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?
2) the distance covered by a falling ball varies as the square of the time taken.if a ball falls through 1800m in 15seconds,calculate the distance when the time taken is 15seconds and the time taken for is distance of 648m.
3)p varies directly as q^2 and inversely as rootR.when q=8 and r=9.,p=16.find p when q=5 and r=9.
#1 done below
#2 s = kt^2
so, s/t^2 is constant. You want
648/t^2 = 1800/15^2
#3 p = kq^2/√r
so, p√r/q^2 is constant
p√9/25 = 16√9/64
Kinda dumb to give you the same r value...
1) For the first question, let's solve it step by step:
A) To find the law connecting the load and force, we can analyze the given information. We know that the force needed to pull a load is partly constant and partly varies with the load itself. In other words, we can express it as:
E = k1 + k2 * load
where E is the force needed, load is the weight of the load, and k1 and k2 are constants.
Using the given information, we can set up two equations:
1.4 = k1 + k2 * 20
2 = k1 + k2 * 30
Simplifying these equations, we get:
k1 + 20k2 = 1.4 (equation 1)
k1 + 30k2 = 2 (equation 2)
Now, we have a system of linear equations that we can solve simultaneously. Subtracting equation 1 from equation 2, we get:
10k2 = 0.6
Dividing both sides by 10, we find:
k2 = 0.06
Substituting the value of k2 into equation 1, we can find k1:
k1 + 20(0.06) = 1.4
k1 + 1.2 = 1.4
k1 = 1.4 - 1.2
k1 = 0.2
So, the law connecting the load and force is:
E = 0.2 + 0.06 * load
B) To find the force for a load of 50g, we can use the law we derived in part A. Substituting the value of load = 50, we get:
E = 0.2 + 0.06 * 50
E = 0.2 + 3
E = 3.2N
Therefore, the force needed for a load of 50g is 3.2N.
2) For the second question:
Given that the distance covered by a falling ball varies as the square of the time taken, we can express it as:
distance = k * time^2
where k is a constant.
Using the given information, we can set up an equation:
1800 = k * 15^2
Simplifying this equation, we get:
1800 = k * 225
k = 1800 / 225
k = 8
Now, we can use the value of k to find the distance when the time taken is 15 seconds and the time taken for a distance of 648m.
When time = 15 seconds:
distance = 8 * 15^2
distance = 8 * 225
distance = 1800m
When distance = 648m:
648 = 8 * time^2
648 / 8 = time^2
81 = time^2
time = √81
time = 9 seconds
Therefore, when the time taken is 15 seconds, the distance covered is 1800m. When the time taken for a distance of 648m is 9 seconds.
3) For the third question:
Given that p varies directly as q^2 and inversely as √r, we can express it as:
p = k * q^2 / √r
where k is a constant.
Using the given information, we can set up an equation:
16 = k * 8^2 / √9
Simplifying this equation, we get:
16 = k * 64 / 3
16 * 3 = k * 64
48 = k * 64
k = 48 / 64
k = 0.75
Now, we can use the value of k to find p when q = 5 and r = 9:
p = 0.75 * 5^2 / √9
p = 0.75 * 25 / 3
p = 18.75 / 3
p = 6.25
Therefore, when q = 5 and r = 9, p is equal to 6.25.