Determine the due date and the amount of interest due at maturity on the following notes: Date of Note Face Amount Interest Rate Term of Note a. January 10* $40,000 5% 90 days b. March 19 18,000 8 180 days c. June 5 90,000 7 30 days d. September 8 36,000 3 90 days e. November 20 27,000 4 60 days * February 2016 has 28 days.

Answer and Explanation:

The computations are shown below:

a.  For January 10

Due date after 90 days would be

= 21 days in January + 28 days in February + 31 days in march + 10 days in April

So 10 April

And the interest would be

= Principal × rate of interest × number of days ÷ (total number of days in a year)  

= $40,000 × 5% × (90 days ÷ 360 days)

= $500

b.  For March 19

Due date after 180 would be

= 12 days in march + 30 days in April + 31 days in May + 30 days in June + 31 days in August + 15 days in September

So 15 September

And the interest would be

= Principal × rate of interest × number of days ÷ (total number of days in a year)  

= $18,000 × 8% × (180 days ÷ 360 days)

= $720

c.  For June 5

Due date after 30 days would be

= 25 days in June + 5 days in July

So 5 July

And the interest would be

= Principal × rate of interest × number of days ÷ (total number of days in a year)  

= $90,000 × 7% × (30 days ÷ 360 days)

= $525

d.  For September 8

Due date after 90 days would be

= 22 days in September + 31 days in October + 30 days in November + 7 days in December

So 7 December

And the interest would be

= Principal × rate of interest × number of days ÷ (total number of days in a year)  

= $36,000 × 3% × (90 days ÷ 360 days)

= $270

e.  For November 20

Due date after 60 days would be

= 10 days in November + 31 days in December + 19 days in January

So 19 January

And the interest would be

= Principal × rate of interest × number of days ÷ (total number of days in a year)  

= $27,000 × 4% × (60 days ÷ 360 days)

= $180

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

d = 0.0969 m

d = 9.69 cm