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Naya [18.7K]
3 years ago
9

Alicia can row 6 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 3 miles/hour faster

than she rows upstream. Find Alicia’s rowing rate each way. Round your answers to the nearest tenth, if necessary.
Physics
2 answers:
m_a_m_a [10]3 years ago
4 0
Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
olga_2 [115]3 years ago
4 0

Full...Solving Rational Equations Quiz part 1.

1.c. n^2-6/n^2-2 ; n = +/- sqrt5, n= +/- sqrt2

2.B.  4a/7b^2 , a = 0, b = 0

3.C. (x-4)^2/(x+3)(x+1) ; x= -4,-3,-2,-1,4

4.B. (x+1)(x-1)(x^2+1)

5.A. 7a-49/(a-8)(a+8)

6.A. 21a-28/(A-6)(a+8)

7.C. 4x/3x^2+10x+3

8.C. 3x^2(y+4)/7y

9.D. -11/3

10.D. 14

11. D. 9 mi/h downstream, 6 mi/h upstream

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3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

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