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polet [3.4K]
2 years ago
5

How many times does lightning strike the empire state building

Physics
1 answer:
Salsk061 [2.6K]2 years ago
3 0

Answer:

Lightning strikes the empire state building at an average of about 23 times a year.

Explanation:

The Empire State Building is one of the tallest buildings in New York. Because of how high it stretches up into the sky, lightning strikes are quite common to it. This is because part of the building touches the clouds which are usually charged during thunder storms.

According to weather reports, and the Empire State Building website, lightning strikes the empire state building about 23 times a year on the average.

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The battery provides electrical energy to run an electric circuit.
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3 years ago
A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the
kotykmax [81]

Answer:

M₂ = M  then L₂ = L

M₂> M  then L₂ = \frac{M}{M_{2}} L

Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

          ∑ τ = 0

           

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

         M L + M₁ 0 - m₂ L₂ = 0

         M L - m₂ L₂ = 0

         L₂ = \frac{M}{M_{2}} L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L

6 0
3 years ago
An object buoyant force and wieght arent the smae thing
Alona [7]

I hope this is true or false



The answer would be TRUE

7 0
2 years ago
A disc with a mass of 1 kg moves horizontally to the right with a speed of 7 m/s on a table with negligible friction when it col
Elodia [21]

Answer:

1.6 m/s

Explanation:

First you need to find the momentums of each disc by multiplying their velocities with mass.

disc 1: 7*1= 7 kg m/s

disc 2: 1*9= 9 kg m/s

Second, you need to find the total momentum of the system by adding the momentums of each sphere.

9+7= 16 kg m/s

Because momentum is conserved, this is equal to the momentum of the composite body.

Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum

16/10=1.6

The velocity of the composite body is 1.6 m/s.

7 0
2 years ago
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
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