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mylen [45]
3 years ago
12

a man can row about 4kmperhr in a still water.he rows the boat 2km up the stream and 2km back to his starting point in 2hr.how f

ast is the stream flowing. ​
Physics
1 answer:
Nat2105 [25]3 years ago
7 0

<u>Answer</u>:

The stream flowing  at a speed of 2.828 \mathrm{km} / \mathrm{hr}

<u>Explanation</u>:

Given:  

Distance = 2km (both in upstream and downstream)  

The speed in still water be x km/hr.  

The speed in upstream = 4-x  

Speed in downstream = 4+x  

Solution:

We know that, Speed = distance/time  

So, Time = distance/speed

Therefore,  

2=\left(\frac{2}{4-x}\right)+\left(\frac{2}{4+x}\right)

2=\frac{2(4+x)+2(4-x)}{(4-x)(4+x)}

2(4-x)(4+x)=2(4+x)+2(4-x)

2(4-x)(4+x)=2(4+x+4-x)

By cancelling 2 on both sides,

16-x^{2}=8

x^{2}=16-8=8

x=\sqrt{8}

x=2.828 \mathrm{km} / \mathrm{hr}

Result:

Thus the speed of the stream is 2.828 \mathrm{km} / \mathrm{hr}

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Electric field

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2 years ago
One solution to minimize resonance with buildings is to ______ the width to span ratio.
Shtirlitz [24]

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Increase

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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
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