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professor190 [17]
3 years ago
5

Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an

d averages 42 mph. What is the distance between his home and school if the total traveling time is 1 hr 15 min?
Physics
1 answer:
Gala2k [10]3 years ago
4 0

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

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Explanation:

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F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

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Width of the pool = w = 50 ft

length of the pool = l= 100 ft

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weight density = ρg = 62.5 lb/ft

Solution:

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F = \frac{pgwh}{2} (2x_{1}+h)

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b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

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1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

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2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

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h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

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F = 25000lb + 150000lb

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d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

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