdx = 2/3 Td
Tdy = 2/3 Td
Tdz = 1/3 Td
Sum of the forces in the x direction:
∑F = ma
Tb − 2/3 Td = 0
Td = 3/2 Tb
Sum of the forces in the y direction:
∑F = ma
2/3 Td − Tc = 0
Td = 3/2 Tc
Sum of the forces in the z direction:
∑F = ma
1/3 Td − mg = 0
Td = 3mg
From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.
8600 N = 3mg
m = 8600 N / (3 × 9.8 m/s²)
m ≈ 292.5 kg
Rounded to three significant figures, the maximum mass of the crate is 293 kg.