Answer:
Part a)

Part B)
percentage increase is
%
Explanation:
Part a)
As we know that the beat frequency is

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase
So we have


Part B)
percentage increase in the tension of the string will be given as


now we have

so we have


so we have

percentage increase is

The solution for the problem is:
1 Watt = 1 Joule per second
1 Watt*second = 1 Joule
a Kilowatt is 1,000 Watts
an hour is 60 seconds times 60 minutes or 3,600 seconds
a Kilowatt * hour is 1,000 Watts in 3,600 seconds
15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000
Watt*second
54,000,000 Watt*second = ? Joules
54,000,000 Joules / second = 54,000,000 Watts
Answer:
a) x(t) = 10t + (2/3)*t^3
b) x*(0.1875) = 10.18 m
Explanation:
Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.
Given:
- v(t) = 10 + 2*t^2 (radar gun)
- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)
Find:
-The position x of horse as a function of time t in radar system.
-The position of the horse at x = 2m in our coordinate system
Solution:
- The position of horse according to radar gun:
v(t) = dx / dt = 10 + 2*t^2
- Separate variables:
dx = (10 + 2*t^2).dt
- Integrate over interval x = 0 @ t= 0
x(t) = 10t + (2/3)*t^3
- time @ x = 2 :
2 = 10t + (2/3)*t^3
0 = 10t + (2/3)*t^3 + 2
- solve for t:
t = 0.1875 s
- Evaluate x* at t = 0.1875 s
x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3
x*(0.1875) = 10.18 m