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kaheart [24]
3 years ago
15

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

utes, she walks 125 meters to the west to meet a friend and then walks 45 meters north with her friend to a cafe for lunch. Overall, she takes 10 minutes to do all this. What is Mary’s average speed?
8 meters/second
1.25 meters/second
25 meters/second
4.5 meters/second
0.42 meters/second
Physics
1 answer:
maksim [4K]3 years ago
6 0

Answer:

0.42 meters/second

Explanation:

The total distance traveled by Mary is: 80 m + 125 m + 45 m = 250 m  

We want the speed in meters/second, so we have to convert the 10 minutes elapsed during the trip into seconds. Then we need a conversion factor, 1 minute is equivalent to 60 seconds.

time = 10 minutes * (60 seconds/1 minute) = 600 seconds (notice that minute is cancelled because is multiplying and dividing)

Now, we can compute the average speed is as follows:

Average speed = total distance/time = 250 m/600 s =  0.42 m/s

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8 0
2 years ago
From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The init
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8 0
3 years ago
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

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.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

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

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

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

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

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

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

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

here

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

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

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}

F = 2187937.5 lb

7 0
3 years ago
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