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Eduardwww [97]
3 years ago
5

If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake​

Physics
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

Remember the relation:

Speed*Time = Distance.

We can estimate that the speed at which an earthquake "moves", in the surface, is:

S = 6km/s  (this is a low estimation actually)

Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.

So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:

Speed*Time = Distance.

to find the distance.

First, we should write the time in seconds

we know that:

1 min = 60 s

then:

12.5 min = 12.5*(60 s) = 750 s

Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:

(6km/s)*750s = 4500 km

The distance between the detector and the site of the quake is around 4500 km.

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A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

8 0
3 years ago
Anyone good with scientific notations? I sure am not
defon
B is the answer that I know of.
7 0
3 years ago
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What is the density of the solid given that its mass is 200
alina1380 [7]

Answer:

a) 2cm³

b) 100 g/cm³

Explanation:

a- 9-7= 2cm³

b- 200 divided by 2= 100 g/cm³

Hope this helps... correct me if i'm wrong

7 0
3 years ago
Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has
a_sh-v [17]

Answer:

  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

         Em₀ = U = m g h

final point. Lower, just before the crash

         Em_f = K = ½ m v_{b}^2

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = \sqrt{2gh}

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ \sqrt{2gh}

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

          Em₀ = K = ½ 3m v²

Final point. Higher

          Em_f = U = (3m) g h'

          Em₀ = Em_f

          ½ 3m v² = 3m g h’

           

we substitute

         h’=  \frac{v^2}{2g}

         h’ =  \frac{1}{3^2} \  \frac{ 2gh}{2g}

         h’ = 1/9 h

6 0
3 years ago
A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e
alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

K.E=\frac{1mv^2}{2}

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m= mass of an electron = 9.109\times 10^{-31}kg

v= velocity of object = ?

Putting in the values in the equation:

0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}

v=466462m/s

The speed necessary for the electron to have this energy is 466462 m/s

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