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skelet666 [1.2K]
3 years ago
15

John has a boat that will travel at the rate of 15 kph in still water. he can go upstream for 35 km in the same time it takes to

go 140 km downstream. how fast is his boat traveling when he goes downstream? 22 kph 24 kph 26 kph
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
5 0
Let F = the downstream speed of the water. 

<span>Then the boat's upstream speed is: 15 - F </span>
<span>The boat's downstream speed is: 15 + F </span>


<span>Assume both the journeys mentioned take T hours, then using "speed x time = distance" we get: </span>

<span>Downstream journey: (15 + F)T = 140 </span>
<span>Upstream journey: (15 - F)T = 35 </span>


<span>Add the two formulae together: </span>

<span>(15 + F)T + (15 - F)T = 140 + 35 </span>

<span>15T + FT + 15T - FT = 175 </span>

<span>30T = 175 </span>

<span>T = 35/6 </span>


<span>Use one of the equations to find F: </span>

<span>(15 + F)T = 140 </span>

<span>15 + F = 140/T </span>
<span>F = 140/T - 15 </span>
<span>F = 140/(35/6) - 15 </span>
<span>F = 24 - 15 </span>
<span>F = 9 </span>

<span>i.e. the downstream speed of the water is 9 kph </span>

<span>Therefore, the boat's speed downstream is 15 + F = 15 + 9 = 24 kph.
the answer is:                       *24kph*</span>
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4 0
2 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\&#10;x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
What is the term for the force exerted by an object when it is pushed by another object?
kolezko [41]
Action-reaction pairs.

This is in reference to Newton’s second law of motion.
5 0
3 years ago
Read 2 more answers
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i
lianna [129]

Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

I_{1} w_{1} = I_{2} w_{2}    ....1

where I_{1} and I_{2} are the initial and final moment of inertia respectively.

and w_{1} and w_{2} are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

I = mr^{2}    ....2

where m is the mass of the rotating body,

and r is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from I_{1} to I_{2} will cause the angular speed of the system to increase from w_{1} to w_{2} .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0
3 years ago
A ball is dropped off the side of a bridge. after 1.55 s, how far has it fallen
givi [52]

Answer:

38.64 feet

Explanation:

x=x0 + vx0t + 1/2axt2

x= 0 +  0  + 1/2 X 32.17 ft/sec2 X 1.55 sec2

x = 38.64 feet

7 0
3 years ago
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