different questions pls! Very urgent!
1) the force E needed to make a machine pull,a load is partly constant and partly varies as the load to be pulled itself.when the load is 20g,the force needed is 1.4n,where as the force needed for 30g load is 2N.
Find the:
A)law connecting the load and force?
B)force for the load of 50g?
E = a + kL
a+20k = 1.4
a+30k = 2.0
Now just solve for a and k, and then go for L=50
A) To determine the law connecting the load and force, we can use the given information about the force needed for different loads. From the information provided, we know that the force needed is partly constant (independent of the load) and partly varies with the load itself.
Let's denote the constant force as Fc and the force that varies with the load as Fv. We are given two data points:
- When the load is 20g, the force needed is 1.4N.
- When the load is 30g, the force needed is 2N.
From these data points, we can set up two equations:
1) Fc + Fv = 1.4N (when the load is 20g)
2) Fc + Fv = 2N (when the load is 30g)
By subtracting equation 1 from equation 2, we can eliminate the constant force Fc:
(2N) - (1.4N) = Fc + Fv - (Fc + Fv)
0.6N = 0
This result tells us that the constant force term cancels out, indicating that Fc = 0. Therefore, the law connecting the load and force is F = k × load, where k is the constant of variation.
B) Now, to find the force for a load of 50g, we can use the law we derived in part A. We have F = k × load, where F is the force and load is the weight of the load (in grams).
Substituting the values we know: F = k × 50g
To solve for k, we can use the previously given data point:
1.4N = k × 20g
Solving for k:
k = 1.4N / 20g = 0.07
Now we can find the force for a 50g load:
F = 0.07 × 50g = 3.5N
Therefore, the force needed for a load of 50g is 3.5N.