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lina2011 [118]
3 years ago
13

Please help :l

Physics
2 answers:
zysi [14]3 years ago
4 0

Answer:

im

Explanation:

lora16 [44]3 years ago
3 0
A Fossil correlation
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Teachers are interested in knowing what study techniques their students are utilizing. The researchers randomly select every 10t
uranmaximum [27]

Answer:

Simple Random Sample (SRS)

Explanation:

5 0
3 years ago
Scientists measure the time between the arrival of an earthquake's _______ and _______ waves to help determine the distance betw
soldier1979 [14.2K]

Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

Explanation:

P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake,  the longer the time difference between the two (2) waves will be.

Using several, at least 3, seismographs located at different geoghraphical locations and detecting earthquakes, geologists can extrapolate the epicenter of an earthquake using the time differences in arrivals of the two waves in each of the seismographs, using the mathematics of triangulation.

Learn More:

For more on P- and S-- waves check out;

brainly.com/question/11915788

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#LearnWithBrainly

7 0
4 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
Help Please!!!!!!!!!!!!!!!!!!
julia-pushkina [17]
The answer is 59.4 degrees.
4 0
3 years ago
Unlike mechanical waves such as sound or earthquake waves, EM waves can travel through empty space. True or false
Alex777 [14]

Answer:

its true

Explanation:

5 0
3 years ago
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