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yanalaym [24]
3 years ago
7

A dolphin's tops speed is 17 m/s. If a dolphin swam at this constant velocity for one hour

Physics
1 answer:
Tpy6a [65]3 years ago
7 0

Answer:

<h3>The answer is 61,200 m</h3>

Explanation:

To find the distance covered by the dolphin , we use the formula

<h3>distance = velocity × time</h3>

From the question

velocity = 17 m/s

time = 3600 s

We have

distance = 17 × 3600

We have the final answer as

<h3>61,200 m</h3>

Hope this helps you

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Where they slide over each other.

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In a transform boundary, neither plate is added to at the boundary nor destroyed.  They are marked in some places by features like  stream beds that have been split in half and the two halves moved in opposite directions.



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Which of the following objects uses two converging lenses to produce two images, the second of which is virtual, magnified, and
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<span>Microscope. A basic microscope is made up of two converging lenses. One reason for using two lenses rather than just one is that it's easier to get higher magnification.</span>
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At the Earth's surface, warm air expands and rises, leaving behind what is known as an area of _____; cold air is dense and sink
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Answer:

Low-pressure zone

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A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H
Wewaii [24]

Answer:

(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s

Explanation:

Given data

v_{o}=7.14m/s\\\alpha =36.5^o

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2

Using equation of motion

v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s

4 0
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