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Alika [10]
3 years ago
12

If you know the distance of an earthquake epicenter from three seismic stations, how can you find the exact location of the epic

enter of the earthquake.
Physics
1 answer:
Contact [7]3 years ago
6 0

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

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Suppose a cheetah running at a velocity of 5 m/s east slows down. After 15 seconds, the cheetah has reached a velocity of 25 m/s
Darina [25.2K]

Answer:

1.33m/s²

Explanation:

Given parameters:

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time taken  = 15seconds

Unknown:

Acceleration  = ?

Solution:

Acceleration is the rate of change of velocity with time.

  Acceleration  = \frac{v - u}{t}  

  v is the final velocity

  u is the initial velocity

  t is the time taken

 So;

    Acceleration  = \frac{25 - 5}{15}    = 1.33m/s²

5 0
3 years ago
Which of the following is not an example of kinetic energy?
ycow [4]
Only first is kinetic.
So 2-3-4 are not
5 0
4 years ago
Read 2 more answers
Si lo deseas cada estudiante puede hablar algún otro aspecto de la pandemia que les parezca importante e interesante
iren2701 [21]

Answer:

Give them assignment on the pandemic.

Explanation:

If you wish, each student can talk about some other aspect of the pandemic that seems important and interesting then you have to give them the assignment of presentation and give reward on them. The students will definitely make the presentation on other aspects of the pandemic and when they present it in the class it enhance the knowledge of other students as well.

4 0
3 years ago
A ball rolls down the street 9 meters in 3.5 seconds, what is it's average speed?
artcher [175]

How do we find the Average Speed ?

The Average Speed is the distance traveled for the object, divided by the elapsed time taken to travel that distance, so the formula is :

<em>S = D / T</em>

where S = Average Speed, D = Distance, and T =  Times

Here the object is the ball, the distance is 9 meters and the time elapsed for the ball to roll down the street is 3.5 seconds. Applying the formula we get :

S = 9 / 3.5 ≈ 2.6 metres per second

The ≈ symbols (which is the <em>wavy equal sign</em>) means we can't write the exact result, because there is an infinite quantity of numbers after the decimal point. So we need to give an approximation, since the result is 2.57142857142857...etc. we choosed to round up the result to 2.6.

7 0
3 years ago
Read 2 more answers
A metal cube, 2.00cm on each side, has a density of 6600 kg/m3. find its apparent weight when it is totally submerged in water.
LekaFEV [45]
Answer: 0.439488 N

Explanation:


The apparent weight of the metal is computed as

apparent weight = weight - weight of the displaced fluid 

To compute the weight of the metal, we use the following formula:

\text{weight} = \rho gV

where 

\rho = \text{density of the metal = 6,600 kg/m}^3
g = \text{gravitational acceleration = 9.81 m/s}^2
V = \text{volume of the metal}

Note that the volume is unknown but we can compute this because the metal is a cube with edge = 2 cm = 0.02 m. So, the volume of the metal is given by

\text{Volume} = \text{edge}^3&#10;\\ = (0.02)^3&#10;\\ \boxed{\text{Volume = 0.000008 m} ^3}

Thus, the weight of the metal is computed as

\text{weight} = \rho gV&#10;\\ = (\text{6,600 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.517968 kg \(\cdot \) m/s}^2&#10;\\ \boxed{\text{weight of the metal} = \text{0.517968 N}}

Next, we compute the displaced weight or buoyancy, which has the following formula

\text{weight of the displaced fluid} = \rho' gV'

where

\rho' = \text{density of the fluid (water) = 1,000 kg/m}^3&#10;\\ g = \text{gravitational acceleration = 9.81 m/s}^2&#10;\\ V' = \text{displaced volume}

Note that the displaced volume is equal to the volume of the submerged metal. Since the metal has a volume of \text{0.000008 m} ^3, the displaced volume is \text{0.000008 m} ^3. 

Thus, the weight of the displaced fluid is calculated as 

\text{weight of the displaced fluid} = \rho' gV'&#10;\\ = (\text{1,000 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.07848 kg \(\cdot \) m/s}^2 \\ \boxed{\text{weight of the displaced fluid} = \text{0.07848 N}}

Therefore,

apparent weight of the metal
= weight of the metal - weight of the displaced fluid 
= 0.517968 N - 0.07848 N
apparent weight of the metal = 0.439488 N

3 0
3 years ago
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