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nevsk [136]
3 years ago
13

Which is used to determine earthquake risk ?

Physics
1 answer:
anastassius [24]3 years ago
6 0

Answer:

it can be determined as

1. loss of life

2. damage of buildings, houses , public properties etc

3. can cause tsunami in countries like japan

4. loss of money

5. people may get homeless

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HELP ASAP <br><br> What happens when the mantle melts?
ankoles [38]

It produces oceanic crust,

3 0
3 years ago
Read 2 more answers
A spinning top initially spins at 16rad/s but slows down to 12rad/s in 18s, due to friction. If the rotational inertia of the to
Alex Ar [27]

Answer:

The  change  in angular momentum is \Delta  L  = 0.0016 \ kgm^2/s

Explanation:

From the question we are told that

      The initial angular velocity of the spinning top is  w_1  =  16 \ rad/s

      The angular velocity after it slow down is  w_2 =  12 \ rad/s

      The time  for it to slow down is  t =  18 \ s

       The rotational inertia due to friction is  I =  0.0004 \ kg \cdot m^2

 Generally the change in the angular momentum is  mathematically represented as  

         \Delta  L  =  I  *(w_1 -w_2)

substituting values  

        \Delta  L  =  0.0004  *(16 -12)

       \Delta  L  = 0.0016 \ kgm^2/s

5 0
3 years ago
6. Given that white light contains all colors of the spectrum, what growth result would you expect under white light?
lesya692 [45]

You could probably expect normal plant growth, as a white light is similar to the sun in the respect that it contains all colors of the spectrum.

4 0
3 years ago
A car travels north for 20 km. The car then travels south for 35 km.
PSYCHO15rus [73]

Answer: -15 km

Explanation:

8 0
3 years ago
Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid
algol13

Answer:

The intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

|E_{1}|=k\frac{q_{1}}{d^{2}}

Where:

  • k is Coulomb's constant
  • q(1) is 20.00 μC or 20*10⁻⁶ C
  • d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}

|E_{1}|=1.8*10^{7}\: N/C

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

|E_{2}|=k\frac{q_{2}}{d^{2}}

|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}

|E_{2}|=7.2*10^{6}\: N/C

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

I hope it helps you!

5 0
3 years ago
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