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3 years ago

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David rowed a boat upstream for three miles and then returned to point he started from. The entire journey took four hours. The

speed of the stream is one mile per hour. Find David's speed in still water. (Hint: speed = distance ÷ time, upstream speed = speed of the boat – speed of the stream, and downstream speed = speed of the boat + speed of the stream)
David's speed in still water is ____ miles per hour.

1 answer:

frez [133]3 years ago

4 0

Upstream speed = S - 1
Downstream speed = S + 1

Average speed = total distance / total time

Average speed = (S - 1) + (S + 1) / 2
= S

S = 6 miles / 4 hours
S = 1.5 miles per hour

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An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is

KatRina [158]

Answer: 89.803 ft

Explanation:

The complete question is written below:

An astronaut on the moon throws a baseball upward. The astronaut is 6ft, 6 in. tall and the initial velocity of the ball is 30 ft per second. The height s of the ball in feet is given by the equation,s=-2.7t^2+30t+6.5, where t is the number of seconds after the ball is thrown.

The ball will never reach a height of 100ft. How can this be determined algebraically?

We have the following equation that expresses the height $s$ as a function of time:

$s=-2.7t^{2}+30t+6.5$ (1)

Now, if we wan to find the maximum height the baseball reaches and prove it is less than 100 ft, we firstly have to find the time $t_{total}$ the whole parabolic movement lasts and then find $t_{smax}=\frac{t_{total}}{2}$ which is the time it takes the baseball to reach its maximum height.

So, if we want to calculate $t_{total}$ , this is fulfilled when $s=0$ , when the baseball hits the ground:

$0=-2.7t^{2}+30t+6.5$ (2)

This is a quadratic equation of the form $0=at^{2}+bt+c$ , and we have to use the quadratic formula if we want to find $t_{total}$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (3)

Where $a=-2.7$ , $b=30$ , $c=6.5$

Substituting the known values and choosing the positive result of the equation:

$t_{total}=11.323 s$ (4)

Then we can calculate $t_{smax}$ :

$t_{smax}=\frac{t_{total}}{2}$ (5)

$t_{smax}=\frac{11.323 s}{2}$

$t_{smax}=5.661 s$ (6)

Substituting (6) in (1):

$s=-2.7(5.661 s)^{2}+30(5.661 s)+6.5$ (7)

$s=89.803 ft$ (8) This is the maximum height the baseball reaches, as we can see it is less than 100 ft

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3 years ago

What are the 6 uses of electromagnet?

Serhud [2]

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3 years ago

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A body cycles continously through a distance of 10km in 5 mins. calculate its average speed

stira [4]

Answer:

distance=10km=10*1000=10000m

time=5min=5*60=300sec

average speed=distance/time

= 10000/300=33.33m/s

Explanation:

3 0

3 years ago

Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area sho

Nady [450]

Answer:

centroid: (x, y) = (81.25 mm, 137.5 mm)
I = 8719.31 mm^2 for unit mass

Explanation:

Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.

The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...

y1 = 0.8x +125

The lower boundary is given in terms of y, but we can solve for y to find ...

100x = y^2

y2 = 10√x

Then our differential of area is ...

dA = (y1 -y2)dx

__

The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.

The area will be ...

$\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx$

The y-coordinate of the centroid is ...

$\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5$

Similarly, the x-coordinate is ...

$\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25$

That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.

__

The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for $\overline{x}$ , but multiplies dA by x^2 instead of x.

The attachment shows that value to be ...

I ≈ 8719.31 mm^2 (normalized by area)

The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.

7 0

3 years ago