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kow [346]
4 years ago
11

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

She hears the echo 3.7s\; s after shouting. The speed of sound in air is 343 m/s\rm m/s. Determine the length of the lake. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Nataliya [291]4 years ago
8 0

To solve the exercise it is necessary to take into account the definition of speed as a function of distance and time, and the speed of air in the sound, as well

v=\frac{d}{t}

Where,

V= Velocity

d= distance

t = time

Re-arrange the equation to find the distance we have,

d=vt

Replacing with our values

d= (343)(3.7)

d= 1269.1m

It is understood that the sound comes and goes across the entire lake therefore, the length of the lake is half the distance found, that is

L_{lake} = \frac{d}{2}

L_{lake} = \frac{1269.1}{2}

L_{lake} = 634.55m

Therefore the length of the lake is 634,55m

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A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and a
NikAS [45]

Answer:

a=1.024m/s

t=15.62s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t                         (1)

{Vf^{2}-Vo^2}/{2.a} =X      (2)

X=Xo+ VoT+0.5at^{2}      (3)

X=(Vf+Vo)T/2                   (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this problem

Vf=16m/s

Vo=0m/s, the cart starts from the rest

X=125m

we can use the ecuation number tow to calculate the acceleration

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{16^{2}-0^2}/{2(125)} =a

a=1.024m/s

to calculate the time we can use the ecuation number 1

Vf=Vo+a.t    

t=(Vf-Vo)/a

t=(16-0)/1.024

t=15.62s

6 0
3 years ago
The Plateau of Tibet is a dry place. Do you think the wide-leafed plant fossils found on the plateau could grow there today? Exp
ludmilkaskok [199]

No, I dont think that they could grow there now in this day and age becuase the Plateu of Tbiet is very dry and also a little bit dusty. They barely even have grass down there. Wide leaf plants susually grow somewhee like the rainforest where it is warm, sunny, and rainy but in a place like Tibet, I don't think that they would be able to survive there.

7 0
4 years ago
You have a ruler that is marked to the nearest half centimeter. Which of these measurements is written with the correct amount o
NeX [460]
The answer would be b 3.14cm

8 0
3 years ago
The greenhouse effect that raises the surface temperature of planets
PSYCHO15rus [73]

Answer:

c

Explanation:

Green house effect:

 Temperature of atmosphere is going to  increased continuously, because infrared light absorbed by gas .This increased in temperature atmosphere    is called green house effect.

The green houses gases are as follows

,Carbon diaoxide,water vapor  ,,Carbon diaoxide ,Methane ,,Carbon diaoxide oxide,CFCs ,ozone.

So the option c is correct.

3 0
3 years ago
The Petronas twin towers in Malaysia and the Chicago Sears tower have heights of about 452 m and 443 m respectively. If objects
fredd [130]

Answer:

0.09 s

Explanation:

From the second equation of motion,

S=ut+\frac{1}{2}at^{2}

Here, u is the initial velocity, a is the acceleration due to gravity, t is time taken, and S is the total displacement or distance.

From the given problem,

initial velocity is zero for both the case.

And the distance of twin  tower of malaysia is, S_{1}=452m

And the distance of Sears tower of Chicago is, S_{1}=443m

Now,rearrange the distance equation for t.

t=\sqrt{\frac{2S}{a} }

So time difference.

\Delta t=\sqrt{\frac{2(452)}{9.8} }-\sqrt{\frac{2(443)}{9.8} }\\\Delta t=\sqrt{92.244898}-\sqrt{90.4081633}  \\\Delta t=9.60 s-9.51s\\\Delta t=0.09 s

Therefore, the difference in time, object will reach the ground is 0.09 s

8 0
3 years ago
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