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kicyunya [14]
2 years ago
8

What is an earthquake​

Physics
2 answers:
VMariaS [17]2 years ago
8 0

Answer:

When a underground suddenly breaks along a fault it cause vibration it is called earthquake.

inysia [295]2 years ago
5 0
An earthquake is when the ground shakes.
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4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C
aniked [119]

Answer:

a)  ΔL/L = F / (E A),  b)   L_{f} = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

     E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

    P = E ΔL / L

The pressure is defined as

    P = F / A

Let's replace

   F / A = E ΔL / L

   F = E A ΔL / L

   ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

    ΔL = L [F / EA]

    L_{f} -L = L (F / EA)

    L_{f} = L + L (F / EA)

    L_{f} = L (1 + L (F / EA))

4 0
3 years ago
When a substance changes state ____ does not change the speed of the molecules?
Vilka [71]
I don't understand what you are looking for. I can tell you that the speed of molecules does change during state changing.
6 0
3 years ago
Name the type of reproduction process as shown in Fig 1 and Fig 2. State one point of difference between the two
likoan [24]

Figure 1= binary fission in amoeba

figure 2= budding in yeast

difference

1.Parent divides to form two daughter cells and itself gets disappeared in binary fission but in budding , a bud gets matured and detaches from the parent

6 0
3 years ago
S.I unit for moment of inertia of a fly wheel​
qaws [65]

Answer:

m is expressed in kilograms and r in metres, with I (moment of inertia) having the dimension kilogram-metre square.

8 0
1 year ago
An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
Liono4ka [1.6K]

Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

4 0
3 years ago
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